{"id":3526,"date":"2022-02-18T02:00:00","date_gmt":"2022-02-17T19:00:00","guid":{"rendered":"https:\/\/example.com\/?p=3526"},"modified":"2024-05-19T15:32:28","modified_gmt":"2024-05-19T08:32:28","slug":"latihan-soal-utbk-kimia","status":"publish","type":"post","link":"https:\/\/staging-blog.sirogu.com\/blog\/latihan-soal-utbk-kimia","title":{"rendered":"Latihan Soal &#038; Pembahasan UTBK 2022: Kimia"},"content":{"rendered":"<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>\n<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>\n<p style=\"text-align: center; font-size: 16px;\">&#8212;<\/p>\n<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>\n<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>\n<p style=\"text-align: justify;\"><strong>Subtopik <\/strong> <strong>: Konsep Kilat Sifat Keperiodikan Unsur<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong style=\"background-color: transparent; font-size: 1rem;\">Perhatikan tabel SPU berikut!<\/strong><\/p>\n<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>\n<p style=\"text-align: justify;\">1. Pernyataan yang tepat untuk sifat keperiodikan kelima unsur tersebut adalah \u2026<\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li>Jari-jari atom unsur A lebih kecil daripada unsur C.<\/li>\n<li>Energi ionisasi unsur B lebih besar daripada unsur C.<\/li>\n<li>Unsur A memiliki keelektronegatifan terbesar.<\/li>\n<li>Afinitas elektron unsur A lebih besar dari unsur C.<\/li>\n<li>Unsur B memiliki jari-jari atom paling besar.<\/li>\n<\/ol>\n<p style=\"font-weight: bold; text-align: justify;\">Kunci Jawaban : B<\/p>\n<p style=\"font-weight: bold; text-align: justify;\">Pembahasan :<\/p>\n<p style=\"text-align: justify;\">Sifat-sifat keperiodikan unsur, yaitu sebagai berikut.<\/p>\n<ul style=\"text-align: justify;\">\n<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 \/>\n<\/strong><\/li>\n<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 \/>\n<\/strong><\/li>\n<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 \/>\n<\/strong><\/li>\n<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. \u00a0Afinitas elektron unsur A lebih kecil dari unsur C. <strong>(Opsi D salah)<\/strong><\/li>\n<\/ul>\n<p style=\"text-align: justify;\"><strong><u>Jawaban yang tepat adalah B.<\/u><\/strong><\/p>\n<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>\n<p>&nbsp;<\/p>\n<h2 style=\"font-size: 16px;\"><strong>Topik<\/strong> <strong>: Ikatan Kimia, Bentuk Molekul, dan Gaya Antarmolekul<\/strong><\/h2>\n<p><strong>Subtopik <\/strong> <strong>: Konsep Kilat Ikatan Kimia dan <a href=\"\/blog\/bentuk-molekul-dan-definisinya\" rel=\"noopener\">Bentuk Molekul<\/a><\/strong><\/p>\n<p><span style=\"background-color: transparent; font-size: 1rem;\">Diketahui data unsur sebagai berikut.<\/span><\/p>\n<p style=\"text-align: center;\">Unsur X : golongan IIA dan periode 3<\/p>\n<p style=\"text-align: center;\">Unsur Y : golongan VIA dan periode 2<\/p>\n<p>2. Jika X dan Y berikatan, ikatan yang mungkin terjadi dan senyawa yang terbentuk secara berurutan adalah &#8230;.<\/p>\n<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>\n<p>&nbsp;<\/p>\n<p style=\"font-weight: bold;\">Kunci Jawaban : A<\/p>\n<p style=\"font-weight: bold;\">Pembahasan :<\/p>\n<p>Unsur X : golongan IIA \u00a0elektron 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>\n<p>Unsur Y : golongan VIA \u00a0elektron 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>\n<p>Senyawa yang terbentuk : \u00a0<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>\n<p>Ikatan yang terbentuk merupakan ikatan ionik karena terjadi serah terima elektron.<\/p>\n<p><strong><u>Jadi, jawaban yang tepat adalah A.<\/u><\/strong><\/p>\n<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>\n<p>&nbsp;<\/p>\n<h2 style=\"text-align: justify; font-size: 16px;\"><strong>Topik<\/strong> <strong>: Kimia Karbon<\/strong><\/h2>\n<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>\n<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 &#8230;.<\/span><\/p>\n<p>&nbsp;<\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<li>1-propanol<\/li>\n<li>Propanal<\/li>\n<li>2-propanon<\/li>\n<li>asam propanoat<\/li>\n<li>metil asetat<\/li>\n<\/ol>\n<p style=\"font-weight: bold; text-align: justify;\">Jawaban: D<\/p>\n<p style=\"font-weight: bold; text-align: justify;\">Pembahasan:<\/p>\n<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>\n<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>\n<p style=\"text-align: justify;\">Jenis senyawa pada opsi jawaban adalah sebagai berikut.<\/p>\n<ol style=\"list-style-type: upper-alpha; text-align: justify;\">\n<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>\n<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>\n<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>\n<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>\n<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>\n<\/ol>\n<p style=\"text-align: justify;\">Dengan demikian, kemungkinan larutan X yang digunakan oleh laboran tersebut adalah <strong>asam propanoat.<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong><u>Jadi, jawaban yang benar adalah D.<\/u><\/strong><\/p>\n<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>\n<p>&nbsp;<\/p>\n<h2 style=\"font-size: 16px;\"><strong>Topik<\/strong> <strong>: <a href=\"\/blog\/latihan-soal-sbmptn-kimia\" rel=\"noopener\">Stoikiometri<\/a><\/strong><\/h2>\n<p><strong>Subtopik <\/strong> <strong>: Konsep Kilat Stoikiometri Reaksi dan Kadar<\/strong><\/p>\n<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>\n<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>\n<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 &#8230;.<\/p>\n<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>\n<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>\n<p style=\"font-weight: bold;\">Kunci Jawaban : E<\/p>\n<p style=\"font-weight: bold;\">Pembahasan :<\/p>\n<p>Pertama, setarakan persamaan reaksinya terlebih dahulu.<\/p>\n<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>\n<p>Lalu, hitung mol urea menggunakan persamaan berikut.<\/p>\n<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>\n<p style=\"text-align: justify;\">Kemudian, hitung mol gas karbon dioksida dan gas nitrogen menggunakan perbandingan koefisien sebagai berikut.<\/p>\n<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>\n<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>\n<p>\u00a0<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>\n<p><strong>Ingat!<\/strong> Gas\u00a0<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\" \/>\u00a0adalah 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>\n<p><strong><u>Jadi, jawaban yang tepat adalah E.<\/u><\/strong><\/p>\n<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>\n<p>&nbsp;<\/p>\n<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>\n<p><strong>Subtopik<\/strong> <strong>: Konsep Kilat Sifat Koligatif Larutan<\/strong><\/p>\n<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 &#8230;.\u00a0<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>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>8,8 g<\/li>\n<li>5,5\u00a0g<\/li>\n<li>4,4\u00a0g<\/li>\n<li>4,0 g<\/li>\n<li>3,0\u00a0g<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p style=\"font-weight: bold;\">Jawaban: A<\/p>\n<p style=\"font-weight: bold;\">Pembahasan:<\/p>\n<p>Isotonik artinya memiliki tekanan osmotik yang sama.<\/p>\n<p><strong>Diketahui:<\/strong><\/p>\n<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>\n<p><strong>Ditanyakan:<\/strong><\/p>\n<p>Massa NaCl yang dilarutkan<\/p>\n<p><strong>Pembahasan<\/strong>:<\/p>\n<p>NaCl merupakan garam yang bersifat elektrolit kuat, artinya dapat terurai menjadi ion-ionnya.<\/p>\n<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>\n<p>Jumlah ion (n) = 2<\/p>\n<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>\n<p>Massa NaCl yang diperlukan untuk membuat larutan infus tersebut dapat ditentukan melalui persamaan berikut.<\/p>\n<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>\n<p>Berdasarkan persamaan di atas, massa dapat ditentukan sebagai berikut.<\/p>\n<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>\n<p><strong><u>Jadi, jawaban yang tepat adalah A.<\/u><\/strong><\/p>\n<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>\n<p>&nbsp;<\/p>\n<p style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Ikatan Kimia dan <a href=\"\/blog\/bentuk-molekul-dan-definisinya\" rel=\"noopener\">Bentuk Molekul<\/a><\/strong><\/p>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Bentuk dan Geometri Molekul<\/strong><\/p>\n<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\u00a0<\/sub>(NA Xe\u00a0= 54, F\u00a0= 9) adalah\u2026<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>segi empat planar, sp<sup>3<\/sup>d<sup>2<\/sup><\/li>\n<li>tetrahedron, sp<sup>3<\/sup>d<\/li>\n<li>segi empat planar, sp<sup>2<\/sup>d<\/li>\n<li>tetrahedron, sp<sup>3<\/sup><\/li>\n<li>bipiramida trigonal, sp<sup>3<\/sup><\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Kunci<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>A<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<p>Untuk membuat hibridisasi yang dilihat adalah atom pusat. Pada senyawa XeF<sub>4<\/sub>\u00a0yang menjadi atom pusat adalah Xe.\u00a0Konfigurasi elektron\u00a0<sub>54<\/sub>Xe = [Kr] 5s<sup>2<\/sup>4d<sup>10<\/sup>\u00a05p<sup>6<\/sup><\/p>\n<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&amp;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&amp;name=1-405577069126.png 255w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/1-405577069126.png?width=509&amp;name=1-405577069126.png 509w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/1-405577069126.png?width=764&amp;name=1-405577069126.png 764w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/1-405577069126.png?width=1018&amp;name=1-405577069126.png 1018w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/1-405577069126.png?width=1273&amp;name=1-405577069126.png 1273w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/1-405577069126.png?width=1527&amp;name=1-405577069126.png 1527w\" alt=\"prediksi soal Unsur utbk sbmptn kimia\" width=\"509\" \/><\/p>\n<p>&nbsp;<\/p>\n<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>\n<p>&nbsp;<\/p>\n<h2 style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong><a href=\"\/blog\/konsep-reaksi-reduksi-oksidasi-kimia-kelas-10\" rel=\"noopener\">Redoks<\/a> dan Elektrokimia<\/strong><\/h2>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/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>\n<p>7. Suatu sel elektrokimia dengan elektroda Zn dan Ag pada keadaan standar menghasilkan arus 0,75 A selama 321 menit. Diketahui nilai:<\/p>\n<p>E<sup>0<\/sup>\u00a0Zn<sup>2+<\/sup>\u00a0|\u00a0Zn = -0,76 V<\/p>\n<p>E<sup>0<\/sup>\u00a0Ag<sup>+<\/sup>\u00a0|\u00a0Ag = +0,8 V<\/p>\n<p>maka pengurangan massa dianoda adalah \u2026\u00a0gram.\u00a0(F = 96.500 C, Ar Zn = 65, Ag = 108)<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>3,250<\/li>\n<li>4,865<\/li>\n<li>6,321<\/li>\n<li>10,850<\/li>\n<li>13,106<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Kunci<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>B<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<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>\n<p>Katoda: Ag<sup>+<\/sup>\u00a0\u00a0+ \u00a0e<sup>&#8211;<\/sup>\u00a0\u2192 \u00a0Ag<\/p>\n<p>Anoda: Zn \u2192 \u00a0Zn<sup>2+<\/sup>\u00a0\u00a0+ \u00a02e<sup>&#8211;<\/sup><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/2-231.png?width=383&amp;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&amp;name=2-231.png 192w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/2-231.png?width=383&amp;name=2-231.png 383w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/2-231.png?width=575&amp;name=2-231.png 575w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/2-231.png?width=766&amp;name=2-231.png 766w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/2-231.png?width=958&amp;name=2-231.png 958w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/2-231.png?width=1149&amp;name=2-231.png 1149w\" alt=\"2-231\" width=\"383\" \/><\/p>\n<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>\n<p>&nbsp;<\/p>\n<h2 style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>: <a href=\"\/blog\/pengertian-dan-faktor-yang-memengaruhi-laju-reaksi\" rel=\"noopener\">Laju Reaksi<\/a><\/strong><\/h2>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Persamaan Laju Reaksi dan Konstanta Laju Reaksi Multi Experiment<\/strong><\/p>\n<p>8. Percobaan penentuan laju reaksi:<\/p>\n<p>2H<sub>2<\/sub>(<em><i>g<\/i><\/em>) \u00a0+ \u00a02NO(<em><i>g<\/i><\/em>)\u00a0<img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=27&amp;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&amp;name=3-86.png 14w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=27&amp;name=3-86.png 27w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=41&amp;name=3-86.png 41w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=54&amp;name=3-86.png 54w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=68&amp;name=3-86.png 68w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=81&amp;name=3-86.png 81w\" alt=\"3-86\" width=\"27\" \/>\u00a02H<sub>2<\/sub>O(<em><i>g<\/i><\/em>) \u00a0+ \u00a0N<sub>2<\/sub>(<em><i>g<\/i><\/em>)<\/p>\n<p>memberikan data sebagai berikut:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/4-54.png?width=421&amp;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&amp;name=4-54.png 211w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/4-54.png?width=421&amp;name=4-54.png 421w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/4-54.png?width=632&amp;name=4-54.png 632w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/4-54.png?width=842&amp;name=4-54.png 842w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/4-54.png?width=1053&amp;name=4-54.png 1053w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/4-54.png?width=1263&amp;name=4-54.png 1263w\" alt=\"prediksi soal utbk sbmptn kimia\" width=\"421\" \/><\/p>\n<p>Harga tetapan laju reaksi dari reaksi tersebut adalah\u2026<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>4 \u00d7 10<sup>8<\/sup><\/li>\n<li>8 \u00d7 10<sup>5<\/sup><\/li>\n<li>8 \u00d7 10<sup>6<\/sup><\/li>\n<li>16 \u00d7 10<sup>5<\/sup><\/li>\n<li>16 \u00d7 10<sup>4<\/sup><\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Kunci<\/strong>\u00a0<strong>: C<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<p>Menentukan orde [NO] menggunakan persamaan 2 dan 3<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/5-74.png?width=217&amp;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&amp;name=5-74.png 109w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/5-74.png?width=217&amp;name=5-74.png 217w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/5-74.png?width=326&amp;name=5-74.png 326w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/5-74.png?width=434&amp;name=5-74.png 434w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/5-74.png?width=543&amp;name=5-74.png 543w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/5-74.png?width=651&amp;name=5-74.png 651w\" alt=\"5-74\" width=\"217\" \/><\/p>\n<p>Menentukan orde [H<sub>2<\/sub>] menggunakan persamaan 1\u00a0dan 2<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/6-78.png?width=212&amp;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&amp;name=6-78.png 106w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/6-78.png?width=212&amp;name=6-78.png 212w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/6-78.png?width=318&amp;name=6-78.png 318w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/6-78.png?width=424&amp;name=6-78.png 424w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/6-78.png?width=530&amp;name=6-78.png 530w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/6-78.png?width=636&amp;name=6-78.png 636w\" alt=\"6-78\" width=\"212\" \/><\/p>\n<p>Maka persamaan laju reaksinya\u00a0<img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/7-51.png?width=125&amp;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&amp;name=7-51.png 63w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/7-51.png?width=125&amp;name=7-51.png 125w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/7-51.png?width=188&amp;name=7-51.png 188w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/7-51.png?width=250&amp;name=7-51.png 250w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/7-51.png?width=313&amp;name=7-51.png 313w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/7-51.png?width=375&amp;name=7-51.png 375w\" alt=\"7-51\" width=\"125\" \/><\/p>\n<p>Mencari nilai tetapan laju reaksi (k) menggunakan percobaan 1.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/8-53.png?width=413&amp;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&amp;name=8-53.png 207w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/8-53.png?width=413&amp;name=8-53.png 413w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/8-53.png?width=620&amp;name=8-53.png 620w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/8-53.png?width=826&amp;name=8-53.png 826w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/8-53.png?width=1033&amp;name=8-53.png 1033w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/8-53.png?width=1239&amp;name=8-53.png 1239w\" alt=\"8-53\" width=\"413\" \/><\/p>\n<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>\n<p>&nbsp;<\/p>\n<p style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Redoks dan Elektrokimia<\/strong><\/p>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Sel Volta<\/strong><\/p>\n<p>9. Diketahui potensial standar beberapa sel volta sebagai berikut:<\/p>\n<p>P | P<sup>2+<\/sup>\u00a0|| Q<sup>2+<\/sup>\u00a0| Q\u00a0 \u00a0 \u00a0E<sup>0<\/sup>\u00a0= 2,46 V<\/p>\n<p>R | R<sup>2+<\/sup>\u00a0|| S<sup>2+<\/sup>\u00a0| S\u00a0 \u00a0 \u00a0E<sup>0<\/sup>\u00a0= 1,1 V<\/p>\n<p>R | R<sup>2+<\/sup>\u00a0|| Q<sup>2+<\/sup>\u00a0| Q\u00a0 \u00a0 E<sup>0<\/sup>\u00a0= 1,56 V<\/p>\n<p>Potensial standar sel P | P<sup>2+<\/sup>\u00a0|| S<sup>2+<\/sup>\u00a0| S adalah\u2026<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>1,1 V<\/li>\n<li>2,00 V<\/li>\n<li>2,66 V<\/li>\n<li>3,56 V<\/li>\n<li>4,02 V<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Kunci<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>B<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<p>P | P<sup>2+<\/sup>\u00a0|| Q<sup>2+<\/sup>\u00a0| Q\u00a0 \u00a0 E<sup>o<\/sup>\u00a0= 2,46 V<\/p>\n<p>R | R<sup>2+<\/sup>\u00a0|| S<sup>2+<\/sup>\u00a0| S\u00a0 \u00a0 E<sup>o<\/sup>\u00a0= 1,1 V<\/p>\n<p>R | R<sup>2+<\/sup>\u00a0|| Q<sup>2+<\/sup>\u00a0| Q\u00a0 \u00a0 E<sup>o<\/sup>\u00a0= 1,56 V (perlu dibalik untuk mengeliminasi R)<\/p>\n<p>Reaksi yang dibalik memiliki potensial sel dengan nilai berlawanan, sehingga:<\/p>\n<p>P | P<sup>2+<\/sup>\u00a0||\u00a0Q<sup>2+<\/sup>\u00a0| Q\u00a0 \u00a0 E<sup>o<\/sup>\u00a0= 2,46 V<\/p>\n<p>R | R<sup>2+<\/sup>|| S<sup>2+<\/sup>\u00a0| S\u00a0 \u00a0 \u00a0E<sup>o<\/sup>\u00a0= 1,1 V<\/p>\n<p><u>Q | Q<\/u><u><sup>2+<\/sup><\/u><u>\u00a0||\u00a0<\/u><u>R<\/u><u><sup>2+<\/sup><\/u><u>\u00a0| R<\/u>\u00a0 \u00a0<u>E<\/u><u><sup>o<\/sup><\/u><u>\u00a0= -1,56 V \u00a0\u00a0<\/u>+<\/p>\n<p>P | P<sup>2+<\/sup>\u00a0|| S<sup>2+<\/sup>\u00a0| S\u00a0 \u00a0 E<sup>o<\/sup>\u00a0= 2,00 V<\/p>\n<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>\n<p>&nbsp;<\/p>\n<h2 style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>:\u00a0<\/strong><a href=\"\/blog\/kesetimbangan-kimia-jenis-jenis-reaksi-dan-tetapan\" rel=\"noopener\"><strong>Kesetimbangan Kimia<\/strong><\/a><\/h2>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Konstanta Kesetimbangan Tekanan (Kp)<\/strong><\/p>\n<p>10. Dalam wadah 2 liter, gas H<sub>2<\/sub>dan gas I<sub>2<\/sub>\u00a0bereaksi membentuk gas HI menurut persamaan reaksi:<\/p>\n<p>H<sub>2<\/sub>(<em><i>g<\/i><\/em>) \u00a0+ \u00a0I<sub>2<\/sub>(<em><i>g<\/i><\/em>)\u00a0<img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=27&amp;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&amp;name=3-86.png 14w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=27&amp;name=3-86.png 27w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=41&amp;name=3-86.png 41w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=54&amp;name=3-86.png 54w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=68&amp;name=3-86.png 68w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=81&amp;name=3-86.png 81w\" alt=\"3-86\" width=\"27\" \/>\u00a02HI(<em><i>g<\/i><\/em>)<\/p>\n<p>Jika pada kesetimbangan reaksi terdapat 0,1 mol H<sub>2<\/sub>, 0,2 mol I<sub>2<\/sub>\u00a0dan 0,5 mol HI, dengan tekanan total sebesar 5\u00a0atm, maka nilai K<sub>p<\/sub>\u00a0reaksi tersebut adalah\u2026<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>7,50<\/li>\n<li>12,5<\/li>\n<li>25,2<\/li>\n<li>32,5<\/li>\n<li>50,0<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Kunci<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>B<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<p>Diketahui: volume = 2 liter; mol total = 0,1 mol + 0,2 mol + 0,5 mol = 0,8 mol<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/9-48.png?width=408&amp;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&amp;name=9-48.png 204w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/9-48.png?width=408&amp;name=9-48.png 408w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/9-48.png?width=612&amp;name=9-48.png 612w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/9-48.png?width=816&amp;name=9-48.png 816w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/9-48.png?width=1020&amp;name=9-48.png 1020w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/9-48.png?width=1224&amp;name=9-48.png 1224w\" alt=\"9-48\" width=\"408\" \/><\/p>\n<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>\n<p>&nbsp;<\/p>\n<h2 style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Stoikiometri<\/strong><\/h2>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Persamaan Kimia; Perhitungan Kimia dalam Suatu Persamaan Reaksi<\/strong><\/p>\n<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=\u00a012 H\u00a0=\u00a01 O\u00a0=\u00a016) adalah\u2026<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>16 gram<\/li>\n<li>30 gram<\/li>\n<li>32 gram<\/li>\n<li>44 gram<\/li>\n<li>60 gram<\/li>\n<\/ol>\n<p><strong>Kunci<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>E<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<p>Metana = CH<sub>4<\/sub>\u00a0(Mr = 16) Massa etana = x gram\u00a0(sebagai komponen yang ditanya)<\/p>\n<p>Etana = C<sub>2<\/sub>H<sub>6<\/sub>\u00a0(Mr = 30) Massa metana = (76 \u2013 x) gram<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/10-39.png?width=257&amp;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&amp;name=10-39.png 129w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/10-39.png?width=257&amp;name=10-39.png 257w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/10-39.png?width=386&amp;name=10-39.png 386w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/10-39.png?width=514&amp;name=10-39.png 514w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/10-39.png?width=643&amp;name=10-39.png 643w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/10-39.png?width=771&amp;name=10-39.png 771w\" alt=\"10-39\" width=\"257\" \/><\/p>\n<p>Reaksi pembakaran etana:<\/p>\n<p>C<sub>2<\/sub>H<sub>6<\/sub>\u00a0\u00a0\u00a0\u00a0+\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0O<sub>2<\/sub>\u00a0\u00a0\u00a0\u2192 \u00a02CO<sub>2<\/sub>\u00a0 \u00a0\u00a0+ \u00a0\u00a0\u00a03H<sub>2<\/sub>O<\/p>\n<p><em><i>\u00a0<img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/11-145.png?width=319&amp;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&amp;name=11-145.png 160w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/11-145.png?width=319&amp;name=11-145.png 319w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/11-145.png?width=479&amp;name=11-145.png 479w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/11-145.png?width=638&amp;name=11-145.png 638w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/11-145.png?width=798&amp;name=11-145.png 798w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/11-145.png?width=957&amp;name=11-145.png 957w\" alt=\"11-145\" width=\"319\" \/><\/i><\/em><\/p>\n<p>Reaksi pembakaran metana:<\/p>\n<p>CH<sub>4<\/sub>\u00a0\u00a0+ \u00a0\u00a0\u00a0 2O<sub>2<\/sub>\u00a0\u2192 \u00a0\u00a0\u00a0CO<sub>2<\/sub>\u00a0 + 2H<sub>2<\/sub>O<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/12-33.png?width=401&amp;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&amp;name=12-33.png 201w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/12-33.png?width=401&amp;name=12-33.png 401w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/12-33.png?width=602&amp;name=12-33.png 602w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/12-33.png?width=802&amp;name=12-33.png 802w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/12-33.png?width=1003&amp;name=12-33.png 1003w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/12-33.png?width=1203&amp;name=12-33.png 1203w\" alt=\"12-33\" width=\"401\" \/><\/p>\n<p>Mol CO<sub>2<\/sub>\u00a0total = 5 mol<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/13-30.png?width=190&amp;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&amp;name=13-30.png 95w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/13-30.png?width=190&amp;name=13-30.png 190w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/13-30.png?width=285&amp;name=13-30.png 285w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/13-30.png?width=380&amp;name=13-30.png 380w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/13-30.png?width=475&amp;name=13-30.png 475w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/13-30.png?width=570&amp;name=13-30.png 570w\" alt=\"13-30\" width=\"190\" \/><\/p>\n<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>\n<p>&nbsp;<\/p>\n<p style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Kesetimbangan Kimia<\/strong><\/p>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Pergeseran Kesetimbangan dan Faktor yang Mempengaruhi\u2026<\/strong><\/p>\n<p>12. Di bawah ini adalah produksi NH<sub>3<\/sub>dalam reaksi:<\/p>\n<p>N<sub>2<\/sub>(<em><i>g<\/i><\/em>) \u00a0+ \u00a03H<sub>2<\/sub>(<em><i>g<\/i><\/em>)\u00a0<img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=27&amp;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&amp;name=3-86.png 14w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=27&amp;name=3-86.png 27w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=41&amp;name=3-86.png 41w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=54&amp;name=3-86.png 54w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=68&amp;name=3-86.png 68w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=81&amp;name=3-86.png 81w\" alt=\"3-86\" width=\"27\" \/>\u00a02NH<sub>3<\/sub>(<em><i>g<\/i><\/em>)\u00a0 \u00a0 \u0394H = -92,2 kJ<\/p>\n<p>Jika diketahui pada suhu 25\u00a0\u2103 nilai K<sub>c<\/sub>\u00a0= 3,5 \u00d7 10<sup>8<\/sup>, maka pernyataan berikut ini yang benar adalah\u2026<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>jika suhu turun, produksi meningkat<\/li>\n<li>ditambah katalis, produksi menurun<\/li>\n<li>jika volume turun, produksi menurun<\/li>\n<li>proses produksinya disebut dengan proses Kontak<\/li>\n<li>pengurangan konsentrasi N<sub>2<\/sub>, produksi meningkat<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Kunci<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>A<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<p>Produksi NH<sub>3<\/sub>\u00a0disebut dengan proses Haber-Bosch dengan reaksi<\/p>\n<p>N<sub>2<\/sub>(<em><i>g<\/i><\/em>) \u00a0+ \u00a03H<sub>2<\/sub>(<em><i>g<\/i><\/em>)\u00a0\u00a0\u00a0<img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=27&amp;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&amp;name=3-86.png 14w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=27&amp;name=3-86.png 27w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=41&amp;name=3-86.png 41w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=54&amp;name=3-86.png 54w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=68&amp;name=3-86.png 68w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/3-86.png?width=81&amp;name=3-86.png 81w\" alt=\"3-86\" width=\"27\" \/>\u00a0\u00a0\u00a02NH<sub>3<\/sub>(<em><i>g<\/i><\/em>)\u00a0 \u00a0 \u00a0\u0394H = -92,2 kJ (reaksi eksoterm)<\/p>\n<p>Maka, pernyataan yang tepat yaitu:<\/p>\n<ul>\n<li>jika suhu turun, akan bergeser ke eksoterm maka produksi meningkat.<\/li>\n<li>ditambah katalis, reaksi akan semakin cepat, tidak berpengaruh pada hasil produksi.<\/li>\n<li>jika volume turun, akan bergeser ke koefisien kecil maka produksi akan meningkat.<\/li>\n<li>pengurangan konsentrasi N<sub>2<\/sub>, akan bergeser ke arah reaktan,jadi\u00a0produksi\u00a0menurun.<\/li>\n<\/ul>\n<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>\n<p>&nbsp;<\/p>\n<p style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Sifat Koligatif Larutan<\/strong><\/p>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Kenaikan\u00a0<\/strong><strong>T<\/strong><strong>itik Didih<\/strong><\/p>\n<p>13. Struktur molekul PVC dapat dinyatakan (\u2013CH2 \u2013CHCl\u2013)<sub>n<\/sub>. Larutan yang mengandung 12,5 gram bahan tersebut dalam 500 gram benzena mendidih pada suhu 80,11\u2103. Jika harga K<sub>b<\/sub>\u00a02,5 \u2103\/molal dan titik didih benzena 80,10 \u2103, maka harga n (Ar H\u00a0= 1, C\u00a0= 12, Cl\u00a0= 35,5) adalah&#8230;<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>50<\/li>\n<li>100<\/li>\n<li>500<\/li>\n<li>1000<\/li>\n<li>1500<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Kunci<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>B<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<p>Diketahui:<\/p>\n<p>massa PVC (terlarut) = 12,5 gram; T<sub>b<\/sub>\u00a0= 80,11\u00a0\u2103; K<sub>b<\/sub>\u00a0= 2,5 \u2103\/molal<\/p>\n<p>massa benzena (pelarut) = 500 gram; \u0394T<sub>b<\/sub>\u00a0= 80,11\u00a0\u2103\u00a0&#8211; 80,10 \u2103 = 0,01 \u2103<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/14-5577069093.png?width=505&amp;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&amp;name=14-5577069093.png 253w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/14-5577069093.png?width=505&amp;name=14-5577069093.png 505w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/14-5577069093.png?width=758&amp;name=14-5577069093.png 758w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/14-5577069093.png?width=1010&amp;name=14-5577069093.png 1010w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/14-5577069093.png?width=1263&amp;name=14-5577069093.png 1263w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/14-5577069093.png?width=1515&amp;name=14-5577069093.png 1515w\" alt=\"14-5577069093\" width=\"505\" \/><\/p>\n<p>(Rumus Empiris)n = Rumus Molekul<\/p>\n<p>(Mr Empiris)n = Mr Molekul<\/p>\n<p>(62,5)n = 6250\u00a0 \u00a0 \u00a0<img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/15-25.png?width=87&amp;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&amp;name=15-25.png 44w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/15-25.png?width=87&amp;name=15-25.png 87w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/15-25.png?width=131&amp;name=15-25.png 131w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/15-25.png?width=174&amp;name=15-25.png 174w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/15-25.png?width=218&amp;name=15-25.png 218w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/15-25.png?width=261&amp;name=15-25.png 261w\" alt=\"15-25\" width=\"87\" \/><\/p>\n<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>\n<p>&nbsp;<\/p>\n<h2 style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>:\u00a0<\/strong><a href=\"\/blog\/persamaan-termokimia-dan-jenis-perubahan-entalpi-standar\" rel=\"noopener\"><strong>Termokimia<\/strong><\/a><\/h2>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Penentuan Perubahan Entalpi<\/strong><\/p>\n<p>14. Diketahui energi rata-rata sebagai berikut:<\/p>\n<p>C \u2013 H = 414 kJ\/mol\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 H \u2013 Cl = 432 kJ\/mol<\/p>\n<p>Cl \u2013 Cl = 244 kJ\/mol\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C \u2013 Cl = 326 kJ\/mol<\/p>\n<p>Perubahan entalpi untuk 3,2 gram metana pada reaksi:<\/p>\n<p>CH<sub>4<\/sub>(<em><i>g<\/i><\/em>) \u00a0+ \u00a0Cl<sub>2<\/sub>(<em><i>g<\/i><\/em>) \u00a0\u00a0\u2192 \u00a0\u00a0CH<sub>3<\/sub>Cl(<em><i>g<\/i><\/em>) \u00a0+ \u00a0HCl(<em><i>g<\/i><\/em>) adalah \u2026 kJ.\u00a0(Ar C\u00a0= 12, H\u00a0= 1)<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>-20<\/li>\n<li>+20<\/li>\n<li>-80<\/li>\n<li>+100<\/li>\n<li>-100<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Kunci<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>A<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<p>Persamaan reaksi:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/16-21.png?width=412&amp;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&amp;name=16-21.png 206w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/16-21.png?width=412&amp;name=16-21.png 412w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/16-21.png?width=618&amp;name=16-21.png 618w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/16-21.png?width=824&amp;name=16-21.png 824w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/16-21.png?width=1030&amp;name=16-21.png 1030w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/16-21.png?width=1236&amp;name=16-21.png 1236w\" alt=\"16-21\" width=\"412\" \/><\/p>\n<p>\u0394H = \u2211H reaktan &#8211; \u2211H produk<\/p>\n<p>\u0394H = 658 \u2013 758 = -100 kJ\/mol<\/p>\n<p>Maka untuk 3,2 gram metana dilepaskan kalor sebesar\u00a0100 kJ\/mol.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/17-18.png?width=195&amp;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&amp;name=17-18.png 98w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/17-18.png?width=195&amp;name=17-18.png 195w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/17-18.png?width=293&amp;name=17-18.png 293w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/17-18.png?width=390&amp;name=17-18.png 390w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/17-18.png?width=488&amp;name=17-18.png 488w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/17-18.png?width=585&amp;name=17-18.png 585w\" alt=\"17-18\" width=\"195\" \/><\/p>\n<p>\u0394H<sub>r<\/sub>\u00a0= 0,2 mol \u00d7 -100 kJ\/mol = -20 kJ<\/p>\n<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>\n<p>&nbsp;<\/p>\n<p style=\"font-size: 16px;\"><strong>Topik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Termokimia<\/strong><\/p>\n<p><strong>Subtopik<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>Penentuan Perubahan Entalpi<\/strong><\/p>\n<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\u2103 menjadi 50\u00a0\u2103?\u00a0Diketahui:<\/p>\n<p>\u2206H<sup>o<\/sup><sub>f<\/sub>\u00a0H<sub>2<\/sub>O(g) = -240 kJ\/mol<\/p>\n<p>\u2206H<sup>o<\/sup><sub>f<\/sub>\u00a0CO<sub>2<\/sub>(g) = -394 kJ\/mol<\/p>\n<p>\u2206H<sup>o<\/sup><sub>f\u00a0<\/sub>\u00a0C<sub>2<\/sub>H<sub>5<\/sub>OH = -277 kJ\/mol<\/p>\n<p>kalor jenis air = 4,2 J\/gK<\/p>\n<p>Ar C = 12, H = 1, O = 16<\/p>\n<ol style=\"list-style-type: upper-alpha;\">\n<li>0,235 gram<\/li>\n<li>0,47 gram<\/li>\n<li>0,94 gram<\/li>\n<li>6,3 gram<\/li>\n<li>12,6 gram<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p><strong>Kunci<\/strong>\u00a0<strong>:\u00a0<\/strong><strong>B<\/strong><\/p>\n<p><strong>Pembahasan<\/strong>\u00a0<strong>:<\/strong><\/p>\n<p>Mr etanol = 46\u00a0g\/mol; \u2206T = 50\u00a0\u2103 \u2013 20\u00a0\u2103 = 30 \u2103 = 30 K<\/p>\n<p>Reaksi pembakaran etanol adalah: C<sub>2<\/sub>H<sub>5<\/sub>OH + 3O<sub>2\u00a0<\/sub>\u2192 2CO<sub>2<\/sub>\u00a0+ 3H<sub>2<\/sub>O<\/p>\n<p>\u2206H =\u00a0<img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=43&amp;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&amp;name=20-15.png 22w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=43&amp;name=20-15.png 43w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=65&amp;name=20-15.png 65w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=86&amp;name=20-15.png 86w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=108&amp;name=20-15.png 108w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=129&amp;name=20-15.png 129w\" alt=\"20-15\" width=\"43\" \/><sub>produk<\/sub>\u00a0&#8211;<img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=43&amp;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&amp;name=20-15.png 22w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=43&amp;name=20-15.png 43w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=65&amp;name=20-15.png 65w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=86&amp;name=20-15.png 86w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=108&amp;name=20-15.png 108w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/20-15.png?width=129&amp;name=20-15.png 129w\" alt=\"20-15\" width=\"43\" \/><sub>reaktan<\/sub><\/p>\n<p>= (2\u00d7(-394)\u00a0+ 3\u00d7(-240)) \u2013 (\u2013277 + 0)<\/p>\n<p>= \u20131231 kJ\/mol<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/19-16.png?width=343&amp;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&amp;name=19-16.png 172w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/19-16.png?width=343&amp;name=19-16.png 343w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/19-16.png?width=515&amp;name=19-16.png 515w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/19-16.png?width=686&amp;name=19-16.png 686w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/19-16.png?width=858&amp;name=19-16.png 858w, https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/\/19-16.png?width=1029&amp;name=19-16.png 1029w\" alt=\"19-16\" width=\"343\" \/><\/p>\n<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>\n<p>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\u00a0<span style=\"font-weight: bold;\">#PelatnasUTBK!<\/span> 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.<\/p>\n<p><a class=\"rg-cta\" style=\"text-align: center;\" href=\"https:\/\/t.me\/PelatnasUTBK2023\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/cta\/08dbc3c1-522d-4ce9-99ad-730dc9c4dace.png\" alt=\"[IDN] Blog Grup Pelatnas UTBK Ruangguru\" width=\"239\" height=\"52\" \/><\/a><\/p>\n<p style=\"text-align: justify;\">Yuk, berlatih lebih banyak soal lagi dengan ikut tryout UTBK di\u00a0<a style=\"font-weight: bold;\" href=\"https:\/\/ruangguru.onelink.me\/blPk\/3b1dd5cf\" target=\"_blank\" rel=\"noopener\">ruanguji<\/a>.\u00a0<em>Psst<\/em>, soal-soal dan sistem penilaiannya sama seperti UTBK aslinya,\u00a0<em>lho<\/em>! Yakin nggak mau cobain?<\/p>\n<p style=\"text-align: justify;\"><a class=\"rg-cta\" style=\"text-align: center;\" href=\"https:\/\/t.me\/PelatnasUTBK2023\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/cta\/08dbc3c1-522d-4ce9-99ad-730dc9c4dace.png\" alt=\"[IDN] Blog Grup Pelatnas UTBK Ruangguru\" width=\"239\" height=\"52\" \/><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Sudah siap hadapi UTBK tahun 2022 nanti? Yuk, tingkatkan persiapanmu dengan latihan soal UTBK 2022 Kimia beserta pembahasannya di bawah ini. Selamat mengerjakan! &#8212; Topik : Struktur Atom dan Sistem Periodik Unsur Subtopik : Konsep Kilat Sifat Keperiodikan Unsur Perhatikan tabel SPU berikut! 1. Pernyataan yang tepat untuk sifat keperiodikan kelima unsur tersebut adalah \u2026 [&hellip;]<\/p>\n","protected":false},"author":22,"featured_media":3526,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_knawatfibu_url":["https:\/\/cdn-web.ruangguru.com\/landing-pages\/assets\/hs\/Latihan%20soal%20utbk%20kimia.jpg"],"_edit_lock":["1716107548:1"],"_edit_last":["1"],"_aioseo_title":[null],"_aioseo_description":["Sudah siap hadapi UTBK tahun 2022? Yuk, tingkatkan persiapanmu dengan latihan soal UTBK 2022 Kimia beserta pembahasannya di bawah ini. Selamat mengerjakan!"],"_aioseo_keywords":[""],"_aioseo_og_title":[null],"_aioseo_og_description":[null],"_aioseo_og_article_section":[""],"_aioseo_og_article_tags":[""],"_aioseo_twitter_title":[null],"_aioseo_twitter_description":[null]},"categories":[563,565],"tags":[55,58],"class_list":["post-3526","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-pojok-kampus","category-seleksi-masuk","tag-pojok-kampus","tag-seleksi-masuk"],"aioseo_notices":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v25.9 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Latihan Soal &amp; Pembahasan UTBK 2022: Kimia - Belajar Gratis di Rumah Kapan Pun! | Blog Ruangguru<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/staging-blog.sirogu.com\/blog\/latihan-soal-utbk-kimia\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Latihan Soal &amp; Pembahasan UTBK 2022: Kimia - Belajar Gratis di Rumah Kapan Pun! | Blog Ruangguru\" \/>\n<meta property=\"og:description\" content=\"Sudah siap hadapi UTBK tahun 2022 nanti? 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