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Subtopik : Konsep Kilat Sifat Keperiodikan Unsur
Perhatikan tabel SPU berikut!
1. Pernyataan yang tepat untuk sifat keperiodikan kelima unsur tersebut adalah …
- Jari-jari atom unsur A lebih kecil daripada unsur C.
- Energi ionisasi unsur B lebih besar daripada unsur C.
- Unsur A memiliki keelektronegatifan terbesar.
- Afinitas elektron unsur A lebih besar dari unsur C.
- Unsur B memiliki jari-jari atom paling besar.
Kunci Jawaban : B
Pembahasan :
Sifat-sifat keperiodikan unsur, yaitu sebagai berikut.
- Jari-jari atom adalah jarak dari inti atom ke kulit terluar yang ditempati elektron. Dalam satu golongan dari atas ke bawah, jari-jari atom cenderung makin besar, sedangkan dalam satu periode dari kiri ke kanan, cenderung makin kecil. Jari-jari atom unsur A lebih besar daripada unsur C. Unsur yang memiliki jari-jari atom terbesar adalah unsur E. (Opsi A dan E salah)
- Energi ionisasi adalah energi minimum yang diperlukan atom dalam keadaan gas untuk melepaskan elektron pada kulit terluarnya. Dalam satu golongan dari atas ke bawah, energi ionisasi cenderung makin kecil, sedangkan dalam satu periode dari kiri ke kanan cenderung makin besar. Oleh karena itu, energi ionisasi unsur B lebih besar daripada unsur C. (Opsi B benar)
- Keelektronegatifan adalah kecenderungan suatu atom untuk menarik elektron dari atom lain dalam suatu ikatan kimia. Dalam satu golongan dari atas ke bawah, keelektronegatifan cenderung makin kecil, sedangkan dalam satu periode dari kiri ke kanan cenderung makin besar. Unsur yang memiliki keelektronegatifan terbesar adalah unsur B. (Opsi C salah)
- Afinitas elektron adalah besarnya energi yang dihasilkan atau dibebaskan oleh atom dalam keadaan gas untuk menangkap satu elektron. Dalam satu golongan dari atas ke bawah, afinitas elektron cenderung makin kecil, sedangkan dalam satu periode dari kiri ke kanan cenderung makin besar. Afinitas elektron unsur A lebih kecil dari unsur C. (Opsi D salah)
Jawaban yang tepat adalah B.
Topik : Ikatan Kimia, Bentuk Molekul, dan Gaya Antarmolekul
Subtopik : Konsep Kilat Ikatan Kimia dan Bentuk Molekul
Diketahui data unsur sebagai berikut.
Unsur X : golongan IIA dan periode 3
Unsur Y : golongan VIA dan periode 2
2. Jika X dan Y berikatan, ikatan yang mungkin terjadi dan senyawa yang terbentuk secara berurutan adalah ….
Kunci Jawaban : A
Pembahasan :
Unsur X : golongan IIA elektron valensi = 2. Unsur X stabil dengan melepas 2 elektron membentuk .
Unsur Y : golongan VIA elektron valensi = 6. Unsur Y stabil dengan menangkap 2 elektron membentuk .
Senyawa yang terbentuk :
Ikatan yang terbentuk merupakan ikatan ionik karena terjadi serah terima elektron.
Jadi, jawaban yang tepat adalah A.
Topik : Kimia Karbon
Subtopik : Konsep Kilat Senyawa Turunan Alkana
3. Seorang laboran akan membuat larutan dengan cara mencampurkan larutan X dengan larutan NaOH dalam gelas kimia. Volume dan konsentrasi dari masing-masing reaktan adalah sama, yaitu 100 mL dan 0,3 M. Diketahui bahwa larutan X diperoleh dari oksidasi larutan Y yang memberikan hasil positif dengan pereaksi Tollens. Kemungkinan larutan X yang digunakan oleh laboran tersebut adalah ….
- 1-propanol
- Propanal
- 2-propanon
- asam propanoat
- metil asetat
Jawaban: D
Pembahasan:
Pereaksi Tollens digunakan untuk mengidentifikasi senyawa aldehid dan keton. Reaksi oksidasi aldehid dengan pereaksi Tollens akan memberikan hasil positif (terbentuk endapan perak), sedangkan senyawa keton memberikan hasil negatif (tidak terbentuk endapan perak).
Larutan Y memberikan hasil positif dengan pereaksi Tollens dan ketika dioksidasi menghasilkan larutan X, artinya Y merupakan senyawa aldehid. Hasil oksidasi aldehid adalah senyawa yang memiliki gugus karboksil atau asam karboksilat. Oleh karena itu, X merupakan senyawa asam karboksilat.
Jenis senyawa pada opsi jawaban adalah sebagai berikut.
- 1-propanol alkohol primer
- propanal aldehid
- 2-propanon keton
- asam propanoat asam karboksilat
- metil asetat ester
Dengan demikian, kemungkinan larutan X yang digunakan oleh laboran tersebut adalah asam propanoat.
Jadi, jawaban yang benar adalah D.
Subtopik : Konsep Kilat Stoikiometri Reaksi dan Kadar
4. Di dalam laboratorium, suatu sampel urin yang mengandung 1,5 gram urea direaksikan dengan asam nitrit berlebih. Urea bereaksi menurut persamaan berikut.
Gas yang dihasilkan lalu dialirkan melalui larutan NaOH dan volume akhir diukur. Volume gas yang telah melewati larutan NaOH pada temperatur dan tekanan ruang adalah ….
Kunci Jawaban : E
Pembahasan :
Pertama, setarakan persamaan reaksinya terlebih dahulu.
Lalu, hitung mol urea menggunakan persamaan berikut.
Kemudian, hitung mol gas karbon dioksida dan gas nitrogen menggunakan perbandingan koefisien sebagai berikut.
Dari reaksi tersebut, terdapat 2 jenis gas, yaitu . Pengukuran sampel dilakukan pada temperatur dan tekanan ruang, yaitu pada suhu 25 derajat dan tekanan 1 atm. Dengan demikian, volume yang dihasilkan dapat dihitung sebagai berikut.
Ingat! Gas adalah gas asam sehingga akan bereaksi ketika dilarutkan ke dalam NaOH. Dengan demikian, volume gas yang telah melewati larutan NaOH (volume gas ) pada kondisi yang ditentukan pada soal adalah .
Jadi, jawaban yang tepat adalah E.
Subtopik : Konsep Kilat Sifat Koligatif Larutan
5. Dalam dunia kedokteran, tekanan osmotik perlu diperhatikan dalam proses pemberian infus kepada pasien. Tekanan osmotik dalam darah berkisar 7,7 atm pada suhu darah 38 derajat celcius. Apabila seseorang memerlukan injeksi cairan nutrisi/infus, tekanan osmotik cairan infus tersebut harus isotonik dengan tekanan osmotik darah. Jumlah NaCl yang diperlukan untuk membuat larutan infus sebanyak 1000 mL yang dibutuhkan oleh pasien tersebut adalah ….
- 8,8 g
- 5,5 g
- 4,4 g
- 4,0 g
- 3,0 g
Jawaban: A
Pembahasan:
Isotonik artinya memiliki tekanan osmotik yang sama.
Diketahui:
Ditanyakan:
Massa NaCl yang dilarutkan
Pembahasan:
NaCl merupakan garam yang bersifat elektrolit kuat, artinya dapat terurai menjadi ion-ionnya.
Jumlah ion (n) = 2
Massa NaCl yang diperlukan untuk membuat larutan infus tersebut dapat ditentukan melalui persamaan berikut.
Berdasarkan persamaan di atas, massa dapat ditentukan sebagai berikut.
Jadi, jawaban yang tepat adalah A.
Topik : Ikatan Kimia dan Bentuk Molekul
Subtopik : Bentuk dan Geometri Molekul
6. Unsur Xe merupakan golongan gas mulia yang sukar berikatan dengan unsur lain. Namun, seiring berkembangnya teknologi, ternyata dapat dibuat senyawa dari unsur Xe dengan unsur paling elektronegatif F yaitu XeF4. Geometri dan tipe hibridisasi pada molekul XeF4 (NA Xe = 54, F = 9) adalah…
- segi empat planar, sp3d2
- tetrahedron, sp3d
- segi empat planar, sp2d
- tetrahedron, sp3
- bipiramida trigonal, sp3
Kunci : A
Pembahasan :
Untuk membuat hibridisasi yang dilihat adalah atom pusat. Pada senyawa XeF4 yang menjadi atom pusat adalah Xe. Konfigurasi elektron 54Xe = [Kr] 5s24d10 5p6
Topik : Redoks dan Elektrokimia
Subtopik : Sel Volta; Hukum Faraday
7. Suatu sel elektrokimia dengan elektroda Zn dan Ag pada keadaan standar menghasilkan arus 0,75 A selama 321 menit. Diketahui nilai:
E0 Zn2+ | Zn = -0,76 V
E0 Ag+ | Ag = +0,8 V
maka pengurangan massa dianoda adalah … gram. (F = 96.500 C, Ar Zn = 65, Ag = 108)
- 3,250
- 4,865
- 6,321
- 10,850
- 13,106
Kunci : B
Pembahasan :
Pada katoda terjadi reaksi reduksi, sedangkan anoda oksidasi. Potensial reduksi Ag lebih besar dari Zn, sehingga pengurangan massa di anoda terjadi dari oksidasi Zn.
Katoda: Ag+ + e– → Ag
Anoda: Zn → Zn2+ + 2e–
Subtopik : Persamaan Laju Reaksi dan Konstanta Laju Reaksi Multi Experiment
8. Percobaan penentuan laju reaksi:
2H2(g) + 2NO(g) 2H2O(g) + N2(g)
memberikan data sebagai berikut:
Harga tetapan laju reaksi dari reaksi tersebut adalah…
- 4 × 108
- 8 × 105
- 8 × 106
- 16 × 105
- 16 × 104
Kunci : C
Pembahasan :
Menentukan orde [NO] menggunakan persamaan 2 dan 3
Menentukan orde [H2] menggunakan persamaan 1 dan 2
Maka persamaan laju reaksinya
Mencari nilai tetapan laju reaksi (k) menggunakan percobaan 1.
Topik : Redoks dan Elektrokimia
Subtopik : Sel Volta
9. Diketahui potensial standar beberapa sel volta sebagai berikut:
P | P2+ || Q2+ | Q E0 = 2,46 V
R | R2+ || S2+ | S E0 = 1,1 V
R | R2+ || Q2+ | Q E0 = 1,56 V
Potensial standar sel P | P2+ || S2+ | S adalah…
- 1,1 V
- 2,00 V
- 2,66 V
- 3,56 V
- 4,02 V
Kunci : B
Pembahasan :
P | P2+ || Q2+ | Q Eo = 2,46 V
R | R2+ || S2+ | S Eo = 1,1 V
R | R2+ || Q2+ | Q Eo = 1,56 V (perlu dibalik untuk mengeliminasi R)
Reaksi yang dibalik memiliki potensial sel dengan nilai berlawanan, sehingga:
P | P2+ || Q2+ | Q Eo = 2,46 V
R | R2+|| S2+ | S Eo = 1,1 V
Q | Q2+ || R2+ | R Eo = -1,56 V +
P | P2+ || S2+ | S Eo = 2,00 V
Subtopik : Konstanta Kesetimbangan Tekanan (Kp)
10. Dalam wadah 2 liter, gas H2dan gas I2 bereaksi membentuk gas HI menurut persamaan reaksi:
H2(g) + I2(g) 2HI(g)
Jika pada kesetimbangan reaksi terdapat 0,1 mol H2, 0,2 mol I2 dan 0,5 mol HI, dengan tekanan total sebesar 5 atm, maka nilai Kp reaksi tersebut adalah…
- 7,50
- 12,5
- 25,2
- 32,5
- 50,0
Kunci : B
Pembahasan :
Diketahui: volume = 2 liter; mol total = 0,1 mol + 0,2 mol + 0,5 mol = 0,8 mol
Topik : Stoikiometri
Subtopik : Persamaan Kimia; Perhitungan Kimia dalam Suatu Persamaan Reaksi
11. Suatu campuran yang hanya mengandung gas metana dan etana dibakar sempurna sehingga dihasilkan 220 gram CO2. Jika massa campuran yang dibakar adalah 76 gram, maka massa gas etana di dalam campuran gas tersebut (Ar C= 12 H = 1 O = 16) adalah…
- 16 gram
- 30 gram
- 32 gram
- 44 gram
- 60 gram
Kunci : E
Pembahasan :
Metana = CH4 (Mr = 16) Massa etana = x gram (sebagai komponen yang ditanya)
Etana = C2H6 (Mr = 30) Massa metana = (76 – x) gram
Reaksi pembakaran etana:
C2H6 + O2 → 2CO2 + 3H2O
Reaksi pembakaran metana:
CH4 + 2O2 → CO2 + 2H2O
Mol CO2 total = 5 mol
Topik : Kesetimbangan Kimia
Subtopik : Pergeseran Kesetimbangan dan Faktor yang Mempengaruhi…
12. Di bawah ini adalah produksi NH3dalam reaksi:
N2(g) + 3H2(g) 2NH3(g) ΔH = -92,2 kJ
Jika diketahui pada suhu 25 ℃ nilai Kc = 3,5 × 108, maka pernyataan berikut ini yang benar adalah…
- jika suhu turun, produksi meningkat
- ditambah katalis, produksi menurun
- jika volume turun, produksi menurun
- proses produksinya disebut dengan proses Kontak
- pengurangan konsentrasi N2, produksi meningkat
Kunci : A
Pembahasan :
Produksi NH3 disebut dengan proses Haber-Bosch dengan reaksi
N2(g) + 3H2(g) 2NH3(g) ΔH = -92,2 kJ (reaksi eksoterm)
Maka, pernyataan yang tepat yaitu:
- jika suhu turun, akan bergeser ke eksoterm maka produksi meningkat.
- ditambah katalis, reaksi akan semakin cepat, tidak berpengaruh pada hasil produksi.
- jika volume turun, akan bergeser ke koefisien kecil maka produksi akan meningkat.
- pengurangan konsentrasi N2, akan bergeser ke arah reaktan,jadi produksi menurun.
Topik : Sifat Koligatif Larutan
Subtopik : Kenaikan Titik Didih
13. Struktur molekul PVC dapat dinyatakan (–CH2 –CHCl–)n. Larutan yang mengandung 12,5 gram bahan tersebut dalam 500 gram benzena mendidih pada suhu 80,11℃. Jika harga Kb 2,5 ℃/molal dan titik didih benzena 80,10 ℃, maka harga n (Ar H = 1, C = 12, Cl = 35,5) adalah…
- 50
- 100
- 500
- 1000
- 1500
Kunci : B
Pembahasan :
Diketahui:
massa PVC (terlarut) = 12,5 gram; Tb = 80,11 ℃; Kb = 2,5 ℃/molal
massa benzena (pelarut) = 500 gram; ΔTb = 80,11 ℃ – 80,10 ℃ = 0,01 ℃
(Rumus Empiris)n = Rumus Molekul
(Mr Empiris)n = Mr Molekul
(62,5)n = 6250
Subtopik : Penentuan Perubahan Entalpi
14. Diketahui energi rata-rata sebagai berikut:
C – H = 414 kJ/mol H – Cl = 432 kJ/mol
Cl – Cl = 244 kJ/mol C – Cl = 326 kJ/mol
Perubahan entalpi untuk 3,2 gram metana pada reaksi:
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) adalah … kJ. (Ar C = 12, H = 1)
- -20
- +20
- -80
- +100
- -100
Kunci : A
Pembahasan :
Persamaan reaksi:
ΔH = ∑H reaktan – ∑H produk
ΔH = 658 – 758 = -100 kJ/mol
Maka untuk 3,2 gram metana dilepaskan kalor sebesar 100 kJ/mol.
ΔHr = 0,2 mol × -100 kJ/mol = -20 kJ
Topik : Termokimia
Subtopik : Penentuan Perubahan Entalpi
15. Etanol, disebut juga etil alkohol, memiliki rumus molekul C2H5OH adalah sejenis cairan yang mudah menguap dan mudah terbakar sehingga etanol dapat digunakan sebagai bahan bakar. Jika spiritus dianggap hanya mengandung etanol, berapa gram spiritus yang harus dibakar untuk menaikkan suhu 100 gram air dari 20℃ menjadi 50 ℃? Diketahui:
∆Hof H2O(g) = -240 kJ/mol
∆Hof CO2(g) = -394 kJ/mol
∆Hof C2H5OH = -277 kJ/mol
kalor jenis air = 4,2 J/gK
Ar C = 12, H = 1, O = 16
- 0,235 gram
- 0,47 gram
- 0,94 gram
- 6,3 gram
- 12,6 gram
Kunci : B
Pembahasan :
Mr etanol = 46 g/mol; ∆T = 50 ℃ – 20 ℃ = 30 ℃ = 30 K
Reaksi pembakaran etanol adalah: C2H5OH + 3O2 → 2CO2 + 3H2O
∆H = produk –reaktan
= (2×(-394) + 3×(-240)) – (–277 + 0)
= –1231 kJ/mol
Nah, itulah beberapa kumpulan latihan soal UTBK Kimia yang bisa kamu jadikan bahan belajarmu untuk persiapan SBMPTN 2022 mendatang. Mau dapat persiapan ekstra untuk UTBK? Kamu bisa bergabung di dalam #PelatnasUTBK! Dapatkan program harian yang dirancang khusus oleh tim Ruangguru untuk membantu persiapanmu. Tidak hanya materi pelajaran, #PelatnasUTBK juga akan memberikan pelatihan kesiapan mental, sehingga fisik dan mentalmu lebih siap dan optimal. Daftar dengan klik di bawah ini ya.
Yuk, berlatih lebih banyak soal lagi dengan ikut tryout UTBK di ruanguji. Psst, soal-soal dan sistem penilaiannya sama seperti UTBK aslinya, lho! Yakin nggak mau cobain?
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<h1 class="content-title">
Latihan Soal & Pembahasan UTBK 2022: Kimia </h1>
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<p class="content-author"><a href="https://www.ruangguru.com/blog/author/shabrina-alfari" class="">Shabrina Alfari</a> </p>
<p class="content-info"><span>February 18, 2022 • </span> <span>10 minutes read</span></p>
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<p style="text-align: justify;"><img decoding="async" style="width: 820px; margin-left: auto; margin-right: auto; display: block;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/Latihan%20soal%20utbk%20kimia.jpg" alt="Latihan soal utbk kimia" /></p>
<p style="text-align: center; font-size: 16px;"><em>Sudah siap hadapi UTBK tahun 2022 nanti? Yuk, tingkatkan persiapanmu dengan latihan soal UTBK 2022 Kimia beserta pembahasannya di bawah ini. Selamat mengerjakan!</em></p>
<p style="text-align: center; font-size: 16px;">—</p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<h2 style="text-align: justify; font-weight: bold; font-size: 16px;">Topik : Struktur Atom dan <a href="/blog/mengenal-sistem-periodik-unsur-kimia" rel="noopener">Sistem Periodik Unsur</a></h2>
<p style="text-align: justify;"><strong>Subtopik </strong> <strong>: Konsep Kilat Sifat Keperiodikan Unsur</strong></p>
<p style="text-align: justify;"><strong style="background-color: transparent; font-size: 1rem;">Perhatikan tabel SPU berikut!</strong></p>
<p style="text-align: justify;"><strong style="background-color: transparent; font-size: 1rem;"><img decoding="async" style="width: 620px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/4-Nov-23-2021-01-40-57-76-AM.png" alt="4-Nov-23-2021-01-40-57-76-AM" width="620" /></strong></p>
<p style="text-align: justify;">1. Pernyataan yang tepat untuk sifat keperiodikan kelima unsur tersebut adalah …</p>
<ol style="list-style-type: upper-alpha; text-align: justify;">
<li>Jari-jari atom unsur A lebih kecil daripada unsur C.</li>
<li>Energi ionisasi unsur B lebih besar daripada unsur C.</li>
<li>Unsur A memiliki keelektronegatifan terbesar.</li>
<li>Afinitas elektron unsur A lebih besar dari unsur C.</li>
<li>Unsur B memiliki jari-jari atom paling besar.</li>
</ol>
<p style="font-weight: bold; text-align: justify;">Kunci Jawaban : B</p>
<p style="font-weight: bold; text-align: justify;">Pembahasan :</p>
<p style="text-align: justify;">Sifat-sifat keperiodikan unsur, yaitu sebagai berikut.</p>
<ul style="text-align: justify;">
<li>Jari-jari atom adalah jarak dari inti atom ke kulit terluar yang ditempati elektron. Dalam satu golongan dari atas ke bawah, jari-jari atom cenderung makin besar, sedangkan dalam satu periode dari kiri ke kanan, cenderung makin kecil. Jari-jari atom unsur A lebih besar daripada unsur C. Unsur yang memiliki jari-jari atom terbesar adalah unsur E. <strong>(Opsi A dan E salah)</strong><strong><br />
</strong></li>
<li>Energi ionisasi adalah energi minimum yang diperlukan atom dalam keadaan gas untuk melepaskan elektron pada kulit terluarnya. Dalam satu golongan dari atas ke bawah, energi ionisasi cenderung makin kecil, sedangkan dalam satu periode dari kiri ke kanan cenderung makin besar. Oleh karena itu, energi ionisasi unsur B lebih besar daripada unsur C. <strong>(Opsi B benar)</strong><strong><br />
</strong></li>
<li>Keelektronegatifan adalah kecenderungan suatu atom untuk menarik elektron dari atom lain dalam suatu ikatan kimia. Dalam satu golongan dari atas ke bawah, keelektronegatifan cenderung makin kecil, sedangkan dalam satu periode dari kiri ke kanan cenderung makin besar. Unsur yang memiliki keelektronegatifan terbesar adalah unsur B. <strong>(Opsi C salah)</strong><strong><br />
</strong></li>
<li>Afinitas elektron adalah besarnya energi yang dihasilkan atau dibebaskan oleh atom dalam keadaan gas untuk menangkap satu elektron. Dalam satu golongan dari atas ke bawah, afinitas elektron cenderung makin kecil, sedangkan dalam satu periode dari kiri ke kanan cenderung makin besar. Afinitas elektron unsur A lebih kecil dari unsur C. <strong>(Opsi D salah)</strong></li>
</ul>
<p style="text-align: justify;"><strong><u>Jawaban yang tepat adalah B.</u></strong></p>
<p style="text-align: justify;"><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<h2 style="font-size: 16px;"><strong>Topik</strong> <strong>: Ikatan Kimia, Bentuk Molekul, dan Gaya Antarmolekul</strong></h2>
<p><strong>Subtopik </strong> <strong>: Konsep Kilat Ikatan Kimia dan <a href="/blog/bentuk-molekul-dan-definisinya" rel="noopener">Bentuk Molekul</a></strong></p>
<p><span style="background-color: transparent; font-size: 1rem;">Diketahui data unsur sebagai berikut.</span></p>
<p style="text-align: center;">Unsur X : golongan IIA dan periode 3</p>
<p style="text-align: center;">Unsur Y : golongan VIA dan periode 2</p>
<p>2. Jika X dan Y berikatan, ikatan yang mungkin terjadi dan senyawa yang terbentuk secara berurutan adalah ….</p>
<p><img decoding="async" style="width: 126px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/5-Nov-23-2021-01-42-44-95-AM.png" alt="5-Nov-23-2021-01-42-44-95-AM" width="126" /></p>
<p> </p>
<p style="font-weight: bold;">Kunci Jawaban : A</p>
<p style="font-weight: bold;">Pembahasan :</p>
<p>Unsur X : golongan IIA elektron valensi = 2. Unsur X stabil dengan melepas 2 elektron membentuk <img decoding="async" style="width: 26px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/6-Nov-23-2021-01-42-45-05-AM.png" alt="6-Nov-23-2021-01-42-45-05-AM" width="26" />.</p>
<p>Unsur Y : golongan VIA elektron valensi = 6. Unsur Y stabil dengan menangkap 2 elektron membentuk <img decoding="async" style="width: 33px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/7-Nov-23-2021-01-42-45-46-AM.png" alt="7-Nov-23-2021-01-42-45-46-AM" width="33" />.</p>
<p>Senyawa yang terbentuk : <img decoding="async" style="width: 109px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/8-Nov-23-2021-01-42-45-09-AM.png" alt="8-Nov-23-2021-01-42-45-09-AM" width="109" /></p>
<p>Ikatan yang terbentuk merupakan ikatan ionik karena terjadi serah terima elektron.</p>
<p><strong><u>Jadi, jawaban yang tepat adalah A.</u></strong></p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<h2 style="text-align: justify; font-size: 16px;"><strong>Topik</strong> <strong>: Kimia Karbon</strong></h2>
<p style="text-align: justify;"><strong>Subtopik</strong> <strong>: Konsep Kilat Senyawa Turunan <a href="/blog/senyawa-turunan-alkana-alkohol" rel="noopener">Alkana</a></strong></p>
<p style="text-align: justify;"><span style="background-color: transparent; font-size: 1rem;">3. Seorang laboran akan membuat larutan dengan cara mencampurkan larutan X dengan larutan NaOH dalam gelas kimia. Volume dan konsentrasi dari masing-masing reaktan adalah sama, yaitu 100 mL dan 0,3 M. Diketahui bahwa larutan X diperoleh dari oksidasi larutan Y yang memberikan hasil positif dengan pereaksi Tollens. </span><span style="background-color: transparent; font-size: 1rem;">Kemungkinan larutan X yang digunakan oleh laboran tersebut adalah ….</span></p>
<p> </p>
<ol style="list-style-type: upper-alpha; text-align: justify;">
<li>1-propanol</li>
<li>Propanal</li>
<li>2-propanon</li>
<li>asam propanoat</li>
<li>metil asetat</li>
</ol>
<p style="font-weight: bold; text-align: justify;">Jawaban: D</p>
<p style="font-weight: bold; text-align: justify;">Pembahasan:</p>
<p style="text-align: justify;">Pereaksi Tollens digunakan untuk mengidentifikasi senyawa aldehid dan keton. Reaksi oksidasi aldehid dengan pereaksi Tollens akan memberikan hasil positif (terbentuk endapan perak), sedangkan senyawa keton memberikan hasil negatif (tidak terbentuk endapan perak).</p>
<p style="text-align: justify;">Larutan Y memberikan hasil positif dengan pereaksi Tollens dan ketika dioksidasi menghasilkan larutan X, artinya <strong>Y merupakan senyawa aldehid</strong>. Hasil oksidasi aldehid adalah senyawa yang memiliki gugus karboksil atau asam karboksilat. Oleh karena itu, <strong>X merupakan senyawa asam karboksilat</strong>.</p>
<p style="text-align: justify;">Jenis senyawa pada opsi jawaban adalah sebagai berikut.</p>
<ol style="list-style-type: upper-alpha; text-align: justify;">
<li>1-propanol <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/image-png-Nov-23-2021-01-47-40-17-AM.png" /> alkohol primer</li>
<li>propanal <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/image-png-Nov-23-2021-01-47-46-25-AM.png" /> aldehid</li>
<li>2-propanon <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/image-png-Nov-23-2021-01-47-48-55-AM.png" /> keton</li>
<li>asam propanoat <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/image-png-Nov-23-2021-01-47-51-12-AM.png" /> asam karboksilat</li>
<li>metil asetat <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/image-png-Nov-23-2021-01-47-58-40-AM.png" />ester</li>
</ol>
<p style="text-align: justify;">Dengan demikian, kemungkinan larutan X yang digunakan oleh laboran tersebut adalah <strong>asam propanoat.</strong></p>
<p style="text-align: justify;"><strong><u>Jadi, jawaban yang benar adalah D.</u></strong></p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<h2 style="font-size: 16px;"><strong>Topik</strong> <strong>: <a href="/blog/latihan-soal-sbmptn-kimia" rel="noopener">Stoikiometri</a></strong></h2>
<p><strong>Subtopik </strong> <strong>: Konsep Kilat Stoikiometri Reaksi dan Kadar</strong></p>
<p style="text-align: justify;"><span style="background-color: transparent; font-size: 1rem;">4. Di dalam laboratorium, suatu sampel urin yang mengandung 1,5 gram urea direaksikan dengan asam nitrit berlebih. Urea bereaksi menurut persamaan berikut.</span></p>
<p><img decoding="async" style="width: 464px; margin-left: auto; margin-right: auto; display: block;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/9-Nov-23-2021-01-52-49-72-AM.png" alt="9-Nov-23-2021-01-52-49-72-AM" width="464" /></p>
<p style="text-align: justify;">Gas yang dihasilkan lalu dialirkan melalui larutan NaOH dan volume akhir diukur. Volume gas yang telah melewati larutan NaOH pada temperatur dan tekanan ruang adalah ….</p>
<p style="text-align: justify;"><img decoding="async" style="width: 486px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/10-Nov-23-2021-01-52-49-81-AM.png" alt="10-Nov-23-2021-01-52-49-81-AM" width="486" /></p>
<p><img decoding="async" style="width: 96px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/11-Nov-23-2021-01-52-49-76-AM.png" alt="11-Nov-23-2021-01-52-49-76-AM" width="96" /></p>
<p style="font-weight: bold;">Kunci Jawaban : E</p>
<p style="font-weight: bold;">Pembahasan :</p>
<p>Pertama, setarakan persamaan reaksinya terlebih dahulu.</p>
<p><img decoding="async" style="width: 396px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/12-Nov-23-2021-01-52-49-78-AM.png" alt="12-Nov-23-2021-01-52-49-78-AM" width="396" /></p>
<p>Lalu, hitung mol urea menggunakan persamaan berikut.</p>
<p><img decoding="async" style="width: 617px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/13-Nov-23-2021-01-52-49-80-AM.png" alt="13-Nov-23-2021-01-52-49-80-AM" width="617" /></p>
<p style="text-align: justify;">Kemudian, hitung mol gas karbon dioksida dan gas nitrogen menggunakan perbandingan koefisien sebagai berikut.</p>
<p><img decoding="async" style="width: 559px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/14-Nov-23-2021-01-52-49-83-AM.png" alt="14-Nov-23-2021-01-52-49-83-AM" width="559" /></p>
<p style="text-align: justify;">Dari reaksi tersebut, terdapat 2 jenis gas, yaitu <img decoding="async" style="width: 148px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/15-Nov-23-2021-01-52-49-99-AM.png" alt="15-Nov-23-2021-01-52-49-99-AM" width="148" />. Pengukuran sampel dilakukan pada temperatur dan tekanan ruang, yaitu pada suhu 25 derajat dan tekanan 1 atm. Dengan demikian, volume <img decoding="async" style="width: 148px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/15-Nov-23-2021-01-52-49-99-AM.png" alt="15-Nov-23-2021-01-52-49-99-AM" width="148" /> yang dihasilkan dapat dihitung sebagai berikut.</p>
<p> <img decoding="async" style="background-color: transparent; font-size: 1rem; width: 628px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/16-Nov-23-2021-01-52-50-24-AM.png" alt="16-Nov-23-2021-01-52-50-24-AM" width="628" /></p>
<p><strong>Ingat!</strong> Gas <img decoding="async" style="width: 28px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/17-Nov-23-2021-01-52-49-89-AM.png" alt="17-Nov-23-2021-01-52-49-89-AM" width="28" /> adalah gas asam sehingga akan bereaksi ketika dilarutkan ke dalam NaOH. Dengan demikian, volume gas yang telah melewati larutan NaOH (volume gas <img decoding="async" style="width: 21px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/18-Nov-23-2021-01-52-50-05-AM.png" alt="18-Nov-23-2021-01-52-50-05-AM" width="21" />) pada kondisi yang ditentukan pada soal adalah <img decoding="async" style="width: 65px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/19-Nov-23-2021-01-52-49-91-AM.png" alt="19-Nov-23-2021-01-52-49-91-AM" width="65" />.</p>
<p><strong><u>Jadi, jawaban yang tepat adalah E.</u></strong></p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<h2 style="font-size: 16px;"><strong>Topik</strong> <strong>: <a href="/blog/mengetahui-pengertian-sifat-koligatif-pada-larutan" rel="noopener">Sifat Koligatif Larutan</a></strong></h2>
<p><strong>Subtopik</strong> <strong>: Konsep Kilat Sifat Koligatif Larutan</strong></p>
<p style="text-align: justify;"><span style="background-color: transparent; font-size: 1rem;">5. Dalam dunia kedokteran, tekanan osmotik perlu diperhatikan dalam proses pemberian infus kepada pasien. Tekanan osmotik dalam darah berkisar 7,7 atm pada suhu darah 38 derajat celcius</span><span style="background-color: transparent; font-size: 1rem;">. Apabila seseorang memerlukan injeksi cairan nutrisi/infus, tekanan osmotik cairan infus tersebut harus isotonik dengan tekanan osmotik darah. Jumlah NaCl yang diperlukan untuk membuat larutan infus sebanyak 1000 mL yang dibutuhkan oleh pasien tersebut adalah …. <img decoding="async" style="width: 298px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/20-Nov-23-2021-02-01-12-01-AM.png" alt="20-Nov-23-2021-02-01-12-01-AM" width="298" /></span></p>
<ol style="list-style-type: upper-alpha;">
<li>8,8 g</li>
<li>5,5 g</li>
<li>4,4 g</li>
<li>4,0 g</li>
<li>3,0 g</li>
</ol>
<p> </p>
<p style="font-weight: bold;">Jawaban: A</p>
<p style="font-weight: bold;">Pembahasan:</p>
<p>Isotonik artinya memiliki tekanan osmotik yang sama.</p>
<p><strong>Diketahui:</strong></p>
<p><img decoding="async" style="width: 366px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/21-Nov-23-2021-02-01-11-87-AM.png" alt="21-Nov-23-2021-02-01-11-87-AM" width="366" /></p>
<p><strong>Ditanyakan:</strong></p>
<p>Massa NaCl yang dilarutkan</p>
<p><strong>Pembahasan</strong>:</p>
<p>NaCl merupakan garam yang bersifat elektrolit kuat, artinya dapat terurai menjadi ion-ionnya.</p>
<p><img decoding="async" style="width: 119px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/22-Nov-23-2021-02-01-12-02-AM.png" alt="22-Nov-23-2021-02-01-12-02-AM" width="119" /></p>
<p>Jumlah ion (n) = 2</p>
<p><img decoding="async" style="width: 271px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/23-Nov-23-2021-02-01-12-15-AM.png" alt="23-Nov-23-2021-02-01-12-15-AM" width="271" /></p>
<p>Massa NaCl yang diperlukan untuk membuat larutan infus tersebut dapat ditentukan melalui persamaan berikut.</p>
<p><img decoding="async" style="width: 356px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/24-Nov-23-2021-02-01-12-05-AM.png" alt="24-Nov-23-2021-02-01-12-05-AM" width="356" /></p>
<p>Berdasarkan persamaan di atas, massa dapat ditentukan sebagai berikut.</p>
<p><img decoding="async" style="width: 577px;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs/25-Nov-23-2021-02-01-12-35-AM.png" alt="25-Nov-23-2021-02-01-12-35-AM" width="577" /></p>
<p><strong><u>Jadi, jawaban yang tepat adalah A.</u></strong></p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<p style="font-size: 16px;"><strong>Topik</strong> <strong>: </strong><strong>Ikatan Kimia dan <a href="/blog/bentuk-molekul-dan-definisinya" rel="noopener">Bentuk Molekul</a></strong></p>
<p><strong>Subtopik</strong> <strong>: </strong><strong>Bentuk dan Geometri Molekul</strong></p>
<p>6. Unsur Xe merupakan golongan gas mulia yang sukar berikatan dengan unsur lain. Namun, seiring berkembangnya teknologi, ternyata dapat dibuat senyawa dari unsur Xe dengan unsur paling elektronegatif F yaitu XeF<sub>4</sub>. Geometri dan tipe hibridisasi pada molekul XeF<sub>4 </sub>(NA Xe = 54, F = 9) adalah…</p>
<ol style="list-style-type: upper-alpha;">
<li>segi empat planar, sp<sup>3</sup>d<sup>2</sup></li>
<li>tetrahedron, sp<sup>3</sup>d</li>
<li>segi empat planar, sp<sup>2</sup>d</li>
<li>tetrahedron, sp<sup>3</sup></li>
<li>bipiramida trigonal, sp<sup>3</sup></li>
</ol>
<p> </p>
<p><strong>Kunci</strong> <strong>: </strong><strong>A</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>Untuk membuat hibridisasi yang dilihat adalah atom pusat. Pada senyawa XeF<sub>4</sub> yang menjadi atom pusat adalah Xe. Konfigurasi elektron <sub>54</sub>Xe = [Kr] 5s<sup>2</sup>4d<sup>10</sup> 5p<sup>6</sup></p>
<p><img decoding="async" style="margin-left: auto; margin-right: auto; display: block;" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//1-405577069126.png?width=509&name=1-405577069126.png" sizes="(max-width: 509px) 100vw, 509px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//1-405577069126.png?width=255&name=1-405577069126.png 255w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//1-405577069126.png?width=509&name=1-405577069126.png 509w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//1-405577069126.png?width=764&name=1-405577069126.png 764w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//1-405577069126.png?width=1018&name=1-405577069126.png 1018w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//1-405577069126.png?width=1273&name=1-405577069126.png 1273w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//1-405577069126.png?width=1527&name=1-405577069126.png 1527w" alt="prediksi soal Unsur utbk sbmptn kimia" width="509" /></p>
<p> </p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<h2 style="font-size: 16px;"><strong>Topik</strong> <strong>: </strong><strong><a href="/blog/konsep-reaksi-reduksi-oksidasi-kimia-kelas-10" rel="noopener">Redoks</a> dan Elektrokimia</strong></h2>
<p><strong>Subtopik</strong> <strong>: </strong><strong><a href="/blog/rangkaian-sel-volta" rel="noopener">Sel Volta</a>; <a href="/blog/mengenal-bunyi-hukum-faraday" rel="noopener">Hukum Faraday</a></strong></p>
<p>7. Suatu sel elektrokimia dengan elektroda Zn dan Ag pada keadaan standar menghasilkan arus 0,75 A selama 321 menit. Diketahui nilai:</p>
<p>E<sup>0</sup> Zn<sup>2+</sup> | Zn = -0,76 V</p>
<p>E<sup>0</sup> Ag<sup>+</sup> | Ag = +0,8 V</p>
<p>maka pengurangan massa dianoda adalah … gram. (F = 96.500 C, Ar Zn = 65, Ag = 108)</p>
<ol style="list-style-type: upper-alpha;">
<li>3,250</li>
<li>4,865</li>
<li>6,321</li>
<li>10,850</li>
<li>13,106</li>
</ol>
<p> </p>
<p><strong>Kunci</strong> <strong>: </strong><strong>B</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>Pada katoda terjadi reaksi reduksi, sedangkan anoda oksidasi. Potensial reduksi Ag lebih besar dari Zn, sehingga pengurangan massa di anoda terjadi dari oksidasi Zn.</p>
<p>Katoda: Ag<sup>+</sup> + e<sup>–</sup> → Ag</p>
<p>Anoda: Zn → Zn<sup>2+</sup> + 2e<sup>–</sup></p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//2-231.png?width=383&name=2-231.png" sizes="(max-width: 383px) 100vw, 383px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//2-231.png?width=192&name=2-231.png 192w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//2-231.png?width=383&name=2-231.png 383w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//2-231.png?width=575&name=2-231.png 575w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//2-231.png?width=766&name=2-231.png 766w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//2-231.png?width=958&name=2-231.png 958w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//2-231.png?width=1149&name=2-231.png 1149w" alt="2-231" width="383" /></p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<h2 style="font-size: 16px;"><strong>Topik</strong> <strong>: <a href="/blog/pengertian-dan-faktor-yang-memengaruhi-laju-reaksi" rel="noopener">Laju Reaksi</a></strong></h2>
<p><strong>Subtopik</strong> <strong>: </strong><strong>Persamaan Laju Reaksi dan Konstanta Laju Reaksi Multi Experiment</strong></p>
<p>8. Percobaan penentuan laju reaksi:</p>
<p>2H<sub>2</sub>(<em><i>g</i></em>) + 2NO(<em><i>g</i></em>) <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=27&name=3-86.png" sizes="(max-width: 27px) 100vw, 27px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=14&name=3-86.png 14w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=27&name=3-86.png 27w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=41&name=3-86.png 41w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=54&name=3-86.png 54w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=68&name=3-86.png 68w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=81&name=3-86.png 81w" alt="3-86" width="27" /> 2H<sub>2</sub>O(<em><i>g</i></em>) + N<sub>2</sub>(<em><i>g</i></em>)</p>
<p>memberikan data sebagai berikut:</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//4-54.png?width=421&name=4-54.png" sizes="(max-width: 421px) 100vw, 421px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//4-54.png?width=211&name=4-54.png 211w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//4-54.png?width=421&name=4-54.png 421w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//4-54.png?width=632&name=4-54.png 632w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//4-54.png?width=842&name=4-54.png 842w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//4-54.png?width=1053&name=4-54.png 1053w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//4-54.png?width=1263&name=4-54.png 1263w" alt="prediksi soal utbk sbmptn kimia" width="421" /></p>
<p>Harga tetapan laju reaksi dari reaksi tersebut adalah…</p>
<ol style="list-style-type: upper-alpha;">
<li>4 × 10<sup>8</sup></li>
<li>8 × 10<sup>5</sup></li>
<li>8 × 10<sup>6</sup></li>
<li>16 × 10<sup>5</sup></li>
<li>16 × 10<sup>4</sup></li>
</ol>
<p> </p>
<p><strong>Kunci</strong> <strong>: C</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>Menentukan orde [NO] menggunakan persamaan 2 dan 3</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//5-74.png?width=217&name=5-74.png" sizes="(max-width: 217px) 100vw, 217px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//5-74.png?width=109&name=5-74.png 109w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//5-74.png?width=217&name=5-74.png 217w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//5-74.png?width=326&name=5-74.png 326w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//5-74.png?width=434&name=5-74.png 434w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//5-74.png?width=543&name=5-74.png 543w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//5-74.png?width=651&name=5-74.png 651w" alt="5-74" width="217" /></p>
<p>Menentukan orde [H<sub>2</sub>] menggunakan persamaan 1 dan 2</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//6-78.png?width=212&name=6-78.png" sizes="(max-width: 212px) 100vw, 212px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//6-78.png?width=106&name=6-78.png 106w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//6-78.png?width=212&name=6-78.png 212w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//6-78.png?width=318&name=6-78.png 318w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//6-78.png?width=424&name=6-78.png 424w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//6-78.png?width=530&name=6-78.png 530w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//6-78.png?width=636&name=6-78.png 636w" alt="6-78" width="212" /></p>
<p>Maka persamaan laju reaksinya <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//7-51.png?width=125&name=7-51.png" sizes="(max-width: 125px) 100vw, 125px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//7-51.png?width=63&name=7-51.png 63w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//7-51.png?width=125&name=7-51.png 125w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//7-51.png?width=188&name=7-51.png 188w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//7-51.png?width=250&name=7-51.png 250w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//7-51.png?width=313&name=7-51.png 313w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//7-51.png?width=375&name=7-51.png 375w" alt="7-51" width="125" /></p>
<p>Mencari nilai tetapan laju reaksi (k) menggunakan percobaan 1.</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//8-53.png?width=413&name=8-53.png" sizes="(max-width: 413px) 100vw, 413px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//8-53.png?width=207&name=8-53.png 207w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//8-53.png?width=413&name=8-53.png 413w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//8-53.png?width=620&name=8-53.png 620w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//8-53.png?width=826&name=8-53.png 826w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//8-53.png?width=1033&name=8-53.png 1033w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//8-53.png?width=1239&name=8-53.png 1239w" alt="8-53" width="413" /></p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<p style="font-size: 16px;"><strong>Topik</strong> <strong>: </strong><strong>Redoks dan Elektrokimia</strong></p>
<p><strong>Subtopik</strong> <strong>: </strong><strong>Sel Volta</strong></p>
<p>9. Diketahui potensial standar beberapa sel volta sebagai berikut:</p>
<p>P | P<sup>2+</sup> || Q<sup>2+</sup> | Q E<sup>0</sup> = 2,46 V</p>
<p>R | R<sup>2+</sup> || S<sup>2+</sup> | S E<sup>0</sup> = 1,1 V</p>
<p>R | R<sup>2+</sup> || Q<sup>2+</sup> | Q E<sup>0</sup> = 1,56 V</p>
<p>Potensial standar sel P | P<sup>2+</sup> || S<sup>2+</sup> | S adalah…</p>
<ol style="list-style-type: upper-alpha;">
<li>1,1 V</li>
<li>2,00 V</li>
<li>2,66 V</li>
<li>3,56 V</li>
<li>4,02 V</li>
</ol>
<p> </p>
<p><strong>Kunci</strong> <strong>: </strong><strong>B</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>P | P<sup>2+</sup> || Q<sup>2+</sup> | Q E<sup>o</sup> = 2,46 V</p>
<p>R | R<sup>2+</sup> || S<sup>2+</sup> | S E<sup>o</sup> = 1,1 V</p>
<p>R | R<sup>2+</sup> || Q<sup>2+</sup> | Q E<sup>o</sup> = 1,56 V (perlu dibalik untuk mengeliminasi R)</p>
<p>Reaksi yang dibalik memiliki potensial sel dengan nilai berlawanan, sehingga:</p>
<p>P | P<sup>2+</sup> || Q<sup>2+</sup> | Q E<sup>o</sup> = 2,46 V</p>
<p>R | R<sup>2+</sup>|| S<sup>2+</sup> | S E<sup>o</sup> = 1,1 V</p>
<p><u>Q | Q</u><u><sup>2+</sup></u><u> || </u><u>R</u><u><sup>2+</sup></u><u> | R</u> <u>E</u><u><sup>o</sup></u><u> = -1,56 V </u>+</p>
<p>P | P<sup>2+</sup> || S<sup>2+</sup> | S E<sup>o</sup> = 2,00 V</p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<h2 style="font-size: 16px;"><strong>Topik</strong> <strong>: </strong><a href="/blog/kesetimbangan-kimia-jenis-jenis-reaksi-dan-tetapan" rel="noopener"><strong>Kesetimbangan Kimia</strong></a></h2>
<p><strong>Subtopik</strong> <strong>: </strong><strong>Konstanta Kesetimbangan Tekanan (Kp)</strong></p>
<p>10. Dalam wadah 2 liter, gas H<sub>2</sub>dan gas I<sub>2</sub> bereaksi membentuk gas HI menurut persamaan reaksi:</p>
<p>H<sub>2</sub>(<em><i>g</i></em>) + I<sub>2</sub>(<em><i>g</i></em>) <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=27&name=3-86.png" sizes="(max-width: 27px) 100vw, 27px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=14&name=3-86.png 14w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=27&name=3-86.png 27w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=41&name=3-86.png 41w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=54&name=3-86.png 54w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=68&name=3-86.png 68w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=81&name=3-86.png 81w" alt="3-86" width="27" /> 2HI(<em><i>g</i></em>)</p>
<p>Jika pada kesetimbangan reaksi terdapat 0,1 mol H<sub>2</sub>, 0,2 mol I<sub>2</sub> dan 0,5 mol HI, dengan tekanan total sebesar 5 atm, maka nilai K<sub>p</sub> reaksi tersebut adalah…</p>
<ol style="list-style-type: upper-alpha;">
<li>7,50</li>
<li>12,5</li>
<li>25,2</li>
<li>32,5</li>
<li>50,0</li>
</ol>
<p> </p>
<p><strong>Kunci</strong> <strong>: </strong><strong>B</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>Diketahui: volume = 2 liter; mol total = 0,1 mol + 0,2 mol + 0,5 mol = 0,8 mol</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//9-48.png?width=408&name=9-48.png" sizes="(max-width: 408px) 100vw, 408px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//9-48.png?width=204&name=9-48.png 204w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//9-48.png?width=408&name=9-48.png 408w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//9-48.png?width=612&name=9-48.png 612w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//9-48.png?width=816&name=9-48.png 816w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//9-48.png?width=1020&name=9-48.png 1020w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//9-48.png?width=1224&name=9-48.png 1224w" alt="9-48" width="408" /></p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<h2 style="font-size: 16px;"><strong>Topik</strong> <strong>: </strong><strong>Stoikiometri</strong></h2>
<p><strong>Subtopik</strong> <strong>: </strong><strong>Persamaan Kimia; Perhitungan Kimia dalam Suatu Persamaan Reaksi</strong></p>
<p>11. Suatu campuran yang hanya mengandung gas metana dan etana dibakar sempurna sehingga dihasilkan 220 gram CO<sub>2</sub>. Jika massa campuran yang dibakar adalah 76 gram, maka massa gas etana di dalam campuran gas tersebut (Ar C= 12 H = 1 O = 16) adalah…</p>
<ol style="list-style-type: upper-alpha;">
<li>16 gram</li>
<li>30 gram</li>
<li>32 gram</li>
<li>44 gram</li>
<li>60 gram</li>
</ol>
<p><strong>Kunci</strong> <strong>: </strong><strong>E</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>Metana = CH<sub>4</sub> (Mr = 16) Massa etana = x gram (sebagai komponen yang ditanya)</p>
<p>Etana = C<sub>2</sub>H<sub>6</sub> (Mr = 30) Massa metana = (76 – x) gram</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//10-39.png?width=257&name=10-39.png" sizes="(max-width: 257px) 100vw, 257px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//10-39.png?width=129&name=10-39.png 129w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//10-39.png?width=257&name=10-39.png 257w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//10-39.png?width=386&name=10-39.png 386w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//10-39.png?width=514&name=10-39.png 514w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//10-39.png?width=643&name=10-39.png 643w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//10-39.png?width=771&name=10-39.png 771w" alt="10-39" width="257" /></p>
<p>Reaksi pembakaran etana:</p>
<p>C<sub>2</sub>H<sub>6</sub> + O<sub>2</sub> → 2CO<sub>2</sub> + 3H<sub>2</sub>O</p>
<p><em><i> <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//11-145.png?width=319&name=11-145.png" sizes="(max-width: 319px) 100vw, 319px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//11-145.png?width=160&name=11-145.png 160w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//11-145.png?width=319&name=11-145.png 319w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//11-145.png?width=479&name=11-145.png 479w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//11-145.png?width=638&name=11-145.png 638w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//11-145.png?width=798&name=11-145.png 798w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//11-145.png?width=957&name=11-145.png 957w" alt="11-145" width="319" /></i></em></p>
<p>Reaksi pembakaran metana:</p>
<p>CH<sub>4</sub> + 2O<sub>2</sub> → CO<sub>2</sub> + 2H<sub>2</sub>O</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//12-33.png?width=401&name=12-33.png" sizes="(max-width: 401px) 100vw, 401px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//12-33.png?width=201&name=12-33.png 201w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//12-33.png?width=401&name=12-33.png 401w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//12-33.png?width=602&name=12-33.png 602w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//12-33.png?width=802&name=12-33.png 802w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//12-33.png?width=1003&name=12-33.png 1003w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//12-33.png?width=1203&name=12-33.png 1203w" alt="12-33" width="401" /></p>
<p>Mol CO<sub>2</sub> total = 5 mol</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//13-30.png?width=190&name=13-30.png" sizes="(max-width: 190px) 100vw, 190px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//13-30.png?width=95&name=13-30.png 95w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//13-30.png?width=190&name=13-30.png 190w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//13-30.png?width=285&name=13-30.png 285w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//13-30.png?width=380&name=13-30.png 380w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//13-30.png?width=475&name=13-30.png 475w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//13-30.png?width=570&name=13-30.png 570w" alt="13-30" width="190" /></p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<p style="font-size: 16px;"><strong>Topik</strong> <strong>: </strong><strong>Kesetimbangan Kimia</strong></p>
<p><strong>Subtopik</strong> <strong>: </strong><strong>Pergeseran Kesetimbangan dan Faktor yang Mempengaruhi…</strong></p>
<p>12. Di bawah ini adalah produksi NH<sub>3</sub>dalam reaksi:</p>
<p>N<sub>2</sub>(<em><i>g</i></em>) + 3H<sub>2</sub>(<em><i>g</i></em>) <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=27&name=3-86.png" sizes="(max-width: 27px) 100vw, 27px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=14&name=3-86.png 14w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=27&name=3-86.png 27w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=41&name=3-86.png 41w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=54&name=3-86.png 54w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=68&name=3-86.png 68w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=81&name=3-86.png 81w" alt="3-86" width="27" /> 2NH<sub>3</sub>(<em><i>g</i></em>) ΔH = -92,2 kJ</p>
<p>Jika diketahui pada suhu 25 ℃ nilai K<sub>c</sub> = 3,5 × 10<sup>8</sup>, maka pernyataan berikut ini yang benar adalah…</p>
<ol style="list-style-type: upper-alpha;">
<li>jika suhu turun, produksi meningkat</li>
<li>ditambah katalis, produksi menurun</li>
<li>jika volume turun, produksi menurun</li>
<li>proses produksinya disebut dengan proses Kontak</li>
<li>pengurangan konsentrasi N<sub>2</sub>, produksi meningkat</li>
</ol>
<p> </p>
<p><strong>Kunci</strong> <strong>: </strong><strong>A</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>Produksi NH<sub>3</sub> disebut dengan proses Haber-Bosch dengan reaksi</p>
<p>N<sub>2</sub>(<em><i>g</i></em>) + 3H<sub>2</sub>(<em><i>g</i></em>) <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=27&name=3-86.png" sizes="(max-width: 27px) 100vw, 27px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=14&name=3-86.png 14w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=27&name=3-86.png 27w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=41&name=3-86.png 41w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=54&name=3-86.png 54w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=68&name=3-86.png 68w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//3-86.png?width=81&name=3-86.png 81w" alt="3-86" width="27" /> 2NH<sub>3</sub>(<em><i>g</i></em>) ΔH = -92,2 kJ (reaksi eksoterm)</p>
<p>Maka, pernyataan yang tepat yaitu:</p>
<ul>
<li>jika suhu turun, akan bergeser ke eksoterm maka produksi meningkat.</li>
<li>ditambah katalis, reaksi akan semakin cepat, tidak berpengaruh pada hasil produksi.</li>
<li>jika volume turun, akan bergeser ke koefisien kecil maka produksi akan meningkat.</li>
<li>pengurangan konsentrasi N<sub>2</sub>, akan bergeser ke arah reaktan,jadi produksi menurun.</li>
</ul>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<p style="font-size: 16px;"><strong>Topik</strong> <strong>: </strong><strong>Sifat Koligatif Larutan</strong></p>
<p><strong>Subtopik</strong> <strong>: </strong><strong>Kenaikan </strong><strong>T</strong><strong>itik Didih</strong></p>
<p>13. Struktur molekul PVC dapat dinyatakan (–CH2 –CHCl–)<sub>n</sub>. Larutan yang mengandung 12,5 gram bahan tersebut dalam 500 gram benzena mendidih pada suhu 80,11℃. Jika harga K<sub>b</sub> 2,5 ℃/molal dan titik didih benzena 80,10 ℃, maka harga n (Ar H = 1, C = 12, Cl = 35,5) adalah…</p>
<ol style="list-style-type: upper-alpha;">
<li>50</li>
<li>100</li>
<li>500</li>
<li>1000</li>
<li>1500</li>
</ol>
<p> </p>
<p><strong>Kunci</strong> <strong>: </strong><strong>B</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>Diketahui:</p>
<p>massa PVC (terlarut) = 12,5 gram; T<sub>b</sub> = 80,11 ℃; K<sub>b</sub> = 2,5 ℃/molal</p>
<p>massa benzena (pelarut) = 500 gram; ΔT<sub>b</sub> = 80,11 ℃ – 80,10 ℃ = 0,01 ℃</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//14-5577069093.png?width=505&name=14-5577069093.png" sizes="(max-width: 505px) 100vw, 505px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//14-5577069093.png?width=253&name=14-5577069093.png 253w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//14-5577069093.png?width=505&name=14-5577069093.png 505w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//14-5577069093.png?width=758&name=14-5577069093.png 758w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//14-5577069093.png?width=1010&name=14-5577069093.png 1010w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//14-5577069093.png?width=1263&name=14-5577069093.png 1263w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//14-5577069093.png?width=1515&name=14-5577069093.png 1515w" alt="14-5577069093" width="505" /></p>
<p>(Rumus Empiris)n = Rumus Molekul</p>
<p>(Mr Empiris)n = Mr Molekul</p>
<p>(62,5)n = 6250 <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//15-25.png?width=87&name=15-25.png" sizes="(max-width: 87px) 100vw, 87px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//15-25.png?width=44&name=15-25.png 44w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//15-25.png?width=87&name=15-25.png 87w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//15-25.png?width=131&name=15-25.png 131w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//15-25.png?width=174&name=15-25.png 174w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//15-25.png?width=218&name=15-25.png 218w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//15-25.png?width=261&name=15-25.png 261w" alt="15-25" width="87" /></p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<h2 style="font-size: 16px;"><strong>Topik</strong> <strong>: </strong><a href="/blog/persamaan-termokimia-dan-jenis-perubahan-entalpi-standar" rel="noopener"><strong>Termokimia</strong></a></h2>
<p><strong>Subtopik</strong> <strong>: </strong><strong>Penentuan Perubahan Entalpi</strong></p>
<p>14. Diketahui energi rata-rata sebagai berikut:</p>
<p>C – H = 414 kJ/mol H – Cl = 432 kJ/mol</p>
<p>Cl – Cl = 244 kJ/mol C – Cl = 326 kJ/mol</p>
<p>Perubahan entalpi untuk 3,2 gram metana pada reaksi:</p>
<p>CH<sub>4</sub>(<em><i>g</i></em>) + Cl<sub>2</sub>(<em><i>g</i></em>) → CH<sub>3</sub>Cl(<em><i>g</i></em>) + HCl(<em><i>g</i></em>) adalah … kJ. (Ar C = 12, H = 1)</p>
<ol style="list-style-type: upper-alpha;">
<li>-20</li>
<li>+20</li>
<li>-80</li>
<li>+100</li>
<li>-100</li>
</ol>
<p> </p>
<p><strong>Kunci</strong> <strong>: </strong><strong>A</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>Persamaan reaksi:</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//16-21.png?width=412&name=16-21.png" sizes="(max-width: 412px) 100vw, 412px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//16-21.png?width=206&name=16-21.png 206w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//16-21.png?width=412&name=16-21.png 412w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//16-21.png?width=618&name=16-21.png 618w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//16-21.png?width=824&name=16-21.png 824w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//16-21.png?width=1030&name=16-21.png 1030w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//16-21.png?width=1236&name=16-21.png 1236w" alt="16-21" width="412" /></p>
<p>ΔH = ∑H reaktan – ∑H produk</p>
<p>ΔH = 658 – 758 = -100 kJ/mol</p>
<p>Maka untuk 3,2 gram metana dilepaskan kalor sebesar 100 kJ/mol.</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//17-18.png?width=195&name=17-18.png" sizes="(max-width: 195px) 100vw, 195px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//17-18.png?width=98&name=17-18.png 98w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//17-18.png?width=195&name=17-18.png 195w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//17-18.png?width=293&name=17-18.png 293w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//17-18.png?width=390&name=17-18.png 390w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//17-18.png?width=488&name=17-18.png 488w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//17-18.png?width=585&name=17-18.png 585w" alt="17-18" width="195" /></p>
<p>ΔH<sub>r</sub> = 0,2 mol × -100 kJ/mol = -20 kJ</p>
<p><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
<p> </p>
<p style="font-size: 16px;"><strong>Topik</strong> <strong>: </strong><strong>Termokimia</strong></p>
<p><strong>Subtopik</strong> <strong>: </strong><strong>Penentuan Perubahan Entalpi</strong></p>
<p>15. Etanol, disebut juga etil alkohol, memiliki rumus molekul C<sub>2</sub>H<sub>5</sub>OH adalah sejenis cairan yang mudah menguap dan mudah terbakar sehingga etanol dapat digunakan sebagai bahan bakar. Jika spiritus dianggap hanya mengandung etanol, berapa gram spiritus yang harus dibakar untuk menaikkan suhu 100 gram air dari 20℃ menjadi 50 ℃? Diketahui:</p>
<p>∆H<sup>o</sup><sub>f</sub> H<sub>2</sub>O(g) = -240 kJ/mol</p>
<p>∆H<sup>o</sup><sub>f</sub> CO<sub>2</sub>(g) = -394 kJ/mol</p>
<p>∆H<sup>o</sup><sub>f </sub> C<sub>2</sub>H<sub>5</sub>OH = -277 kJ/mol</p>
<p>kalor jenis air = 4,2 J/gK</p>
<p>Ar C = 12, H = 1, O = 16</p>
<ol style="list-style-type: upper-alpha;">
<li>0,235 gram</li>
<li>0,47 gram</li>
<li>0,94 gram</li>
<li>6,3 gram</li>
<li>12,6 gram</li>
</ol>
<p> </p>
<p><strong>Kunci</strong> <strong>: </strong><strong>B</strong></p>
<p><strong>Pembahasan</strong> <strong>:</strong></p>
<p>Mr etanol = 46 g/mol; ∆T = 50 ℃ – 20 ℃ = 30 ℃ = 30 K</p>
<p>Reaksi pembakaran etanol adalah: C<sub>2</sub>H<sub>5</sub>OH + 3O<sub>2 </sub>→ 2CO<sub>2</sub> + 3H<sub>2</sub>O</p>
<p>∆H = <img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=43&name=20-15.png" sizes="(max-width: 43px) 100vw, 43px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=22&name=20-15.png 22w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=43&name=20-15.png 43w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=65&name=20-15.png 65w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=86&name=20-15.png 86w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=108&name=20-15.png 108w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=129&name=20-15.png 129w" alt="20-15" width="43" /><sub>produk</sub> –<img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=43&name=20-15.png" sizes="(max-width: 43px) 100vw, 43px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=22&name=20-15.png 22w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=43&name=20-15.png 43w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=65&name=20-15.png 65w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=86&name=20-15.png 86w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=108&name=20-15.png 108w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//20-15.png?width=129&name=20-15.png 129w" alt="20-15" width="43" /><sub>reaktan</sub></p>
<p>= (2×(-394) + 3×(-240)) – (–277 + 0)</p>
<p>= –1231 kJ/mol</p>
<p><img decoding="async" src="https://cdn-web.ruangguru.com/landing-pages/assets/hs//19-16.png?width=343&name=19-16.png" sizes="(max-width: 343px) 100vw, 343px" srcset="https://cdn-web.ruangguru.com/landing-pages/assets/hs//19-16.png?width=172&name=19-16.png 172w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//19-16.png?width=343&name=19-16.png 343w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//19-16.png?width=515&name=19-16.png 515w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//19-16.png?width=686&name=19-16.png 686w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//19-16.png?width=858&name=19-16.png 858w, https://cdn-web.ruangguru.com/landing-pages/assets/hs//19-16.png?width=1029&name=19-16.png 1029w" alt="19-16" width="343" /></p>
<p style="text-align: justify;"><img decoding="async" class="aligncenter size-medium" src="https://cdn-web-2.ruangguru.com/landing-pages/assets/c71c4e5b-fbae-434f-b2ce-35b757534bc6.png" alt="Separator Latihan Soal UTBK/SMA" /></p>
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debug: false,
plugins: [analyticsTrackable()],
});
/* Additional context tracker pageView */
const post_id = 3526;
const tags_id = [55,58];
const category_id = 563;
const sub_category_id = 565;
const customData = {
tags_id: tags_id,
category_id: category_id,
sub_category_id: sub_category_id,
post_id: post_id,
};
Analytics.plugins.trackable.context(JSON.stringify(customData));
Analytics.ready(function () {
Analytics.page();
});
$(document).on('click', '[data-tracking-event-type]', function (event) {
const eventType = $(this).data('tracking-event-type');
const context = $(this).data('tracking-context');
Analytics.track(eventType, context);
});
</script>
<footer class="footer">
<div class="container">
<div class="row">
<div class="col-lg-3">
<div class="office">
<a href="https://www.ruangguru.com/">
<img src="https://cdn-web-2.ruangguru.com/static/RG_LOGO_WHITE.png?convert=auto" alt="Logo Ruangguru" loading="lazy" />
</a>
<address>
<p class="hq">RUANGGURU HQ</p>
<p>
Gedung Sinarmas MSIG Tower Lt. 33. Jl. Jenderal Sudirman Kav. 21, Karet Kuningan, Setiabudi, Jakarta Selatan, Jakarta 12920
</p>
</address>
</div>
<nav class="footer-menu call-us">
<p class="title">Hubungi Kami</p>
<address>
<a href="mailto:info@ruangguru.com">
<img class="icon-call-us" src="https://cdn-web-2.ruangguru.com/file-uploader/lp/4739a139-8f09-42ea-b517-d842c5d1c28d.png" loading="lazy" alt="Logo Email Footer" />
info@ruangguru.com</a>
<a href="tel:02130930000">
<img class="icon-call-us" src="https://cdn-web-2.ruangguru.com/file-uploader/lp/6f4bb6c8-52c9-4acb-859b-47a10a7eec75.png" loading="lazy" alt="Logo Phone Footer" />(021) 3093 0000</a>
<a href="https://wa.me/6281574410000">
<img class="icon-call-us" src="https://cdn-web-2.ruangguru.com/file-uploader/lp/4b58468a-b2f9-40e4-867f-d3242a065dcf.png" loading="lazy" alt="Logo Whatsapp Footer" />0815 7441 0000</a>
</address>
</nav>
<nav class="footer-menu social d-none d-lg-block">
<p class="title">Ikuti Kami</p>
<a target="_blank" href="https://www.instagram.com/ruangguru/">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/instagram.png?height=60&convert=webp" type="image/webp">
<source srcset="https://cdn-web-2.ruangguru.com/static/instagram.png?height=60" type="image/png">
<img src="https://cdn-web-2.ruangguru.com/static/instagram.png" loading="lazy" alt="Logo Instagram Footer" width="32px" height="32px" />
</picture>
</a>
<a target="_blank" href="https://www.facebook.com/ruanggurucom/">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/facebook.png?height=60&convert=webp" type="image/webp">
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<img src="https://cdn-web-2.ruangguru.com/static/facebook.png" loading="lazy" alt="Logo Facebook Footer" width="32px" height="32px" />
</picture>
</a>
<a target="_blank" href="https://twitter.com/ruangguru/">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/twitter.png?height=60&convert=webp" type="image/webp">
<source srcset="https://cdn-web-2.ruangguru.com/static/twitter.png?height=60" type="image/png">
<img src="https://cdn-web-2.ruangguru.com/static/twitter.png" loading=" lazy" alt="Logo Twitter Footer" width="32px" height="32px" />
</picture>
</a>
<a target="_blank" href="https://www.youtube.com/ruangguru">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/youtube.png?height=60&convert=webp" type="image/webp">
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<img src="https://cdn-web-2.ruangguru.com/static/youtube.png" loading=" lazy" alt="Logo Youtube Footer" width="32px" height="32px" />
</picture>
</a>
<a target="_blank" href="https://www.linkedin.com/company/ruangguru-com">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/linkedin.png?height=60&convert=webp" type="image/webp">
<source srcset="https://cdn-web-2.ruangguru.com/static/linkedin.png?height=60" type="image/png">
<img src="https://cdn-web-2.ruangguru.com/static/linkedin.png" loading=" lazy" alt="Logo LinkedIn Footer" width="32px" height="32px" />
</picture>
</a>
<a target="_blank" href="https://www.tiktok.com/@ruangguru">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/tiktok.png?height=60&convert=webp" type="image/webp">
<source srcset="https://cdn-web-2.ruangguru.com/static/tiktok.png?height=60" type="image/png">
<img src="https://cdn-web-2.ruangguru.com/static/tiktok.png" loading=" lazy" alt="Logo TikTok Footer" width="32px" height="32px" />
</picture>
</a>
</nav>
</div>
<div class="col-lg-9">
<div class="row">
<div class="col-6 col-md-3">
<nav class="footer-menu">
<p class="title">Produk Ruangguru</p>
<ul type="none">
<li><a href="https://roboguru.ruangguru.com" target="_blank">Roboguru</a></li>
<li><a href="https://www.ruangguru.com/roboguruplus">Roboguru Plus</a></li>
<li><a href="https://www.ruangguru.com/dafa-lulu">Dafa dan Lulu</a></li>
<li><a href="https://www.ruangguru.com/for-kids/kursus">Kursus for Kids</a></li>
<li><a href="https://www.ruangguru.com/for-kids">Ruangguru for Kids</a></li>
<li><a href="https://www.ruangguru.com/bisnis">Ruangguru for Business</a></li>
<li><a href="https://www.ruangguru.com/ruanguji">Ruanguji</a></li>
<li><a href="https://www.ruangguru.com/blog">Ruangbaca</a></li>
<li><a href="https://www.ruangguru.com/ruangkelas">Ruangkelas</a></li>
<li><a href="https://www.ruangguru.com/ruangbelajar">Ruangbelajar</a></li>
<li><a href="https://pengajar.ruangguru.com/">Ruangpengajar</a></li>
<li><a href="https://www.ruangguru.com/privat">Ruangguru Privat</a></li>
<li><a href="https://www.ruangpeduli.org/" target="_blank">Ruangpeduli</a></li>
</ul>
</nav>
</div>
<div class="col-6 col-md-3">
<nav class="footer-menu">
<p class="title">Bantuan & Panduan</p>
<ul type="none">
<li><a href="https://www.ruangguru.com/company-credentials">Company Credentials</a></li>
<li><a href="https://www.ruangguru.com/beasiswa">Beasiswa Ruangguru</a></li>
<li><a href="https://www.ruangguru.com/cicilan">Cicilan Ruangguru</a></li>
<li><a href="https://www.ruangguru.com/promo">Promo Ruangguru</a></li>
<li><a href="https://ticketing.ruangguru.com/help?request_type_id=459" target="_blank">Vulnerability Report</a></li>
<li><a href="https://www.ruangguru.com/bantuan#layanan">Layanan Pengaduan</a></li>
<li><a href="https://www.ruangguru.com/terms-conditions">Syarat & Ketentuan</a></li>
<li><a href="https://www.ruangguru.com/privacy-policy">Kebijakan Privasi</a></li>
<li><a href="https://www.ruangguru.com/about-us">Tentang Kami</a></li>
<li><a href="https://www.ruangguru.com/contact-us">Kontak Kami</a></li>
<li><a href="https://www.ruangguru.com/press">Press Kit</a></li>
<li><a href="https://www.ruangguru.com/bantuan">Bantuan</a></li>
<li><a href="https://career.ruangguru.com/">Karier</a></li>
</ul>
</nav>
</div>
<div class="col-6 col-md-3">
<nav class="footer-menu">
<p class="title">Produk Lainnya</p>
<ul type="none">
<li><a href="https://www.brainacademy.id/" target="_blank">Brain Academy</a></li>
<li><a href="https://www.english-academy.id/" target="_blank">English Academy</a></li>
<li><a href="https://skillacademy.com/" target="_blank">Skill Academy</a></li>
<li><a href="https://www.ruangkerja.id/" target="_blank">Ruangkerja</a></li>
<li><a href="https://www.schoters.com/id/" target="_blank">Schoters</a></li>
<li><a href="https://www.kalananti.id/" target="_blank">Kalananti</a></li>
</ul>
</nav>
</div>
<div class="col-6 col-md-3">
<nav class="footer-menu">
<p class="title"> Coba GRATIS Aplikasi Ruangguru</p>
<div class="download-list">
<ul>
<li>
<a target="_blank" rel="nofollow" href="https://apps.apple.com/id/app/ruangguru-solusi-belajar/id1099742206?l=id">
<img src="https://cdn-web-2.ruangguru.com/static/appstore.png" loading="lazy" alt="Button App Store Footer " />
</a>
</li>
<li>
<a target="_blank" rel="nofollow" href="https://play.google.com/store/apps/details?id=com.ruangguru.livestudents&hl=in">
<img src="https://cdn-web-2.ruangguru.com/static/googleplay.png" loading="lazy" alt="Button Google Play Footer " />
</a>
</li>
<li>
<a target="_blank" rel="nofollow" href="https://appgallery.huawei.com/#/app/C102352375">
<img src="https://cdn-web-2.ruangguru.com/static/huawei.png" loading="lazy" alt="Button AppGallery Footer " />
</a>
</li>
</ul>
</div>
</nav>
</div>
</div>
</div>
</div>
<div class="d-lg-none text-center">
<nav class="footer-menu social">
<p class="title">Ikuti Kami</p>
<a target="_blank" href="https://www.instagram.com/ruangguru/">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/instagram.png?height=60&convert=webp" type="image/webp">
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<img src="https://cdn-web-2.ruangguru.com/static/instagram.png" loading="lazy" alt="Logo Instagram Footer" width="32px" height="32px" />
</picture>
</a>
<a target="_blank" href="https://www.facebook.com/ruanggurucom/">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/facebook.png?height=60&convert=webp" type="image/webp">
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<img src="https://cdn-web-2.ruangguru.com/static/facebook.png" loading="lazy" alt="Logo Facebook Footer" width="32px" height="32px" />
</picture>
</a>
<a target="_blank" href="https://twitter.com/ruangguru/">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/twitter.png?height=60&convert=webp" type="image/webp">
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</picture>
</a>
<a target="_blank" href="https://www.youtube.com/ruangguru">
<picture>
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</a>
<a target="_blank" href="https://www.linkedin.com/company/ruangguru-com">
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</a>
<a target="_blank" href="https://www.tiktok.com/@ruangguru">
<picture>
<source srcset="https://cdn-web-2.ruangguru.com/static/tiktok.png?height=60&convert=webp" type="image/webp">
<source srcset="https://cdn-web-2.ruangguru.com/static/tiktok.png?height=60" type="image/png">
<img src="https://cdn-web-2.ruangguru.com/static/tiktok.png" loading=" lazy" alt="Logo TikTok Footer" width="32px" height="32px" />
</picture>
</a>
</nav>
</div>
<p class="footer-copyright">
© 2026 All Rights Reserved PT. Ruang Raya Indonesia
</p>
</div>
<style>
.bantuan-container {
position: relative;
}
.bantuan-element {
position: fixed;
z-index: 999;
display: flex;
align-items: flex-end;
justify-content: flex-end;
cursor: pointer;
transition: all 0.6s ease;
}
.bantuan-top {
bottom: 11rem;
right: 2rem;
}
.bantuan-bottom {
bottom: 2rem;
right: 2rem;
}
.bantuan-top-2 {
bottom: 15rem;
right: 2rem;
}
.bantuan-top-3 {
bottom: 19rem;
right: 2rem;
}
.bantuan-bottom-2 {
bottom: 5rem;
right: 2rem;
}
.bantuan-bottom-3 {
bottom: 8rem;
right: 2rem;
}
@media (min-width: 768px) {
.bantuan-element img {
width: auto;
transition: all 0.4s ease;
}
.bantuan-top img {
width: 140px;
}
.bantuan-bottom img {
width: 140px;
}
.bantuan-top-2 img,
.bantuan-top-3 img {
width: 140px;
}
.bantuan-bottom-2 img,
.bantuan-bottom-3 img {
width: 140px;
}
.bantuan-element.clicked {
bottom: 9rem;
right: 2.5rem;
}
.bantuan-element.clicked img {
width: auto;
height: 64px;
}
}
@media (max-width: 767px) {
.bantuan-top {
bottom: 8rem;
right: 1rem;
align-items: center;
}
.bantuan-bottom {
bottom: 2rem;
right: 1.25rem;
align-items: center;
}
.bantuan-top-2 {
bottom: 12rem;
right: 1rem;
align-items: center;
}
.bantuan-top-3 {
bottom: 16rem;
right: 1rem;
align-items: center;
}
.bantuan-bottom-2 {
bottom: 5rem;
right: 1.25rem;
align-items: center;
}
.bantuan-bottom-3 {
bottom: 8rem;
right: 1.25rem;
align-items: center;
}
.bantuan-element img {
width: auto;
}
.bantuan-top img,
.bantuan-top-2 img,
.bantuan-top-3 img {
width: 96px;
}
.bantuan-bottom img,
.bantuan-bottom-2 img,
.bantuan-bottom-3 img {
width: 96px;
}
}
.bantuan-element:hover {
transform: scale(1.05);
}
.bantuan-element img {
max-width: 100%;
height: auto;
}
</style>
<div id="bantuan-divs-container" style="position: relative" data-product="ruangguru" data-table="rg_floating_icon"></div>
<script>
class FloatingDivsComponent {
constructor(product = 'ruangguru', table = 'rg_floating_icon') {
this.container = document.getElementById('bantuan-divs-container');
this.floatingElements = [];
this.product = product;
this.table = table;
// Configuration
this.config = {
baseApiUrl: 'https://gw.ruangguru.com/api/v3/rdb/p_ddgy8hjx5dxsqx/rg_banner/views/',
apiKey: 'vLyK3dJrDBRmOwueX2OJXibMkgsUbBk0WS4ua2p1',
maxElements: 2,
zIndexBase: 100,
};
// Build dynamic API URL based on table parameter
this.config.apiUrl = this.buildApiUrl();
this.init();
}
buildApiUrl() {
const baseUrl = this.config.baseApiUrl;
const tableName = this.table || 'rg_floating_icon';
const queryParams =
'offset=0&limit=99999&sort=rankPosition&where=(type,eq,floating)~and(showIn,anyof,blog)~and(hide,is,false)~and(product,eq,' +
this.product +
')&nested%5Bcta%5D%5Bfields%5D=Id,name,ctaId,imageUrl,ctaLink,targetUrl';
return `${baseUrl}${tableName}?${queryParams}`;
}
async init() {
try {
await this.fetchFloatingElements();
this.renderFloatingElements();
} catch (error) {
console.error('Failed to initialize floating divs component:', error);
}
}
async fetchFloatingElements() {
try {
const options = {
method: 'GET',
headers: {
'xc-token': this.config.apiKey || 'vLyK3dJrDBRmOwueX2OJXibMkgsUbBk0WS4ua2p1',
},
};
const urlNoco = this.config.apiUrl;
const response = await fetch(urlNoco, options);
if (!response.ok) {
throw new Error(`HTTP error! status: ${response.status}`);
}
const data = await response.json();
this.floatingElements = this.processApiResponse(data);
} catch (error) {
console.error('Error fetching floating elements:', error);
throw error;
}
}
processApiResponse(data) {
if (!data || !data.list || !Array.isArray(data.list)) {
return [];
}
console.log(data.list);
// Ensure we only get maximum 2 elements
const limitedElements = data.list.slice(0, this.config.maxElements);
return limitedElements;
}
renderFloatingElements() {
if (!this.container || this.floatingElements.length === 0) {
return;
}
// Additional safeguard: ensure we never render more than maxElements
const elementsToRender = this.floatingElements.slice(0, this.config.maxElements);
// Clean up existing elements and their event listeners
this.cleanupExistingElements();
// Clear existing content
this.container.innerHTML = '';
elementsToRender.forEach((element, index) => {
const floatingDiv = this.createFloatingDiv(element, index);
this.container.appendChild(floatingDiv);
});
}
cleanupExistingElements() {
// Clear existing elements (no need for event listener cleanup with CSS approach)
const existingElements = this.container.querySelectorAll('[data-floating-element]');
existingElements.forEach(element => {
element.remove();
});
}
createFloatingDiv(element, index) {
const div = document.createElement('div');
const cta = element.cta[0]; // Use the first CTA
// Add data attribute for cleanup tracking
div.setAttribute('data-floating-element', 'true');
// Apply CSS classes based on rankPosition
const positionClass = this.getPositionClass(element.rankPosition, index);
div.className = `bantuan-element ${positionClass}`;
div.style.zIndex = this.config.zIndexBase + index;
// Create the link and image
const link = document.createElement('a');
link.href = 'https://cta.ruangguru.com/' + cta.ctaId || '#';
link.target = '_blank';
// Add tracking attributes
link.setAttribute('data-tracking-event-type', 'landing_page__core__item_click');
link.setAttribute(
'data-tracking-context',
JSON.stringify({
product_name: this.product,
cta_action: 'redirection',
cta_action_detail: link.href,
item_type: 'floating',
item_name: element.name || 'Floating element',
item_category: 'payment',
item_position: index,
item_serial: null,
section_type: 'floating_item',
section_position: 'body',
section_name: 'floating_item',
})
);
const img = document.createElement('img');
img.src = element.imageUrl + '?h=200&convert=webp';
img.alt = element.name || 'Floating element';
img.loading = 'lazy';
// Add click tracking
link.addEventListener('click', () => {
this.trackClick(element, cta);
});
link.appendChild(img);
div.appendChild(link);
return div;
}
getPositionClass(rankPosition, index) {
// Define positioning classes based on rankPosition
if (rankPosition === 1) {
return 'bantuan-top'; // Top position
} else if (rankPosition === 2) {
return 'bantuan-bottom'; // Bottom position
} else {
// For additional elements, distribute them evenly
return `bantuan-${index % 2 === 0 ? 'top' : 'bottom'}-${Math.floor(index / 2) + 1}`;
}
}
trackClick(element, cta) {
// Implement click tracking logic here
console.log('Floating element clicked:', {
elementId: element.Id,
elementName: cta.name,
ctaId: cta.ctaId,
targetUrl: cta.targetUrl,
});
}
// Public method to manually refresh
async refresh() {
try {
await this.fetchFloatingElements();
this.renderFloatingElements();
} catch (error) {
console.error('Failed to refresh floating elements:', error);
throw error;
}
}
// Public method to get current floating elements
getFloatingElements() {
return [...this.floatingElements];
}
}
// Initialize the component when DOM is ready
document.addEventListener('DOMContentLoaded', () => {
// Get product and table parameters from data attributes
const container = document.getElementById('bantuan-divs-container');
const product = container.getAttribute('data-product') || 'ruangguru';
const table = container.getAttribute('data-table') || 'rg_floating_icon';
new FloatingDivsComponent(product, table);
});
// Export for use in other modules if needed
if (typeof module !== 'undefined' && module.exports) {
module.exports = FloatingDivsComponent;
}
</script> <div id="wp-install-nudge" style="
position: fixed;
bottom: 0rem;
right: 0;
z-index: 999;
cursor: pointer;
"></div>
<script>
(function (w, d, s, o, f, js, fjs) {
w['Sales-Nudge-Widget'] = o; w[o] = w[o] || function () { (w[o].q = w[o].q || []).push(arguments) };
js = d.createElement(s), fjs = d.getElementsByTagName(s)[0];
js.id = o; js.src = f; js.async = 1; fjs.parentNode.insertBefore(js, fjs);
}(window, document, 'script', 'wpInstallNudge', 'https://cdn-web-2.ruangguru.com/sales-nudge/6198a95130/dist/widget.js'));
wpInstallNudge('init', { targetElementId: 'wp-install-nudge', clientTarget: 'ruangguru_blog' });
</script>
<script>
// note: disabled due to auto reload page
// var lastScrollTop = 0;
// $(window).scroll(function(event){
// var st = $(this).scrollTop();
// if (st > lastScrollTop){
// $('#wp-install-nudge').slideDown( "slow", function() {});
// } else {
// $('#wp-install-nudge').slideUp( "slow", function() {});
// }
// lastScrollTop = st;
// });
// $(document).ready(function () {
// $("#wp-install-nudge").hide();
// });
$('#wp-install-nudge').click(function(){
Analytics.track("promotion__install_nudge__click",
{
user_email: "ruangguru@blog.com",
cta_link: widgetData.dataConfig.target_link
}).catch((err) => {
console.log("tracker error", err);
});
});
</script>
</footer>
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async function getCtaDetails(ctaIds) {
const endpoint =
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const ctaMap = new Map();
for (const ctaData of ctaList) {
ctaMap.set(ctaData.ctaId, ctaData);
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return ctaMap;
} catch (error) {
console.error("Error fetching CTA details:", error);
return new Map();
}
}
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const regex = /{{cta\((‘(.*?)’|“(.*?)”)\)}}/g;
const bodyHTML = document.body.innerHTML;
const matches = Array.from(bodyHTML.matchAll(regex));
const ctaIds = matches.map((match) => {
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return ctaId;
});
const ctaMap = await getCtaDetails(ctaIds);
const replacedHTML = bodyHTML.replace(regex, (match, _inner, ctaId1, ctaId2) => {
const ctaId = ctaId1 ?? ctaId2;
const ctaData = ctaMap.get(ctaId);
if (ctaData) {
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} else {
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}
});
document.body.innerHTML = replacedHTML;
}
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