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Chemical Nomenclature and Equations - Practice Questions (37)

Question 1: 1. Nitrous oxide ${ } ^ { \left( \mathrm { N } _ { 2 } \mathrm { O } \right) }$ adalah obat bius yan...

1. Nitrous oxide ${ } ^ { \left( \mathrm { N } _ { 2 } \mathrm { O } \right) }$ adalah obat bius yang memberikan analgesia cepat saat dihirup dalam jumlah kecil tanpa merusak jantung, paru-paru, hati, atau ginjal. Nitrous oxide ${ } ^ { \left( \mathrm { N } _ { 2 } \mathrm { O } \right) }$ adalah obat bius. Nitrous oxide $\left( \mathrm { N } _ { 2 } \mathrm { O } \right) _ { \text {中氮元素的化合价为( )} }$ adalah obat bius.

  • A. A. - 1
  • B. B. - 2
  • C. C. + 1
  • D. D. + 2

Answer: C

Solution: Dalam nitro oksida, unsur oksigen umumnya menunjukkan valensi -2. Misalkan valensi unsur nitrogen adalah $x$, dan sesuai dengan fakta bahwa jumlah aljabar dari valensi positif dan negatif adalah nol pada senyawa, maka dapat diperoleh sebagai: $2 x + ( - 2 ) = 0$, maka valensi $x = + 1$. Oleh karena itu Pilihan: C.

Question 2: 2. Istilah-istilah kimia dasar berikut ini dinyatakan dengan benar ( ) <img class="imgSvg" id = "mi...

2. Istilah-istilah kimia dasar berikut ini dinyatakan dengan benar ( ) <img class="imgSvg" id = "mi1mzfvmlf7f2kus5mf" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16ZnZtbGY3ZjJrdXM1bWYiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDE3MCAxMjAuNzUwMDMwNjA4MTE4MzIiIHN0eWxlPSJ3aWR0aDogMTcwLjA1OTYwMDQzODM2NTQ2cHg7IGhlaWdodDogMTIwLjc1MDAzMDYwODExODMycHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnZtbGY3ZjJrdXM1bWYtMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI5Ni41NTk1NjUwOTUxMTczNiIgeTE9Ijk5Ljc1MDAzMDYwODExODMyiB4Mj0iOTYuNTU5NjAwNDM4Mzg1MjkiIHkyPSI2OC4yNTAwMzA2MDgxMzgxNiI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ 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  • A. A. Rumus elektronik dari $\mathrm { CCl } _ { 4 }$:
  • B. B. Rumus struktur HClO: $\mathrm { H } - \mathrm { Cl } - \mathrm { O }$
  • C. C. Model struktur molekul $\mathrm { CO } _ { 2 }$: $\square$ D. Struktur skematik ${ } ^ { 34 } \mathrm {~S} ^ { 2 - }$:
  • D. D. Model struktur molekul $\mathrm { CO } _ { 2 }$: $\square$ D. ${ } ^ { 34 } \mathrm {~S} ^ { 2 - }$

Answer: D

Solution: A. Atom klorin memiliki tujuh elektron di lapisan terluar, empat atom klorin dan atom karbon membentuk pasangan elektron yang sama, atom karbon tetraklorida Cl di lapisan terluar dari struktur 8-elektron, A kesalahan; B. Dalam HClO, atom O dan atom $\mathrm { Cl } , \mathrm { H }$ berbagi satu pasang elektron, dan rumus strukturnya adalah $\mathrm { H } - \mathrm { O } - \mathrm { Cl }$, salah; C. Jari-jari atom C seharusnya lebih besar daripada atom O, modelnya salah, C salah; D. ${ } ^ { 34 } \mathrm {~S}$ atom memiliki 16 elektron di luar inti, dapatkan dua elektron untuk membentuk ${ } ^ { 34 } \mathrm {~S} ^ { 2 - }$, struktur $( + 16 )$ (考攵) $\left. { } ^ { 8 } \right| _ { \mid } ^ { 8 }$, D benar; menyimpulkan jawaban untuk D.

Question 3: 3. Penguasaan istilah kimia yang baik adalah dasar untuk mempelajari kimia, ungkapan yang relevan be...

3. Penguasaan istilah kimia yang baik adalah dasar untuk mempelajari kimia, ungkapan yang relevan berikut ini adalah benar ( ) ![](/images/questions/chem-nomenclature/image-001.jpg) ![](/images/questions/chem-nomenclature/image-002.jpg) <img class="imgSvg" id = "mi1mzfvqq3sqgls7l1" src = "data:image/svg+xml;base64, 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 + 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 + PC90ZXh0Pjx0ZXh0IHg9IjczLjQ5OTk4NTg2MjY4OTY0IiB5PSI4My45OTk5OTk5OTk5OTM2NSIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgIj48L3RleHQ +PC9nPjwvc3ZnPg== "/>

  • A. A. Formula struktural etilena: $\mathrm { CH } _ { 2 } = \mathrm { CH } _ { 2 }$
  • B. B. Model pengisian ruang untuk metana:
  • C. C. Struktur skematis atom belerang:
  • D. D. Formula elektronik untuk karbon tetraklorida:

Answer: C

Solution: A. Diagram menunjukkan bentuk pendek struktural etilena, rumus strukturalnya adalah <img class="imgSvg" id = "mi1mzfx2ww1z8l82l5h" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16Zngyd3cxejhsODJsNWgiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDExNSA0Mi4wMDAwMTQxMzczMDcxNzYiIHN0eWxlPSJ3aWR0aDogMTE1LjQ5OTk5OTk5OTk5NjgycHg7IGhlaWdodDogNDIuMDAwMDE0MTM3MzA3MTc2cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM + PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16Zngyd3cxejhsODJsNWgtMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI0Mi4wMDAwMDEyNzIzNTc2NSIgeTE9IjE4LjE2NTAwMDAwMDAwMDI4NyIgeDI9IjczLjUwMDAwMTI3MjM1NDQ3IiB5Mj0iMTguMTY1MDE0MTM3MzA3NDY2Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4Mnd3MXo4bDgybDVoLTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNDEuOTk5OTk4NzI3NjQyMzUiIHkxPSIyMy44MzQ5OTk5OTk5OTk3MTMiIHgyPSI3My40OTk5OTg3Mjc2MzkxOCIgeTI9IjIzLjgzNTAxNDEzNzMwNjg5MiI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+PC9saW5lYXJHcmFkaWVudD48L2RlZnM+ PG1hc2sgaWQ9InRleHQtbWFzay1taTFtemZ4Mnd3MXo4bDgybDVoIj48cmVjdCB4PSIwIiB5PSIwIiB3aWR0aD0iMTAwJSIgaGVpZ2h0PSIxMDAlIiBmaWxsPSJ3aGl0ZSI +PC9yZWN0PjwvbWFzaz48c3R5bGU+ CiAgICAgICAgICAgICAgICAuZWxlbWVudC1taTFtemZ4Mnd3MXo4bDgybDVoIHsKICAgICAgICAgICAgICAgICAgICBmb250OiAxNHB4IEhlbHZldGljYSwgQXJpYWwsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICAgICAgICAgYWxpZ25tZW50LWJhc2VsaW5lOiAnbWlkZGxlJzsKICAgICAgICAgICAgICAgIH0KICAgICAgICAgICAgICAgIC5zdWItbWkxbXpmeDJ3dzF6OGw4Mmw1aCB7CiAgICAgICAgICAgICAgICAgICAgZm9udDogOC40cHggSGVsdmV0aWNhLCBBcmlhbCwgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgIH0KICAgICAgICAgICAgPC9zdHlsZT48ZyBtYXNrPSJ1cmwoI3RleHQtbWFzay1taTFtemZ4Mnd3MXo4bDgybDVoKSI + PGxpbmUgeDE9IjQyLjAwMDAwMTI3MjM1NzY1IiB5MT0iMTguMTY1MDAwMDAwMDAwMjg3IiB4Mj0iNzMuNTAwMDAxMjcyMzU0NDciIHkyPSIxOC4xNjUwMTQxMzczMDc0NjYiIHN0eWxlPSJzdHJva2UtbGluZWNhcDpyb3VuZDtzdHJva2UtZGFzaGFycmF5Om5vbmU7c3Ryb2tlLXdpZHRoOjEuMjYiIHN0cm9rZT0idXJsKCcjbGluZS1taTFtemZ4Mnd3MXo4bDgybDVoLTEnKSI + PC9saW5lPjxsaW5lIHgxPSI0MS45OTk5OTg3Mjc2NDIzNSIgeTE9IjIzLjgzNDk5OTk5OTk5OTcxMyIgeDI9IjczLjQ5OTk5ODcyNzYzOTE4IiB5Mj0iMjMuODM1MDE0MTM3MzA2ODkyIiBzdHlsZT0ic3Ryb2tlLWxpbmVjYXA6cm91bmQ7c3Ryb2tlLWRhc2hhcnJheTpub25lO3N0cm9rZS13aWR0aDoxLjI2IiBzdHJva2U9InVybCgnI2xpbmUtbWkxbXpmeDJ3dzF6OGw4Mmw1aC0zJykiPjwvbGluZT48L2c +PGc+ PHRleHQgeD0iNzMuNDk5OTk5OTk5OTk2ODIiIHk9IjIxLjAwMDAxNDEzNzMwNzE4IiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAiPjwvdGV4dD48dGV4dCB4PSI0MiIgeT0iMjEiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI +PC90ZXh0PjwvZz48L3N2Zz4="/> Kesalahan; B. Diagram menunjukkan model bola-dan-tongkat metana, dengan tipe pengisian ruang ![](/images/questions/chem-nomenclature/image-003.jpg) Kesalahan B ; C.Diagram tersebut menunjukkan struktur atom belerang, C benar; D.Seharusnya ada 8 elektron di sekitar setiap atom klorin dalam diagram, D salah;

Question 4: 5. Istilah-istilah kimia yang relevan berikut ini dinyatakan dengan benar.

5. Istilah-istilah kimia yang relevan berikut ini dinyatakan dengan benar.

  • A. A. Struktur skematik $\mathrm { Cl } ^ { - }$: (+17) 288
  • B. B. Massa molar ${ } ^ { 2 } \mathrm { H } _ { 2 }$: $2 \mathrm {~g} \cdot \mathrm {~mol} ^ { - 1 }$
  • C. C. Rumus kimia bahan aktif dalam pemutih: NaClO
  • D. D. Persamaan ionisasi NaOH: $\mathrm { NaOH } = \mathrm { Na } ^ { + } + \mathrm { O } ^ { 2 - } + \mathrm { H } ^ { + }$

Answer: A

Solution: A. $\mathrm { Cl } ^ { - }$ memiliki muatan nuklir 17 dan jumlah elektron di luar inti 18, dan strukturnya ditunjukkan sebagai (+17) $\begin{aligned} & 1 ) \\ & 288 \\ & 11 ) \end{aligned}$, A benar; B. Massa molar ${ } ^ { 2 } \mathrm { H } _ { 2 }$ adalah $4 \mathrm {~g} \cdot \mathrm {~mol} ^ { - 1 }$, B salah; C. Rumus kimia bahan aktif pemutih adalah $\mathrm { Ca } ( \mathrm { ClO } ) _ { 2 }$, C salah; D. Persamaan untuk ionisasi NaOH adalah $\mathrm { NaOH } = \mathrm { Na } ^ { + } + \mathrm { OH } ^ { - }$, D salah. Pilihan jawaban A.

Question 5: 6. Pernyataan berikut ini tidak benar ![](/images/questions/chem-nomenclature/image-004.jpg)

6. Pernyataan berikut ini tidak benar ![](/images/questions/chem-nomenclature/image-004.jpg)

  • A. A. Rumus eksperimental untuk benzena: $\mathrm { C } _ { 6 } \mathrm { H } _ { 6 }$
  • B. B. Struktur sederhana asetaldehida: $\mathrm { CH } _ { 3 } \mathrm { CHO }$
  • C. C. Rumus elektronik gugus amino: $\underset { \bullet } { \stackrel { H } { N } } : \mathrm { H }$
  • D. D. Model kelelawar Ethane:

Answer: A

Solution: A. Benzena memiliki rumus molekul: $\mathrm { C } _ { 6 } \mathrm { H } _ { 6 }$ dan rumus percobaannya: $\mathrm { CH } , \mathrm { A }$ salah; B. Asetaldehida mengandung gugus aldehida dan rumus strukturnya adalah: $\mathrm { CH } _ { 3 } \mathrm { CHO }$, B benar; C. Jumlah elektron pada lapisan terluar atom nitrogen adalah 5, dan jumlah elektron pada atom hidrogen adalah 1, maka rumus elektronik gugus aminonya adalah: $\underset { . } { \stackrel { H } { \mathrm { H } } } : \mathrm { H }$, C benar; D. Bentuk pendek struktural etana adalah: $\mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 }$ dan model bola-dan-tongkatnya adalah: ![](/images/questions/chem-nomenclature/image-006.jpg) D adalah benar;

Question 6: 7. Istilah kimia dapat mewakili proses kimia. Istilah-istilah kimia berikut ini ditulis dengan benar...

7. Istilah kimia dapat mewakili proses kimia. Istilah-istilah kimia berikut ini ditulis dengan benar ![](/images/questions/chem-nomenclature/image-006.jpg)

  • A. A. Rumus elektronik untuk $\mathrm { NH } _ { 3 }$: $\underset { \mathrm { H } : } { \stackrel { \mathrm { H } } { \mathrm { N } } } : \mathrm { H }$
  • B. B. Diagram skematik struktur $\mathrm { S } ^ { 2 - }$:
  • C. C. Si berada di Kelompok VI A pada Siklus Ketiga.
  • D. D. $\mathrm { NaHSO } _ { 4 }$ Persamaan ionisasi dalam air: $\mathrm { NaHSO } _ { 4 } = \mathrm { Na } ^ { + } + \mathrm { HSO } _ { 4 } ^ { - }$

Answer: B

Solution: A. Terdapat 8 elektron di sekitar atom N, dan rumus elektronik $\mathrm { NH } _ { 3 }$ seharusnya adalah: $\underset { \mathrm { H } : ( \underset { . } { \mathrm { N } } : \mathrm { H } } { \stackrel { \mathrm { H } } { \mathrm { H } } } , \mathrm { A }$ Salah; B. S adalah unsur 16, dan strukturnya ditampilkan sebagai (+16) 288, yang benar; $\mathrm { S } ^ { 2 - }$ C. Si adalah unsur 14, terletak pada golongan IV A pada siklus ketiga, C salah; D. Persamaan ionisasi $\mathrm { NaHSO } _ { 4 }$ dalam air seharusnya $\mathrm { NaHSO } _ { 4 } = \mathrm { Na } ^ { + } + \mathrm { H } ^ { + } + \mathrm { SO } _ { 4 } ^ { 2 - }$, D salah;

Question 7: 8. Ilmu kimia perlu dijelaskan dalam bahasa khusus kimia. Istilah-istilah kimia berikut ini tidak di...

8. Ilmu kimia perlu dijelaskan dalam bahasa khusus kimia. Istilah-istilah kimia berikut ini tidak dinyatakan dengan benar <img class="imgSvg" id = "mi1mzfvsp36nxcywo6p" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16ZnZzcDM2bnhjeXdvNnAiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDE5MyAxMDUuMDAwMDM2NzI5Nzg3MTciIHN0eWxlPSJ3aWR0aDogMTkzLjExOTIwMDg3NjgxNDU3cHg7IGhlaWdodDogMTA1LjAwMDAzNjcyOTc4NzE3cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnZzcDM2bnhjeXdvNnAtMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSIxMjMuODM5MzkwMDU0NjMwNTQiIHkxPSIzNi43NTAwNTUwOTQ2OTg2MSIgeDI9IjE1MS4xMTkyMDA4NzY4MTQ1NyIgeTI9IjIxLjAwMDA3MzQ1OTYwMjkzIj48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ2c3AzNm54Y3l3bzZwLTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTE4LjE2OTM3MDk2OTI2NzE1IiB5MT0iNjUuMTAwMDUxMjc3NjE5MjMiIHgyPSIxMTguMTY5Mzg3OTM0MDM1NzUiIHkyPSIzOS45MDAwNTEyNzc2MjQ5NTQiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnZzcDM2bnhjeXdvNnAtNSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSIxMjMuODM5MzY4ODQ4NjY5NzgiIHkxPSI2OC4yNTAwNTUwOTQ2OTE0NiIgeDI9IjEyMy44MzkzOTAwNTQ2MzA1NCIgeTI9IjM2Ljc1MDA1NTA5NDY5ODYxIj48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + 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  • A. A. Rumus elektronik untuk hidroksil: ${ } _ { \mathrm { g } \mathrm { g } } ^ { \mathrm { g } \mathrm { g } } : \mathrm { H }$
  • B. B. Rumus kimia propanol: $\mathrm { C } _ { 3 } \mathrm { H } _ { 8 } \mathrm { O }$
  • C. C. Formula struktural etil format: $\mathrm { HCOOCH } _ { 2 } \mathrm { CH } _ { 3 }$
  • D. D. Methylphenol:

Answer: A

Solution: A. Rumus elektronik gugus hidroksil: ${ } _ { \mathrm { g } \mathrm { g } } ^ { \mathrm { g } \mathrm { g } } \mathrm { H }$, jadi A salah; B. Rumus kimia (rumus molekul) propanol: $\mathrm { C } _ { 3 } \mathrm { H } _ { 8 } \mathrm { O }$, jadi B benar; C. Rumus struktur etil format: $\mathrm { HCOOCH } _ { 2 } \mathrm { CH } _ { 3 }$, jadi C benar; D. m-Metilfenol, gugus metil dan hidroksil pada posisi meso. <img class="imgSvg" id = "mi1mzfx55aaunrwbbgv" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16Zng1NWFhdW5yd2JiZ3YiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDE5MyAxMDUuMDAwMDM2NzI5Nzg3MTciIHN0eWxlPSJ3aWR0aDogMTkzLjExOTIwMDg3NjgxNDU3cHg7IGhlaWdodDogMTA1LjAwMDAzNjcyOTc4NzE3cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16Zng1NWFhdW5yd2JiZ3YtMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSIxMjMuODM5MzkwMDU0NjMwNTQiIHkxPSIzNi43NTAwNTUwOTQ2OTg2MSIgeDI9IjE1MS4xMTkyMDA4NzY4MTQ1NyIgeTI9IjIxLjAwMDA3MzQ1OTYwMjkzIj48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + 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Kesimpulannya, jawabannya adalah A.

Question 8: 9. Ilmu kimia perlu dijelaskan dengan bantuan bahasa khusus kimia, dan istilah-istilah berikut ini b...

9. Ilmu kimia perlu dijelaskan dengan bantuan bahasa khusus kimia, dan istilah-istilah berikut ini benar untuk kimia <img class="imgSvg" id = "mi1mzfvuu94ks3qk2pd" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16ZnZ1dTk0a3MzcWsycGQiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDE3NCAxMDQuOTk5OTk5OTk5OTc0NiIgc3R5bGU9IndpZHRoOiAxNzQuMjc5ODE0MzU2NDgwNnB4OyBoZWlnaHQ6IDEwNC45OTk5OTk5OTk5NzQ2cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI + PGRlZnM + PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnZ1dTk0a3MzcWsycGQtMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSIxMDQuOTk5OTk5OTk5OTc0NjIiIHkxPSI1Mi41MDAwMjgyNzQ2MDE2NiIgeDI9IjEyMC43NDk5NzU1MTM0MzM5NSIgeTI9Ijc5Ljc3OTg0MjYzMTEwNzY4Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + 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  • A. A. Rumus elektronik untuk $\mathrm { CO } _ { 2 }$: $\ddot { \mathrm { O } } : \ddot { \mathrm { C } } : \ddot { \mathrm { O } } :$
  • B. B. $\mathrm { CH } _ { 4 }$ Model skala molekul:
  • C. C. Rumus elektronik untuk $\mathrm { NH } _ { 4 } \mathrm { Cl }$: $\left[ \underset { \ddot { \mathrm { H } } } { \stackrel { \mathbf { H } } { = } } \mathbf { H } ^ { + } \mathbf { C l } ^ { - } \right.$
  • D. D. Atom oksigen dengan 10 neutron di dalam nukleusnya: ${ } _ { 8 } ^ { 18 } \mathrm { O }$

Answer: D

Solution: A salah, seharusnya: $\ddot { \mathrm { O } } : \mathbf { x } _ { \mathbf { x } } ^ { \mathbf { x } } \mathrm { C } \mathbf { x } : \ddot { \mathrm { O } } :$; B salah, ini seharusnya model bola dan tongkat; C Salah, seharusnya $\left[ \begin{array} { c } \mathrm { H } : \underset { \ddot { \mathrm { H } } } { \ddot { \mathrm { N } } } : \mathrm { H } \end{array} \right] ^ { + } [ : \ddot { \mathrm { C } 1 } : ] ^ { - }$, jadi jawabannya adalah D. 考点:考查常见化学用语的正误判断 Komentar: Soal ini termasuk soal yang cukup sulit, tetapi juga merupakan tipe soal yang umum dan merupakan poin penting dalam ujian masuk perguruan tinggi. Pertanyaannya bersifat dasar, cukup sulit, terutama untuk menguji keakraban siswa dengan tingkat penguasaan istilah-istilah kimia yang umum. Pertanyaan-pertanyaan harus jelas bahwa istilah-istilah kimia yang umum terutama mencakup simbol elemen, rumus kimia, valensi, rumus elektronik, diagram struktur atom, rumus struktur, kesederhanaan struktur, serta persamaan dan berbagai model, dll., yang perlu diingat oleh siswa.

Question 9: 10. Istilah-istilah kimia berikut ini dinyatakan dengan benar.

10. Istilah-istilah kimia berikut ini dinyatakan dengan benar.

  • A. A. Rumus elektronik untuk $\mathrm { CH } _ { 4 }$: $\mathrm { H } : \stackrel { \text { H } } { \underset { \rightarrow } { \ddot { C } } } : \mathrm { H }$
  • B. B. Rumus molekul propana: $\mathrm { CH } _ { 3 } \mathrm { CHCH } _ { 2 }$
  • C. C. Struktur sederhana asetaldehida: $\mathrm { CH } _ { 3 } \mathrm { COH }$
  • D. D. Rumus struktur propilena yang disederhanakan: $\mathrm { CH } _ { 3 } \mathrm { CHCH } _ { 2 }$

Answer: A

Solution: Atom karbon dalam $\mathrm { A } . \mathrm { CH } _ { 4 }$ membentuk 4 ikatan $\mathrm { C } - \mathrm { H }$, dan rumus elektroniknya adalah $\mathrm { H } : \stackrel { \text { H } } { \stackrel { \mathrm { H } } { \mathrm { C } } } : \mathrm { H } , \mathrm { A }$ yang benar; H B. Rumus struktur propana adalah $\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { CH } _ { 3 }$ dan rumus molekulnya adalah $\mathrm { C } _ { 3 } \mathrm { H } _ { 8 }$, B salah; C. Rumus struktur asetaldehida adalah $\mathrm { CH } _ { 3 } \mathrm { CHO }$, C salah; D. Rumus struktur propilena adalah $\mathrm { CH } _ { 3 } \mathrm { CH } = \mathrm { CH } _ { 2 } , \mathrm { D }$, yang salah;

Question 10: 11. Istilah-istilah kimia berikut ini dinyatakan dengan benar. ![](/images/questions/chem-nomenclat...

11. Istilah-istilah kimia berikut ini dinyatakan dengan benar. ![](/images/questions/chem-nomenclature/image-007.jpg)

  • A. A. $\mathrm { CH } _ { 4 }$ Model pengisian ruang dari sebuah molekul:
  • B. B. Struktur atom fluor: + $+ 9 ) 28$
  • C. C. Simbol nuklida dengan 6 proton dan 8 neutron: ${ } _ { 6 } ^ { 12 } \mathrm { C } 8$
  • D. D. Gugus fungsi etilena: C = C

Answer: A

Solution: A. $\mathrm { CH } _ { 4 }$ adalah molekul berbentuk ortotetrahedral, model molekul yang mengisi ruang: ![](/images/questions/chem-nomenclature/image-003.jpg) B. Atom fluor tidak bermuatan dan memiliki tujuh elektron di lapisan terluar atom; C. Unsur dengan 6 proton adalah C. Dengan 8 neutron, nomor massanya adalah 14, sehingga lambang untuk nuklida dengan 6 proton dan 8 neutron adalah: ${ } ^ { 14 } \mathrm { C }$, dan C salah; D. Gugus fungsional etilena: <img class="imgSvg" id = "mi1mzfx655vfte8ckvp" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16Zng2NTV2ZnRlOGNrdnAiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDE2NiAxMjAuNzQ5OTgxNjM1MDgxNDEiIHN0eWxlPSJ3aWR0aDogMTY1LjgzOTM5MDA1NDYzMDU0cHg7IGhlaWdodDogMTIwLjc0OTk4MTYzNTA4MTQxcHg7IG92ZXJmbG93OiB2aXNpYmxlOyI + PGRlZnM + PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16Zng2NTV2ZnRlOGNrdnAtMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI5Ni41NTk2MDA0Mzg0MDcyNyIgeTE9IjUyLjQ5OTk5OTk5OTk5Mjg2NiIgeDI9IjEyMy44MzkzOTAwNTQ2MzA1NCIgeTI9IjY4LjI1MDAxODM2NDg5MDA0Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + 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Question 11: 12. Representasi atau pernyataan berikut ini adalah benar ![](/images/questions/chem-nomenclature/...

12. Representasi atau pernyataan berikut ini adalah benar ![](/images/questions/chem-nomenclature/image-009.jpg)

  • A. A. Diagram skematik struktur atom belerang:
  • B. B. $\mathrm { CS } _ { 2 }$ untuk model kelelawar:
  • C. C. Rumus terpendek untuk metil format: $\mathrm { CH } _ { 2 } \mathrm { O }$
  • D. D. Rumus molekul kalsium klorida: $\mathrm { CaCl } _ { 2 }$

Answer: C

Solution: A. Menurut nomor atom atom belerang adalah 16, jumlah proton di dalam inti adalah 16, dan jumlah elektron di luar inti juga 16, jadi A salah; B. Menurut penilaian jenis hibridisasi, karbon disulfida dalam hibridisasi karbon sp, konfigurasi ruangnya linier, jadi B salah; C. Rumus struktur metil format adalah $\mathrm { HCOOCH } _ { 3 }$, dan rumus yang paling sederhana adalah $\mathrm { CH } _ { 2 } \mathrm { O }$, jadi C benar; D. Kalsium klorida adalah senyawa ion, bukan molekul, jadi D salah;

Question 12: 13. Istilah atau ungkapan kimia berikut ini tidak tepat.

13. Istilah atau ungkapan kimia berikut ini tidak tepat.

  • A. A. Bentuk pendek dari metil: $- \mathrm { CH } _ { 3 }$
  • B. B. Rumus paling sederhana dari molekul benzena: CH
  • C. C. Rantai karbon n-pentana berbentuk linier
  • D. D. Rumus kimia $\mathrm { CH } _ { 2 } \mathrm { Br } _ { 2 }$ hanya dapat mewakili satu zat.

Answer: C

Solution: A. Metil memiliki rumus struktur: $- \mathrm { CH } _ { 3 }$ , A benar; B. Rumus kimia molekul benzena adalah: $\mathrm { C } _ { 6 } \mathrm { H } _ { 6 }$, dan rumus molekul benzena yang paling sederhana adalah: CH, B benar; C. Kelima atom C dalam molekul n-pentana adalah koplanar - tersusun dalam pola gigi gergaji - tetapi tidak dalam satu garis yang sama, C salah; D. Rumus kimia $\mathrm { CH } _ { 2 } \mathrm { Br } _ { 2 }$ hanya dapat mewakili satu zat, dan tidak ada isomer, D benar;

Question 13: 14. Istilah kimia berikut ini benar <img class="imgSvg" id = "mi1mzfvwvjmfvkggpre" src = "data:imag...

14. Istilah kimia berikut ini benar <img class="imgSvg" id = "mi1mzfvwvjmfvkggpre" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16ZnZ3dmptZnZrZ2dwcmUiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDExNSA2OS4yNzk4MTA4MjIxODQiIHN0eWxlPSJ3aWR0aDogMTE1LjQ5OTk5OTk5OTk5Mjg3cHg7IGhlaWdodDogNjkuMjc5ODEwODIyMTg0cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnZ3dmptZnZrZ2dwcmUtMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI1Ny43NDk5ODE2MzUwOTU3MSIgeTE9IjQ4LjI3OTgxMDgyMjE4NDAxIiB4Mj0iNzMuNDk5OTk5OTk5OTkyODciIHkyPSIyMS4wMDAwMjEyMDU5NjA3NTMiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnZ3dmptZnZrZ2dwcmUtMyIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI0MiIgeTE9IjIxIiB4Mj0iNzMuNDk5OTk5OTk5OTkyODciIHkyPSIyMS4wMDAwMjEyMDU5NjA3NTMiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnZ3dmptZnZrZ2dwcmUtNSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI0NS4yNzM1NzcyOTg2NjIwNzYiIHkxPSIxOS4xMTAwMDIyMDM3ODg1MTUiIHgyPSI2MS4wMjM1NTg5MzM3NTc4IiB5Mj0iNDYuMzg5ODEzMDI1OTcyNTMiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnZ3dmptZnZrZ2dwcmUtNyIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSIzOC43MjY0MjI3MDEzMzc5MjQiIHkxPSIyMi44ODk5OTc3OTYyMTE0ODIiIHgyPSI1NC40NzY0MDQzMzY0MzM2MyIgeTI9IjUwLjE2OTgwODYxODM5NTQ5Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ2d3ZqbWZ2a2dncHJlLTkiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNDIiIHkxPSIyMSIgeDI9IjU3Ljc0OTk4MTYzNTA5NTcxIiB5Mj0iNDguMjc5ODEwODIyMTg0MDEiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+PC9kZWZzPjxtYXNrIGlkPSJ0ZXh0LW1hc2stbWkxbXpmdnd2am1mdmtnZ3ByZSI+ PHJlY3QgeD0iMCIgeT0iMCIgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0id2hpdGUiPjwvcmVjdD48L21hc2s+ PHN0eWxlPgogICAgICAgICAgICAgICAgLmVsZW1lbnQtbWkxbXpmdnd2am1mdmtnZ3ByZSB7CiAgICAgICAgICAgICAgICAgICAgZm9udDogMTRweCBIZWx2ZXRpY2EsIEFyaWFsLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAgICAgICAgIGFsaWdubWVudC1iYXNlbGluZTogJ21pZGRsZSc7CiAgICAgICAgICAgICAgICB9CiAgICAgICAgICAgICAgICAuc3ViLW1pMW16ZnZ3dmptZnZrZ2dwcmUgewogICAgICAgICAgICAgICAgICAgIGZvbnQ6IDguNHB4IEhlbHZldGljYSwgQXJpYWwsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICAgICB9CiAgICAgICAgICAgIDwvc3R5bGU + PGcgbWFzaz0idXJsKCN0ZXh0LW1hc2stbWkxbXpmdnd2am1mdmtnZ3ByZSkiPjxsaW5lIHgxPSI1Ny43NDk5ODE2MzUwOTU3MSIgeTE9IjQ4LjI3OTgxMDgyMjE4NDAxIiB4Mj0iNzMuNDk5OTk5OTk5OTkyODciIHkyPSIyMS4wMDAwMjEyMDU5NjA3NTMiIHN0eWxlPSJzdHJva2UtbGluZWNhcDpyb3VuZDtzdHJva2UtZGFzaGFycmF5Om5vbmU7c3Ryb2tlLXdpZHRoOjEuMjYiIHN0cm9rZT0idXJsKCcjbGluZS1taTFtemZ2d3ZqbWZ2a2dncHJlLTEnKSI + PC9saW5lPjxsaW5lIHgxPSI0MiIgeTE9IjIxIiB4Mj0iNzMuNDk5OTk5OTk5OTkyODciIHkyPSIyMS4wMDAwMjEyMDU5NjA3NTMiIHN0eWxlPSJzdHJva2UtbGluZWNhcDpyb3VuZDtzdHJva2UtZGFzaGFycmF5Om5vbmU7c3Ryb2tlLXdpZHRoOjEuMjYiIHN0cm9rZT0idXJsKCcjbGluZS1taTFtemZ2d3ZqbWZ2a2dncHJlLTMnKSI + PC9saW5lPjxsaW5lIHgxPSI0NS4yNzM1NzcyOTg2NjIwNzYiIHkxPSIxOS4xMTAwMDIyMDM3ODg1MTUiIHgyPSI2MS4wMjM1NTg5MzM3NTc4IiB5Mj0iNDYuMzg5ODEzMDI1OTcyNTMiIHN0eWxlPSJzdHJva2UtbGluZWNhcDpyb3VuZDtzdHJva2UtZGFzaGFycmF5Om5vbmU7c3Ryb2tlLXdpZHRoOjEuMjYiIHN0cm9rZT0idXJsKCcjbGluZS1taTFtemZ2d3ZqbWZ2a2dncHJlLTUnKSI + 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 + PC9saW5lPjwvZz48Zz48dGV4dCB4PSI3My40OTk5OTk5OTk5OTI4NyIgeT0iMjEuMDAwMDIxMjA1OTYwNzUzIiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAiPjwvdGV4dD48dGV4dCB4PSI1Ny43NDk5ODE2MzUwOTU3MSIgeT0iNDguMjc5ODEwODIyMTg0MDEiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI + PC90ZXh0Pjx0ZXh0IHg9IjQyIiB5PSIyMSIgY2xhc3M9ImRlYnVnIiBmaWxsPSIjZmYwMDAwIiBzdHlsZT0iCiAgICAgICAgICAgICAgICAgICBmb250OiA1cHggRHJvaWQgU2Fucywgc2Fucy1zZXJpZjsKICAgICAgICAgICAgICAgIj48L3RleHQ +PC9nPjwvc3ZnPg== "/>

  • A. A. Formula elektronik etilena:
  • B. B. Rumus struktur metana: $\mathrm { CH } _ { 4 }$
  • C. C. Model skala molekul asam asetat:
  • D. D. Model bola-dan-tongkat dari molekul etana:

Answer: C

Solution:

Question 14: 15. Konversi $\mathrm { CO } _ { 2 }$ menjadi bahan bakar melalui hidrogenasi merupakan strategi pen...

15. Konversi $\mathrm { CO } _ { 2 }$ menjadi bahan bakar melalui hidrogenasi merupakan strategi penting untuk mencapai pengurangan emisi $\mathrm { CO } _ { 2 }$, dan para ilmuwan di Cina telah membuat kemajuan baru dalam mengatur selektivitas reaksi hidrogenasi $\mathrm { CO } _ { 2 }$, dan prosesnya diilustrasikan sebagai berikut. Analisis berikut ini adalah benar () ![](/images/questions/chem-nomenclature/image-004.jpg)

  • A. A. Proses ini memutus ikatan polar dalam molekul $\mathrm { H } _ { 2 }$
  • B. B. Rumus elektronik untuk $\mathrm { CO } _ { 2 }$ adalah $\because \mathrm { C } : \mathrm { O } :$.
  • C. C. Dalam reaksi ini, $\mathrm { CO } _ { 2 }$ bertindak sebagai agen pereduksi.
  • D. D. Persamaan kimia untuk reaksi ini adalah $\mathrm { CO } _ { 2 } + 4 \mathrm { H } _ { 2 } \xlongequal { \text { katalis } } \mathrm { CH } _ { 4 } + 2 \mathrm { H } _ { 2 } \mathrm { C }$

Answer: D

Solution: A. Proses ini melibatkan reaksi kimia $\mathrm { CO } _ { 2 } + 4 \mathrm { H } _ { 2 } \xlongequal { \text { katalis } } \mathrm { CH } _ { 4 } + 2 \mathrm { H } _ { 2 } \mathrm { O }$, di mana ikatan lama diputuskan dan ikatan baru terbentuk, tetapi molekul $\mathrm { H } _ { 2 }$ adalah molekul yang ikatan nonpolarnya diputuskan, jadi A salah; B. Rumus elektronik dari $\mathrm { CO } _ { 2 }$ adalah $\ddot { \mathbf { g } } = \mathbf { c } = \mathbf { = } \ddot { \mathbf { o } }$; C. Pada reaksi $\mathrm { CO } _ { 2 }$, valensi karbon berubah dari +4 menjadi -4, yang mengurangi valensi dan bertindak sebagai zat pengoksidasi, maka C salah; D. Menurut analisis di atas, persamaan kimia dari reaksi tersebut adalah: $\mathrm { CO } _ { 2 } + 4 \mathrm { H } _ { 2 } \xlongequal { \text { katalis } } \mathrm { CH } _ { 4 } + 2 \mathrm { H } _ { 2 } \mathrm { O }$, D benar;

Question 15: 17. Istilah atau diagram kimia berikut ini tidak dinyatakan dengan benar () ![](/images/questions/c...

17. Istilah atau diagram kimia berikut ini tidak dinyatakan dengan benar () ![](/images/questions/chem-nomenclature/image-011.jpg)

  • A. A. Model $\mathrm { SO } _ { 3 } ^ { 2 - }$ dari pasangan elektron lapisan valensi yang saling tolak-menolak (VSEPR):
  • B. B. Rumus elektronik untuk $\mathrm { NF } _ { 3 }$:: $\underset { . \ddot { \mathrm { F } } : \ddot { \mathrm { F } } : \ddot { \mathrm { F } } : } { \ddot { \mathrm { F } } : }$
  • C. C. Pembentukan ikatan $\sigma$ dalam molekul HCl $: \stackrel { \mathrm { H } } { \bullet } \rightarrow + \overbrace { } ^ { \mathrm { Cl } } \rightarrow \overbrace { } ^ { \mathrm { H } } \rightarrow \overbrace { } ^ { \mathrm { Cl } }$
  • D. D. Susunan elektron valensi dari atom germanium (Ge) tingkat dasar adalah: $4 s ^ { 2 } 4 p ^ { 2 }$

Answer: B

Solution: A. Atom pusat S dari $\mathrm { SO } _ { 3 } ^ { 2 - }$ memiliki jumlah pasangan elektron valensi sebesar $3 + \frac { 1 } { 2 } ( 6 + 2 - 2 \times 3 ) = 4$, dan model VSEPR adalah tetrahedral, jadi A benar; B. Jumlah elektron terluar dari $N$ dalam ${ } ^ { N F _ { 3 } }$ adalah 5, dan ada sepasang elektron tunggal setelah pembentukan 3 pasang elektron, dan rumus elektronik ${ } ^ { N F _ { 3 } }$ adalah $: \ddot { \mathrm { F } } : \ddot { \mathrm { N } } : \ddot { \mathrm { F } } :$. Rumus elektronik $: \ddot { \mathrm { F } } : \ddot { \mathrm { N } } : \ddot { \mathrm { F } } :$, B salah; C. $\mathrm { s } - \mathrm { p }$ dari $\sigma$ berikatan di dalam molekul HCl, membentuk diagram yang benar, C benar; D. Germanium (Ge) berada pada golongan IVA pada siklus keempat, dan susunan elektron valensinya adalah $4 \mathrm {~s} ^ { 2 } 4 \mathrm { p } ^ { 2 }$, D benar;

Question 16: 19. Ungkapan atau pernyataan berikut ini benar ().

19. Ungkapan atau pernyataan berikut ini benar ().

  • A. A. Model skala dapat mewakili molekul metana dan karbon tetraklorida
  • B. B. - OH dan $\cdot \mathrm { O } _ { \bullet \bullet } \mathbf { H } \quad$ keduanya mewakili gugus hidroksil.
  • C. C. Rumus struktur asetaldehida adalah: $\mathrm { CH } _ { 3 } \mathrm { COH }$
  • D. D. Rumus terpendek untuk benzena: $( \mathrm { CH } ) _ { 6 }$

Answer: B

Solution: A. Jari-jari atom Cl lebih besar daripada atom C, dan model skala molekul CTC salah. Model skala harus konsisten dengan ukuran atom, jadi A salah; B. - OH adalah rumus struktur hidroksil, $\quad \ddot { \mathrm { O } } : \mathrm { H }$ adalah rumus elektronik hidroksil, jadi B benar; C. Molekul asetaldehida mengandung gugus aldehida (- CHO), dan rumus strukturalnya adalah: $\mathrm { CH } _ { 3 } \mathrm { CHO }$, jadi C salah; D, rumus paling sederhana juga disebut rumus eksperimental, dinyatakan dalam molekul jumlah atom dari setiap elemen rasio bilangan bulat paling sederhana dari rumus tersebut, rumus paling sederhana benzena untuk CH, jadi D salah.

Question 17: 20. Istilah kimia berikut ini salah () ![](/images/questions/chem-nomenclature/image-012.jpg)

20. Istilah kimia berikut ini salah () ![](/images/questions/chem-nomenclature/image-012.jpg)

  • A. A. Model skala benzena:
  • B. B. Model kelelawar propana:
  • C. C. Nuklida ${ } ^ { \mathrm { Cl } }$ dengan jumlah neutron 20: ${ } ^ { 20 } \mathrm { Cl }$
  • D. D. Rumus struktur asam hipoklorit: $\mathrm { H } - \mathrm { O } - \mathrm { Cl }$

Answer: C

Solution: A. Benzena memiliki struktur planar, terdapat 6 H dan 6 C dalam molekul benzena, semua ikatan karbon-karbon identik, dan rasio benzena terhadap karbon sama. Contoh model untuk benzena adalah [IMAGE_5]] Oleh karena itu, A adalah benar; B. Model yang diwakili oleh bola dan tongkat adalah model bola dan tongkat, rumus struktur propana adalah $\mathrm { CH } _ { 3 } \mathrm { CH } _ { 2 } \mathrm { CH } _ { 3 }$, jari-jari atom karbon lebih besar dari pada atom hidrogen, model bola dan tongkat adalah <img class="imgSvg" id = "mi1mzfx7pg3ufw2jndp" src = "data:image/svg+xml;base64, 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 + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4N3BnM3VmdzJqbmRwLTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTM1LjM2OTIxOTI0MTcxODg0IiB5MT0iMjEuMDAwMDM2NzI5ODAxNDgiIHgyPSIxNTEuMTE5MjAwODc2ODE0NTUiIHkyPSI0OC4yNzk4NDc1NTE5ODU1Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4N3BnM3VmdzJqbmRwLTUiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTUxLjExOTIwMDg3NjgxNDU1IiB5MT0iNDguMjc5ODQ3NTUxOTg1NSIgeDI9IjE2Ni44NjkyMTkyNDE3MTE3IiB5Mj0iMjEuMDAwMDU3OTM1NzYyMjQzIj48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4N3BnM3VmdzJqbmRwLTciIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTIzLjgzOTM5MDA1NDYzMDUxIiB5MT0iNjQuMDI5ODI5MTg3MDgxMTgiIHgyPSIxNTEuMTE5MjAwODc2ODE0NTUiIHkyPSI0OC4yNzk4NDc1NTE5ODU1Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4N3BnM3VmdzJqbmRwLTkiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iOTYuNTU5NjAwNDM4NDA3MjciIHkxPSI0OC4yNzk4MTA4MjIxODQwMiIgeDI9IjEyMy44MzkzOTAwNTQ2MzA1MSIgeTI9IjY0LjAyOTgyOTE4NzA4MTE4Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + 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Nomor massa ${ } ^ { \mathrm { Cl } }$ dengan nomor neutron 20 $= 20 + 17 = 37$, nuklida tersebut dapat dinyatakan sebagai ${ } ^ { 37 } \mathrm { Cl }$, jadi C salah; D. Asam hipoklorit termasuk senyawa kovalen, atom pusatnya adalah atom O, rumus strukturnya adalah $\mathrm { H } - \mathrm { O } - \mathrm { Cl }$, jadi D benar.

Question 18: 21. Istilah kimia berikut ini digunakan dengan benar (). $H : O ̈ : : O ̈ : H$

21. Istilah kimia berikut ini digunakan dengan benar (). $H : O ̈ : : O ̈ : H$

  • A. A. Rumus struktural untuk $\mathrm { SiO } _ { 2 }$: $\mathrm { O } - \mathrm { Si } = \mathrm { O }$
  • B. B. Rumus elektronik dari $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$:
  • C. C. Persamaan ionisasi untuk $\mathrm { BaSO } _ { 4 }$: $\mathrm { BaSO } _ { 4 } = { } ^ { \mathrm { SO } _ { 4 } ^ { 2 - } } + \mathrm { Ba } ^ { 2 + }$
  • D. D. ${ } ^ { 12 } \mathrm { C }$ dan ${ } ^ { 14 } \mathrm { C }$ adalah sinonim satu sama lain.

Answer: C

Solution: A. $\mathrm { SiO } _ { 2 }$ adalah kristal kovalen, di mana setiap atom Si membentuk 4 $\mathrm { Si } - \mathrm { O }$ ikatan dengan 4 atom O, dan $\mathrm { Si } -$ ikatan O membentuk struktur orto-tetrahedral, dan setiap atom O membentuk $\mathrm { Si } - \mathrm { O }$ ikatan tunggal dengan 2 atom Si. RUMUS_3]] ikatan, A salah; B. Dalam molekul $\mathrm { Si } - \mathrm { O }$, 2 atom O membentuk ikatan $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$, dan setiap atom O membentuk ikatan $\mathrm { O } - \mathrm { O }$ dengan atom H. Setiap atom O kemudian membentuk ikatan $\mathrm { H } - \mathrm { O }$ dengan atom H, menghasilkan struktur yang stabil dengan dua atau delapan elektron pada lapisan terluar molekul, dengan rumus elektronik $\mathrm { H } \% { } ^ { \circ } { } _ { \circ } ^ { \circ } \times { } _ { \times } ^ { \times \times } \times \mathrm { H }$. Rumus elektroniknya adalah $\mathrm { H } \% { } ^ { \circ } { } _ { \circ } ^ { \circ } \times { } _ { \times } ^ { \times \times } \times \mathrm { H }$; C. $\mathrm { BaSO } _ { 4 }$ sulit larut dalam air, tetapi bagian yang terlarut dalam air terionisasi sempurna untuk menghasilkan $\mathrm { Ba } ^ { 2 + } , \mathrm { SO } _ { 4 } ^ { 2 - }$ yang bergerak bebas, sehingga persamaan ionisasinya adalah $\mathrm { BaSO } _ { 4 } = \mathrm { SO } _ { 4 } ^ { 2 - } + \mathrm { Ba } ^ { 2 + }$, C benar; D. Jumlah proton dalam ${ } ^ { 12 } \mathrm { C }$ dan ${ } ^ { 14 } \mathrm { C }$ adalah 6, dan jumlah neutron dalam $6 , 8$ adalah $6 , 8$, yang menunjukkan bahwa keduanya memiliki jumlah proton yang sama namun jumlah neutronnya berbeda. Oleh karena itu, keduanya merupakan isotop satu sama lain; Jawabannya adalah C.

Question 19: 22. Reaksi utama dalam reformasi pirolitik dari $\mathrm { H } _ { 2 } \mathrm {~S} , \mathrm { CH }...

22. Reaksi utama dalam reformasi pirolitik dari $\mathrm { H } _ { 2 } \mathrm {~S} , \mathrm { CH } _ { 4 }$ menjadi $\mathrm { H } _ { 2 }$ adalah $2 \mathrm { H } _ { 2 } \mathrm {~S} = 2 \mathrm { H } _ { 2 } + \mathrm { S } _ { 2 } , 2 \mathrm { H } _ { 2 } \mathrm {~S} + \mathrm { CH } _ { 4 } = \mathrm { CS } _ { 2 } + 4 \mathrm { H } _ { 2 }$. Pernyataan-pernyataan berikut ini benar ( ) ![](/images/questions/chem-nomenclature/image-004.jpg)

  • A. A. $\mathrm { CH } _ { 4 }$ Model molekul kelelawar :
  • B. B. $\mathrm { S } _ { 2 }$ adalah sebuah congener dari $\mathrm { S } _ { 8 }$.
  • C. C. Molekul $\mathrm { CS } _ { 2 }$ berbentuk sudut
  • D. D. Atom belerang dengan jumlah neutron 16: ${ } ^ { 32 } \mathrm {~S}$

Answer: D

Solution: Model kelelawar untuk molekul $\mathrm { A } . \mathrm { CH } _ { 4 }$ adalah $\bigcirc$. B. $\mathrm { S } _ { 2 }$ dan $\mathrm { S } _ { 8 }$ adalah dua monomer belerang yang berbeda dan merupakan isomer satu sama lain; C. Struktur $\mathrm { CS } _ { 2 }$ mirip dengan $\mathrm { CO } _ { 2 }$, yang merupakan molekul linier; D.S Jumlah proton adalah 16, dan jika jumlah neutron adalah 16, nomor massanya adalah 32, D benar; Oleh karena itu, jawabannya adalah D.

Question 20: 23. Pernyataan berikut tentang etanol adalah benar () ![](/images/questions/chem-nomenclature/image...

23. Pernyataan berikut tentang etanol adalah benar () ![](/images/questions/chem-nomenclature/image-015.jpg)

  • A. A. Model bola-dan-tongkat dari molekul tersebut adalah :
  • B. B. Cairan tidak berwarna dan tidak berbau pada suhu kamar
  • C. C. Diisomerisasi dengan asam asetat
  • D. D. Mengubah lakmus ungu menjadi merah.

Answer: A

Solution: A. Rumus struktur etanol adalah ![](/images/questions/chem-nomenclature/image-003.jpg) , A adalah benar; B. Etanol adalah cairan tak berwarna dengan bau khas pada suhu kamar, B salah; C. Rumus molekul etanol adalah $\mathrm { C } _ { 2 } \mathrm { H } _ { 5 } \mathrm { OH }$, dan rumus molekul asam asetat adalah $\mathrm { CH } _ { 3 } \mathrm { COOH }$, kedua rumus molekul tersebut berbeda dan bukan isomer satu sama lain, C salah; D. Etanol bersifat netral dan tidak dapat membuat larutan lakmus ungu menjadi merah, D salah;

Question 21: 25. Istilah-istilah kimia berikut ini benar. ![](/images/questions/chem-nomenclature/image-017.jpg)...

25. Istilah-istilah kimia berikut ini benar. ![](/images/questions/chem-nomenclature/image-017.jpg) <img class="imgSvg" id = "mi1mzfvx68krowfpmi" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16ZnZ4Njhrcm93ZnBtaSIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgMTI5IDg2LjU0NzcyNzIxNDc1MTM4IiBzdHlsZT0id2lkdGg6IDEyOC41NDc3MjcyMTQ3NTEzOHB4OyBoZWlnaHQ6IDg2LjU0NzcyNzIxNDc1MTM4cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnZ4Njhrcm93ZnBtaS0xIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjYyLjQ5MTk1ODExNzU1NjUxIiB5MT0iMjcuMjM2NjgxNDEwMjAxNzYiIHgyPSI4MC4zMTEwNDUwMDQ4MjI3NiIgeTI9IjQ1LjA1NTc3NjI5NDczNjYyIj48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + 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  • A. A. Struktur skematis ion klorida:
  • B. B. Rumus elektronik dari gugus aldehida (- CHO ):
  • C. C. $\mathrm { CH } _ { 4 }$ untuk model kelelawar:
  • D. D. Rumus struktural molekul etilena: $\mathrm { CH } _ { 2 } = \mathrm { CH } _ { 2 }$

Answer: B

Solution: A. Klorin adalah unsur 17, dan struktur ion klorida ditunjukkan secara skematis: ![](/images/questions/chem-nomenclature/image-019.jpg) B. Ikatan rangkap karbon-oksigen terbentuk antara atom karbon dan oksigen dalam gugus aldehida, sehingga rumus elektroniknya adalah: ![](/images/questions/chem-nomenclature/image-018.jpg) C. ![](/images/questions/chem-nomenclature/image-019.jpg) adalah model skala metana dan model bola dan tongkat metana adalah <img class="imgSvg" id = "mi1mzfxapvxj28fwncb" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16ZnhhcHZ4ajI4ZnduY2IiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDEzOSA4OS4yNTAwMDYxMjE2MzIzOCIgc3R5bGU9IndpZHRoOiAxMzguNTU5NjAwNDM4NDE4MjdweDsgaGVpZ2h0OiA4OS4yNTAwMDYxMjE2MzIzOHB4OyBvdmVyZmxvdzogdmlzaWJsZTsiPjxkZWZzPjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4YXB2eGoyOGZ3bmNiLTEiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNjkuMjc5Nzk2Njg0ODgyMzQiIHkxPSIzNi43NTAwMDYxMjE2MzMxOCIgeDI9Ijk2LjU1OTYwMDQzODQxODI3IiB5Mj0iMjEuMDAwMDEyMjQzMjY3MTU3Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + 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Rumus struktur etilena, semua pasangan elektron bersama harus ditunjukkan dengan garis pendek seperti <img class="imgSvg" id = "mi1mzfxb1cgeu0cockni" src = "data:image/svg+xml;base64, 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 +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnhiMWNnZXUwY29ja25pLTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNDEuOTk5OTk4NzI3NjQyMzUiIHkxPSIyMy44MzQ5OTk5OTk5OTk3MTMiIHgyPSI3My40OTk5OTg3Mjc2MzkxOCIgeTI9IjIzLjgzNTAxNDEzNzMwNjg5MiI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+PC9saW5lYXJHcmFkaWVudD48L2RlZnM+ PG1hc2sgaWQ9InRleHQtbWFzay1taTFtemZ4YjFjZ2V1MGNvY2tuaSI+ PHJlY3QgeD0iMCIgeT0iMCIgd2lkdGg9IjEwMCUiIGhlaWdodD0iMTAwJSIgZmlsbD0id2hpdGUiPjwvcmVjdD48L21hc2s+ 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 + PGxpbmUgeDE9IjQxLjk5OTk5ODcyNzY0MjM1IiB5MT0iMjMuODM0OTk5OTk5OTk5NzEzIiB4Mj0iNzMuNDk5OTk4NzI3NjM5MTgiIHkyPSIyMy44MzUwMTQxMzczMDY4OTIiIHN0eWxlPSJzdHJva2UtbGluZWNhcDpyb3VuZDtzdHJva2UtZGFzaGFycmF5Om5vbmU7c3Ryb2tlLXdpZHRoOjEuMjYiIHN0cm9rZT0idXJsKCcjbGluZS1taTFtemZ4YjFjZ2V1MGNvY2tuaS0zJykiPjwvbGluZT48L2c +PGc+ PHRleHQgeD0iNzMuNDk5OTk5OTk5OTk2ODIiIHk9IjIxLjAwMDAxNDEzNzMwNzE4IiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAiPjwvdGV4dD48dGV4dCB4PSI0MiIgeT0iMjEiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI +PC90ZXh0PjwvZz48L3N2Zz4="/> Pilihan jawaban B.

Question 22: 26. Istilah-istilah kimia berikut ini dinyatakan dengan benar.

26. Istilah-istilah kimia berikut ini dinyatakan dengan benar.

  • A. A. Rumus kimia untuk kolagog: $\mathrm { CuSO } _ { 4 }$
  • B. B. Volume molar gas adalah $22.4 \mathrm {~L} / \mathrm { mol }$
  • C. C. Ti: nomor neutron 26
  • D. D. Struktur skematis ion natrium:

Answer: C

Solution: A. Kolekalsiferol adalah kristal tembaga sulfat dengan rumus kimia $\mathrm { CuSO } _ { 4 } \cdot 5 \mathrm { H } _ { 2 } \mathrm { O }$, jadi A salah; B. Volume molar gas dalam kondisi standar adalah sekitar $22.4 \mathrm {~L} / \mathrm { mol }$, kondisinya tidak diketahui, belum tentu $22.4 \mathrm {~L} / \mathrm { mol }$, jadi B salah; C. Jumlah proton + neutron $=$, maka jumlah neutron adalah $48 - 22 = 26$, jadi C benar; D. Jumlah proton ion natrium adalah 11, jumlah elektron di luar inti adalah 10, dan struktur ionnya adalah (+11) 28 /, jadi D salah;

Question 23: 27. Istilah-istilah kimia berikut ini dinyatakan dengan benar sebagai berikut ![](/images/questio...

27. Istilah-istilah kimia berikut ini dinyatakan dengan benar sebagai berikut ![](/images/questions/chem-nomenclature/image-022.jpg)

  • A. A. Rumus struktur sederhana etilena: $\mathrm { CH } _ { 2 } \mathrm { CH } _ { 2 }$
  • B. B. Rumus elektronik untuk karbon tetraklorida: $\mathrm { Cl } : \underset { \underset { \mathrm { Cl } } { \mathrm { Cl } } } { \stackrel { \mathrm { Cl } } { \mathrm { Cl } } } \mathrm { Cl }$
  • C. C. Model skala $\mathrm { CO } _ { 2 }$:
  • D. D. Rumus struktur HClO: $\mathrm { H } - \mathrm { O } - \mathrm { Cl }$

Answer: D

Solution: A. Gugus fungsi dalam etilena adalah ikatan rangkap karbon-karbon, dan bentuk pendek struktural etilena adalah $\mathrm { CH } _ { 2 } = \mathrm { CH } _ { 2 } , \mathrm {~A}$ Salah; B. Rumus elektronik dari $\mathrm { CCl } _ { 4 }$ adalah $\mathrm { CCl } _ { 4 }$ karena penghilangan pasangan elektron tunggal pada atom Cl. C. Jari-jari atom karbon lebih besar daripada jari-jari atom oksigen; D. Rumus struktur HClO adalah $\mathrm { H } - \mathrm { O } - \mathrm { Cl }$, dan D benar; Pilihan jawaban D.

Question 24: 28. Nitrogen dan senyawanya diubah seperti yang ditunjukkan dalam diagram. Analisis berikut ini masu...

28. Nitrogen dan senyawanya diubah seperti yang ditunjukkan dalam diagram. Analisis berikut ini masuk akal ![](/images/questions/chem-nomenclature/image-023.jpg)

  • A. A. Di bawah aksi katalis a, ${ } ^ { \mathrm { N } _ { 2 } }$ dan ${ } ^ { \mathrm { H } _ { 2 } }$ dapat dikonversi menjadi $\mathrm { NH } _ { 3 }$ 100%.
  • B. B. Katalis a mempercepat laju reaksi kimia dengan menurunkan energi ikatan ${ } ^ { \mathrm { N } _ { 2 } }$
  • C. C. Terjadi perubahan valensi unsur ketika $\mathrm { N } - \mathrm { O }$ ikatan terbentuk pada permukaan katalis b
  • D. D. Kedua langkah tersebut adalah fiksasi nitrogen

Answer: C

Solution: A. Reaksi antara nitrogen dan hidrogen adalah reaksi reversibel dan reaktan tidak dapat sepenuhnya diubah menjadi produk, sehingga ${ } ^ { \mathrm { N } _ { 2 } }$ dan ${ } ^ { \mathrm { H } _ { 2 } }$ tidak dapat diubah dari $100 \%$ menjadi ${ } ^ { \mathrm { NH } _ { 3 } }$ dengan katalisator a, A salah; B. Katalis a mempercepat laju reaksi dengan menurunkan energi aktivasi reaksi, tetapi tidak menurunkan energi ikatan zat, yaitu tidak menurunkan energi ikatan ${ } ^ { \mathrm { N } _ { 2 } }$; C. Ketika ikatan nitrogen-oksigen terbentuk pada permukaan katalis b, valensi unsur N berubah dari -3 menjadi +2, dan terjadi perubahan valensi unsur, C benar; D. Fiksasi nitrogen mengacu pada proses pengubahan nitrogen bebas menjadi senyawa nitrogen, sehingga langkah pertama reaksi termasuk dalam fiksasi nitrogen, dan langkah kedua reaksi tidak termasuk dalam fiksasi nitrogen, dan D salah;

Question 25: 29. Istilah-istilah kimia berikut ini digunakan dengan benar. ![](/images/questions/chem-nomenclat...

29. Istilah-istilah kimia berikut ini digunakan dengan benar. ![](/images/questions/chem-nomenclature/image-024.jpg) Berdasarkan informasi dari percobaan berikut, selesaikan sub-pertanyaan berikut: Percobaan dilakukan dengan menggunakan alat yang ditunjukkan di bawah ini untuk memverifikasi bahwa $\mathrm { SO } _ { 2 } , \mathrm { CO } _ { 2 }$ dan $\mathrm { H } _ { 2 } \mathrm { O }$ terdapat dalam gas yang diperoleh dengan memanaskan asam sulfat pekat dengan arang. Selama percobaan diamati bahwa tembaga sulfat anhidrat berubah menjadi biru, larutan magenta memudar, larutan kalium permanganat asam menjadi lebih terang warnanya dan kekeruhan putih muncul pada air kapur yang telah dijernihkan. ![](/images/questions/chem-nomenclature/image-025.jpg)

  • A. A. Persamaan ionisasi untuk abu soda: $\mathrm { NaHCO } _ { 3 } = \mathrm { Na } ^ { + } + \mathrm { HCO } ^ { 3 }$
  • B. B. Struktur skematis atom klorin:
  • C. C. Rumus struktural untuk $\mathrm { H } _ { 2 } \mathrm {~S}$: $\mathrm { H } - \mathrm { S } - \mathrm { H }$
  • D. D. Rumus elektronik untuk hidrogen peroksida: $\mathrm { H } ^ { + } \left( : \ddot { \mathrm { O } } : \ddot { \mathrm { O } } ^ { : } \right) ^ { 2 - } \mathrm { H } ^ { + }$

Answer: C

Solution: A. Soda abu adalah natrium karbonat, dan persamaan ionisasinya adalah: $\mathrm { Na } _ { 2 } \mathrm { CO } _ { 3 } = 2 \mathrm { Na } ^ { + } + \mathrm { CO } ^ { 3 - }$, jadi A salah; B. Jumlah proton klorin 17, skema struktur atom: (+17) 287 / B salah; C. Dalam $\mathrm { H } _ { 2 } \mathrm {~S}$, S dan dua H dihubungkan oleh satu ikatan, dan rumus strukturnya adalah: $\mathrm { H } - \mathrm { S } - \mathrm { H }$, jadi C benar; D. Hidrogen peroksida adalah senyawa kovalen dengan rumus elektronik $\mathrm { H } : \ddot { \mathrm { O } } : \ddot { \mathrm { O } } : \mathrm { H }$, jadi D salah;

Question 26: 30. Istilah-istilah kimia berikut ini dinyatakan dengan benar sebagai berikut $\mathrm { S } ^ { ...

30. Istilah-istilah kimia berikut ini dinyatakan dengan benar sebagai berikut $\mathrm { S } ^ { 2 - }$ ![](/images/questions/chem-nomenclature/image-003.jpg) ${ } _ { 6 } ^ { 8 } \mathrm { C }$

  • A. A. Rumus elektronik untuk $\mathrm { H } _ { 2 } \mathrm { O }$ adalah $\mathrm { H } : \mathrm { O } : \mathrm { H }$.
  • B. B. $\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$ memiliki massa molar sebesar $98 \mathrm {~g} \cdot \mathrm {~mol} ^ { - 1 }$
  • C. C. Struktur + ditampilkan secara skematis
  • D. D. Atom karbon dengan nomor neutron 8 dapat dinyatakan sebagai

Answer: B

Solution:

Question 27: 31. Pernyataan-pernyataan berikut ini tidak benar

31. Pernyataan-pernyataan berikut ini tidak benar

  • A. A. Hanya ikatan kovalen yang ada dalam $\mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$
  • B. B. Menyinari larutan $\mathrm { KMnO } _ { 4 }$ dengan cahaya akan menciptakan jalur cahaya.
  • C. C. $\mathrm { SO } _ { 2 }$ adalah oksida asam
  • D. D. Partikel-partikel yang terdispersi menjadi lebih besar ketika air kapur yang telah dijernihkan menjadi keruh.

Answer: B

Solution:

Question 28: 32. Analisis fenomena eksperimental berikut ini tidak benar Ada $\mathrm { CO } _ { 2 }$ dalam ...

32. Analisis fenomena eksperimental berikut ini tidak benar Ada $\mathrm { CO } _ { 2 }$ dalam gas Ming.

  • A. A. Tembaga sulfat anhidrat berubah menjadi biru untuk membuktikan adanya $\mathrm { H } _ { 2 } \mathrm { O } ( \mathrm { g } )$ di dalam gas.
  • B. B. Perubahan warna larutan magenta membuktikan adanya $\mathrm { SO } _ { 2 }$ di dalam gas.
  • C. C. Pencerahan warna larutan kalium permanganat asam membuktikan $\mathrm { SO } _ { 2 }$ memiliki sifat pemutihan
  • D. D. Bukti keruh putih dalam air kapur yang telah dijernihkan

Answer: C

Solution:

Question 29: 33. Reaksi berikut ini diwakili dengan benar

33. Reaksi berikut ini diwakili dengan benar

  • A. A. Reaksi arang dengan asam sulfat pekat: $2 \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 } ($ pekat $) + \mathrm { C } \triangleq 2 \mathrm { SO } _ { 2 } \uparrow + \mathrm { CO } _ { 2 } \uparrow + 2 \mathrm { H } _ { 2 } \mathrm { O }$
  • B. B. Sulfur dioksida yang larut dalam air: $\mathrm { SO } _ { 2 } + \mathrm { H } _ { 2 } \mathrm { O } \rightleftharpoons \mathrm { H } _ { 2 } \mathrm { SO } _ { 4 }$
  • C. C. Larutan kalium permanganat asam berwarna cerah: $2 \mathrm { MnO } _ { 4 } ^ { - } + 5 \mathrm { SO } _ { 2 } + 4 \mathrm { OH } ^ { - } = 2 \mathrm { Mn } ^ { 2 + } + 5 \mathrm { SO } _ { 4 } ^ { 2 - } + 2 \mathrm { H } _ { 2 } \mathrm { O }$
  • D. D. Kekeruhan putih pada air kapur yang telah dijernihkan : $2 \mathrm { OH } ^ { - } + \mathrm { CO } _ { 2 } = \mathrm { CO } _ { 3 } ^ { 2 - } + \mathrm { H } _ { 2 } \mathrm { O }$

Answer: A

Solution:

Question 30: 34. Istilah-istilah kimia berikut ini tidak benar. $$ \mathrm { CH } _ { 2 } = \mathrm { CH } _ { ...

34. Istilah-istilah kimia berikut ini tidak benar. $$ \mathrm { CH } _ { 2 } = \mathrm { CH } _ { 2 } $$ $\mathrm { CH } _ { 2 } = \mathrm { CH } _ { 2 }$

  • A. A. Formula struktural etilena
  • B. B. (b) Representasi skematis dari struktur: +17) 288
  • C. C. Ikatan kovalen dalam molekul asetilena terdiri dari 3 ikatan $\sigma$ dan 2 ikatan $\pi$
  • D. D. Rumus elektronik untuk $\mathrm { CO } _ { 2 }$: $\because \cdots \cdots 0$.

Answer: D

Solution: A. Ikatan rangkap karbon-karbon tidak dapat dihilangkan, dan rumus struktur etilena adalah: $\mathrm { CH } _ { 2 } = \mathrm { CH } _ { 2 }$, jadi A benar; B. Cl diketahui sebagai unsur 17, ion klorin memiliki 18 elektron di luar inti, sehingga struktur ion klorin benar, B benar; C. Rumus struktur molekul asetilena adalah $\mathrm { H } - \mathrm { C } \equiv \mathrm { C } - \mathrm { H }$, sehingga ikatan kovalennya terdiri dari 3 ikatan $\sigma$ dan 2 ikatan $\pi$, jadi C benar; D. Rumus elektronik dari $\mathrm { CO } _ { 2 }$ adalah: $\because \therefore \mathrm { C } \because \because$, jadi D salah;

Question 31: 35. Istilah atau diagram kimia berikut ini dinyatakan dengan benar. ## $\stackrel { \mathrm { O } }...

35. Istilah atau diagram kimia berikut ini dinyatakan dengan benar. ## $\stackrel { \mathrm { O } } { : }$ <br> A. Rumus elektronik gugus aldehida: $\because \dot { \mathrm { C } } : \mathrm { H }$ <img class="imgSvg" id = "mi1mzfw0h1vqo48lf3f" src = "data:image/svg+xml;base64, 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 +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ 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  • A. A. $\stackrel { \mathrm { O } } { : }$ <br> A. Rumus elektronik gugus aldehida: $\because \dot { \mathrm { C } } : \mathrm { H }$
  • B. B. Rumus garis ikatan untuk 2-butilena:
  • C. C. Pemodelan bola-dan-tongkat cis-2-butena:
  • D. D. Model pengisian ruang untuk bromoetana:

Answer: D

Solution: A. Rumus elektronik gugus aldehida adalah $\stackrel { : \mathrm { O } : } { : \stackrel { : } { \mathrm { C } } : \mathrm { H } }$, jadi A salah; B. Molekul 2-butil dalam 4 atom karbon garis umum, rumus garis ikatan: Oleh karena itu, B salah; C.Dua gugus metil dari cis-2-butena berada di sisi yang sama, dan model bola-dan-tongkat: ![](/images/questions/chem-nomenclature/image-028.jpg) C. Kedua gugus metil cis-2-butena berada pada sisi yang sama, dan model kelelawarnya adalah: ![](/images/questions/chem-nomenclature/image-028.jpg) D. Atom bromin dalam etil bromida memiliki jari-jari terbesar, dan model pengisian ruangnya adalah: ![](/images/questions/chem-nomenclature/image-029.jpg) Jadi, D adalah benar; Pilihan jawaban D.

Question 32: 36. Pada tahun 2019, Lei Xiaoguang, seorang ahli kimia muda di Tiongkok, terpilih sebagai juru bicar...

36. Pada tahun 2019, Lei Xiaoguang, seorang ahli kimia muda di Tiongkok, terpilih sebagai juru bicara untuk nitrogen dalam Tabel Periodik Unsur untuk Ahli Kimia Muda. Pernyataan berikut ini yang terkait dengan unsur nitrogen adalah benar.

  • A. A. ${ } ^ { 14 } \mathrm {~N}$ & ${ } ^ { 14 } \mathrm { C } _ { \text {互为同位素 } }$
  • B. B. N dalam $\mathrm { Si } _ { 3 } \mathrm {~N} _ { 4 }$ adalah harga dari ${ } ^ { + 3 }$.
  • C. C. $\mathrm { NH } _ { 3 }$ lebih stabil secara termal daripada HF.
  • D. D. Rumus elektronik untuk $\mathrm { N } _ { 2 }$ adalah: $\mathrm { N } : : \mathrm { N }$:

Answer: D

Solution: A. Nuklida yang berbeda dari unsur yang sama adalah isotop satu sama lain, ${ } ^ { 14 } \mathrm {~N}$ dan ${ } ^ { 14 } \mathrm { C }$ adalah nuklida yang berbeda dari unsur yang berbeda, bukan isotop. A. Salah; B. Menurut prinsip penulisan rumus kimia dan kekuatan non-logam, N pada $\mathrm { Si } _ { 3 } \mathrm {~N} _ { 4 }$ adalah valensi ${ } ^ { - 3 }$, B salah; C. Semakin kuat non-logam suatu unsur, semakin kuat stabilitas termal hidrida yang sesuai. Non-logam nitrogen lebih kecil daripada fluor, sehingga stabilitas termal $\mathrm { NH } _ { 3 }$ lebih lemah daripada HF; D. Nitrogen membentuk ikatan rangkap tiga nitrogen-nitrogen dengan rumus elektronik $\mathrm { N } : : \mathrm { N } :$, D benar;

Question 33: 37. Pernyataan berikut ini adalah benar ![](/images/questions/chem-nomenclature/image-031.jpg)

37. Pernyataan berikut ini adalah benar ![](/images/questions/chem-nomenclature/image-031.jpg)

  • A. A. ${ } ^ { 14 } \mathrm {~N} _ { 2 }$ dan ${ } ^ { 15 } \mathrm {~N} _ { 2 }$ adalah isotop satu sama lain.
  • B. B. Model bola-dan-tongkat n-butana:
  • C. C. Rumus elektronik NaClO: $\mathrm { Na } : \ddot { \mathrm { O } } : \ddot { \mathrm { C } } \mathrm { l } :$
  • D. D. $\mathrm { NaHCO } _ { 3 }$ Persamaan ionik untuk hidrolisis: $\mathrm { HCO } ^ { 3 } + \mathrm { H } _ { 2 } \mathrm { O } ^ { \hat { \ddagger } } + \mathrm { H } _ { 3 } \mathrm { O } ^ { + } + \mathrm { CO } ^ { 3 - }$

Answer: B

Solution: A. ${ } ^ { 14 } \mathrm {~N} _ { 2 }$ dan ${ } ^ { 15 } \mathrm {~N} _ { 2 }$ keduanya merepresentasikan molekul, bukan atom, dan bukan isotop satu sama lain, jadi A salah; B. Molekul n-butana mengandung 2 gugus metil dan 2 gugus metilena. [IMAGE_7]] B. Molekul n-butana mengandung 2 gugus metil dan 2 gugus metilena, dan model kelelawarnya adalah ![](/images/questions/chem-nomenclature/image-032.jpg). adalah benar; C. NaClO adalah senyawa ionik yang terdiri dari ion $\mathrm { Na } ^ { + }$ dan ion $\mathrm { ClO } ^ { - }$, dan rumus elektronik NaClO adalah $\mathrm { Na } + [ \ddot { \mathrm { O } } : \ddot { \mathrm { C } } \mathrm { l } : ] ^ { - }$, jadi C salah; D. Persamaan ionik untuk hidrolisis $\mathrm { NaHCO } _ { 3 }$ adalah $\mathrm { HCO } ^ { 3 } + \mathrm { H } _ { 2 } \mathrm { O } \rightleftharpoons \mathrm { H } _ { 2 } \mathrm { CO } _ { 3 } + \mathrm { OH } ^ { - }$, jadi D salah;

Question 34: 38 . Rumus elektronik dari zat-zat berikut ini ditulis dengan benar $[ \mathrm { H } : \ddot { O ...

38 . Rumus elektronik dari zat-zat berikut ini ditulis dengan benar $[ \mathrm { H } : \ddot { O } : ] _ { 2 }$

  • A. A. $\mathrm { HCl } \quad \mathrm { H } ^ { + } [ : \ddot { C l } : ] ^ { - }$
  • B. B. $\mathrm { HClO } \quad \mathrm { H } : \ddot { \mathrm { Cl } } : \ddot { \mathrm { O } }$:
  • C. C. $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$
  • D. D. $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 } \mathrm { Na } ^ { + } [ : O : O : ] ^ { 2 - } \mathrm { Na } ^ { + }$

Answer: D

Solution: A. HCl adalah senyawa kovalen, dan rumus elektroniknya adalah $\mathrm { H } \dot { \times } \dot { \mathrm { Cl } } \dot { \cdot }$, jadi A salah; B. Elektron dari HClO B. Rumus elektronik HClO adalah $\left. H \cdot \dot { \sim } _ { \cdot } ^ { \cdot } \dot { C ^ { \prime } } \right] _ { \cdot } ^ { \cdot }$, jadi B salah. B. Rumus elektronik dari HClO adalah $\left. H \cdot \dot { \sim } _ { \cdot } ^ { \cdot } \dot { C ^ { \prime } } \right] _ { \cdot } ^ { \cdot }$, jadi B salah; C. Rumus elektronik dari $\mathrm { H } _ { 2 } \mathrm { O } _ { 2 }$ adalah ${ } _ { \mathrm { H } : \dddot { \mathrm { o } } _ { \cdot } : \ddot { \mathrm { o } } _ { \cdot \mathrm { H } } }$, jadi C salah. Rumus elektronik dari $\mathrm { Na } _ { 2 } \mathrm { O } _ { 2 }$ adalah $\mathrm { Na } ^ { + } \left[ \begin{array} { l l l } \because & 0 & 0 \end{array} \right] ^ { 2 - } \mathrm { Na } ^ { + }$, jadi D benar. Poin : Dalam proses penyelesaian soal-soal semacam ini, Anda harus fokus pada hal-hal berikut: (1) Berikan perhatian khusus pada aspek-aspek berikut saat menulis rumus elektronik: Anion dan kation polinuklear harus ditambahkan "[ ]" dan menunjukkan muatannya, menulis rumus elektronik senyawa kovalen, tidak boleh menggunakan "[ ]", tidak ada elektron valensi yang terikat juga tidak boleh ditulis. (2) Ketika menulis rumus struktur dan rumus sederhana, pertama, urutan ikatan antar atom harus jelas (misalnya HClO harus $\mathrm { H } - \mathrm { O } - \mathrm { Cl }$, bukan $\mathrm { H } - \mathrm { Cl } - \mathrm { O }$), dan kedua, ketika menulis rumus struktur, ikatan rangkap dua karbon-karbon dan ikatan rangkap tiga karbon-karbon harus dituliskan. (3) Model skala dan model bola-dan-tongkat harus mencerminkan ukuran relatif atom dan struktur spasial molekul.

Question 35: 41. Istilah kimia berikut ini dinyatakan dengan benar ( ). ᄃ. Rumus elektronik HCl: $\mathrm { H }...

41. Istilah kimia berikut ini dinyatakan dengan benar ( ). ᄃ. Rumus elektronik HCl: $\mathrm { H } ^ { + } [ : \ddot { \mathrm { C } } 1 : ] ^ { - }$ ![](/images/questions/chem-nomenclature/image-034.jpg)

  • A. A. Skema struktur atom Na: (+11) 281
  • B. B. $\mathrm { N } _ { 2 }$ Rumus struktur molekul: $\mathrm { N } = \mathrm { N }$
  • C. C. Rumus elektronik HCl: $\mathrm { H } ^ { + } [ : \ddot { \mathrm { C } } 1 : ] ^ { - }$
  • D. D. $\mathrm { H } _ { 2 } \mathrm { O }$ Model molekul kelelawar :

Answer: A

Solution: A. Nomor atom Na adalah 11, dan struktur atomnya adalah (+11)281/, A benar; B. Ada ikatan rangkap tiga dalam molekul $\mathrm { N } _ { 2 }$, dan rumus strukturalnya adalah $\mathrm { N } \equiv \mathrm { N } , \mathrm { B }$, yang salah; $\mathrm { C } . \mathrm { HCl }$ adalah senyawa kovalen, dan rumus elektroniknya adalah $\mathrm { H } \times \ddot { \mathrm { C } } \mathrm { l } :$, C salah; D. Model skala molekul $\mathrm { H } _ { 2 } \mathrm { O }$ adalah $\mathrm { H } \times \ddot { \mathrm { C } } \mathrm { l } :$, C salah; D. Model skala molekul $\mathrm { H } _ { 2 } \mathrm { O }$ adalah ![](/images/questions/chem-nomenclature/image-006.jpg) , D salah, dan jawabannya adalah A. 考点:考查化学用语判断

Question 36: 42. Istilah-istilah kimia berikut ini dipahami dengan benar (). <img class="imgSvg" id = "mi1mzfw1y...

42. Istilah-istilah kimia berikut ini dipahami dengan benar (). <img class="imgSvg" id = "mi1mzfw1yqxeudhajd" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16ZncxeXF4ZXVkaGFqZCIgeG1sbnM9Imh0dHA6Ly93d3cudzMub3JnLzIwMDAvc3ZnIiB2aWV3Qm94PSIwIDAgMTQ3IDk2LjU1OTYyODcxMzAxMTk5IiBzdHlsZT0id2lkdGg6IDE0Ni45OTk5OTk5OTk5NzQ2NXB4OyBoZWlnaHQ6IDk2LjU1OTYyODcxMzAxMTk5cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI +PGRlZnM+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZncxeXF4ZXVkaGFqZC0xIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9Ijg1LjkxNDYxNjQ2NzM1ODk3IiB5MT0iNjkuOTk2NjQ1Njk3MjQ3OCIgeDI9Ijk4LjUxNDYzNjA1NjU4MTM0IiB5Mj0iNDguMTcyODE2ODMxNzM0NDgiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ 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PHRleHQgeD0iNDIiIHk9IjQ4LjI3OTc4NjA4MTg5MTY2IiBjbGFzcz0iZGVidWciIGZpbGw9IiNmZjAwMDAiIHN0eWxlPSIKICAgICAgICAgICAgICAgIGZvbnQ6IDVweCBEcm9pZCBTYW5zLCBzYW5zLXNlcmlmOwogICAgICAgICAgICAiPjwvdGV4dD48dGV4dCB4PSI1Ny43NTAwMjQ0ODY1Mjc5NSIgeT0iMjEiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICI + PC90ZXh0Pjx0ZXh0IHg9Ijg5LjI1MDAyNDQ4NjUxNTI0IiB5PSIyMS4wMDAwMjgyNzQ2MTQzMjYiIGNsYXNzPSJkZWJ1ZyIgZmlsbD0iI2ZmMDAwMCIgc3R5bGU9IgogICAgICAgICAgICAgICAgICAgZm9udDogNXB4IERyb2lkIFNhbnMsIHNhbnMtc2VyaWY7CiAgICAgICAgICAgICAgICAgICI +PC90ZXh0PjwvZz48L3N2Zz4="/> ![](/images/questions/chem-nomenclature/image-035.jpg) ![](/images/questions/chem-nomenclature/image-036.jpg) Dapat mewakili molekul metana dan karbon tetraklorida

  • A. A. Rumus molekul benzena
  • B. B. Pemodelan bola-dan-tongkat dari molekul propana
  • C. C. Formula struktural etil format $\mathrm { CH } _ { 3 } \mathrm { COOCH } _ { 3 }$
  • D. D. Model pengisian ruang

Answer: B

Solution: A. Rumus molekul benzena adalah $\mathrm { C } _ { 6 } \mathrm { H } _ { 6 }$ dan rumus struktur benzena adalah <img class="imgSvg" id = "mi1mzfxe6x8ntm57kok" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16ZnhlNng4bnRtNTdrb2siIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDE0NyA5Ni41NTk2Mjg3MTMwMTE5OSIgc3R5bGU9IndpZHRoOiAxNDYuOTk5OTk5OTk5OTc0NjVweDsgaGVpZ2h0OiA5Ni41NTk2Mjg3MTMwMTE5OXB4OyBvdmVyZmxvdzogdmlzaWJsZTsiPjxkZWZzPjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4ZTZ4OG50bTU3a29rLTEiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iODUuOTE0NjE2NDY3MzU4OTciIHkxPSI2OS45OTY2NDU2OTcyNDc4IiB4Mj0iOTguNTE0NjM2MDU2NTgxMzQiIHkyPSI0OC4xNzI4MTY4MzE3MzQ0OCI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbXpmeGU2eDhudG01N2tvay0zIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9Ijg5LjI0OTk3NTUxMzQ0NjY3IiB5MT0iNzUuNTU5NjI4NzEzMDExOTkiIHgyPSIxMDQuOTk5OTk5OTk5OTc0NjMiIHkyPSI0OC4yNzk4NDI2MzExMjAzNDYiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnhlNng4bnRtNTdrb2stNSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI4OS4yNTAwMjQ0ODY1MTUyNCIgeTE9IjIxLjAwMDAyODI3NDYxNDMyNiIgeDI9IjEwNC45OTk5OTk5OTk5NzQ2MyIgeTI9IjQ4LjI3OTg0MjYzMTEyMDM0NiI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. 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Model bola-dan-tongkat dari molekul propana adalah ![](/images/questions/chem-nomenclature/image-037.jpg) , B adalah benar; C. Bentuk pendek struktural dari etil format adalah $\mathrm { HCOOCH } _ { 2 } \mathrm { CH } _ { 3 } \mathrm { CH } _ { 3 } \mathrm { COOH } _ { 3 } { } _ { \text {是乙酸甲酯 } }$, dan C salah; D. Jari-jari atom klorin lebih besar daripada atom karbon, dan model pengisian ruang tidak dapat mewakili molekul karbon tetraklorida. D. Model pengisian ruang tidak dapat mewakili molekul karbon tetraklorida; Jawabannya adalah B.

Question 37: 43. Istilah kimia berikut ini digunakan dengan benar (). <img class="imgSvg" id = "mi1mzfw29pijnkzf...

43. Istilah kimia berikut ini digunakan dengan benar (). <img class="imgSvg" id = "mi1mzfw29pijnkzfp7d" src = "data:image/svg+xml;base64, PHN2ZyBpZD0ic21pbGVzLW1pMW16ZncyOXBpam5remZwN2QiIHhtbG5zPSJodHRwOi8vd3d3LnczLm9yZy8yMDAwL3N2ZyIgdmlld0JveD0iMCAwIDIwOCAxNDUuMDg3NDExNzY1NzE2ODciIHN0eWxlPSJ3aWR0aDogMjA3LjkzNjIyODIzODA0NTY2cHg7IGhlaWdodDogMTQ1LjA4NzQxMTc2NTcxNjg3cHg7IG92ZXJmbG93OiB2aXNpYmxlOyI + PGRlZnM + PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZncyOXBpam5remZwN2QtMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSIxNTIuMzU3MjY2OTkyODkzNDEiIHkxPSIyMSIgeDI9IjE2NS45MzYyMjgyMzgwNDU2NiIgeTI9IjUyLjQ0MDgyMjI5MzY3NDMxNiI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. 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  • A. A. Awan elektron dari $\sigma$ ikatan antara atom karbon dalam molekul etilen:
  • B. B. Model molekul karbon dioksida yang mengisi ruang:
  • C. C. Struktur $\mathrm { H } ^ { - }$ secara skematis ditunjukkan sebagai (+1) $\frac { 1 } { 2 }$
  • D. D. Rumus elektroniknya adalah $: \overline { \mathrm { F } } : \overline { \mathrm { B } } : \overline { \mathrm { F } } :$ | $: \overline { \mathrm { F } } : \overline { \mathrm { B } } : \overline { \mathrm { F } } :$

Answer: C

Solution: A ... <img class="imgSvg" id = "mi1mzfxfqgefush9xq" src = "data:image/svg+xml;base64, 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 +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnhmcWdlZnVzaDl4cS0zIiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjExMi45MzE5MjA2ODY2NDU5IiB5MT0iMTIwLjE2MjM3MTcyNDY1MjA1IiB4Mj0iMTIyLjY2NTk0OTI4Njc2NzQ0IiB5Mj0iOTAuMjA0MDg5Mjc3MDIxMzUiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnhmcWdlZnVzaDl4cS01IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9IjkxLjg1NDI5ODM5Mzk5NzYxIiB5MT0iODMuNjU0ODc3OTMwNDQ4NjYiIHgyPSIxMjIuNjY1OTQ5Mjg2NzY3NDQiIHkyPSI5MC4yMDQwODkyNzcwMjEzNSI + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbXpmeGZxZ2VmdXNoOXhxLTciIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTIyLjY2NTk0OTI4Njc2NzQ0IiB5MT0iOTAuMjA0MDg5Mjc3MDIxMzUiIHgyPSIxMzguNDE1OTQzMTY1MTMzNDYiIHkyPSI2Mi45MjQyODU1MjM0ODU0MyI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbXpmeGZxZ2VmdXNoOXhxLTkiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iNjMuMDc3NjE5MzUzMzM2ODQiIHkxPSI5Ni40NjcwODg2NDQ4NzI0MiIgeDI9IjkxLjg1NDI5ODM5Mzk5NzYxIiB5Mj0iODMuNjU0ODc3OTMwNDQ4NjYiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnhmcWdlZnVzaDl4cS0xMSIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI5MS44NTQyOTgzOTM5OTc2MSIgeTE9IjgzLjY1NDg3NzkzMDQ0ODY2IiB4Mj0iOTUuMTQ2OTUyMDE2ODU5MzkiIHkyPSIxMTQuOTgyMzE2ODk1NjczIj48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4ZnFnZWZ1c2g5eHEtMTMiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iODguNTYxNjQ0NzcxMTM1ODMiIHkxPSI1Mi4zMjc0Mzg5NjUyMjQzMjQiIHgyPSI5MS44NTQyOTgzOTM5OTc2MSIgeTI9IjgzLjY1NDg3NzkzMDQ0ODY2Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4ZnFnZWZ1c2g5eHEtMTUiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iMTE3LjMzODMyMzgxMTc5NjYxIiB5MT0iMzkuNTE1MjI4MjUwODAwNTc1IiB4Mj0iMTM4LjQxNTk0MzE2NTEzMzQ2IiB5Mj0iNjIuOTI0Mjg1NTIzNDg1NDMiPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIyMCUiPjwvc3RvcD48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMTAwJSI +PC9zdG9wPjwvbGluZWFyR3JhZGllbnQ+ PGxpbmVhckdyYWRpZW50IGlkPSJsaW5lLW1pMW16ZnhmcWdlZnVzaDl4cS0xNyIgZ3JhZGllbnRVbml0cz0idXNlclNwYWNlT25Vc2UiIHgxPSI0MiIgeTE9IjczLjA1ODAzMTM3MjE4NzU2IiB4Mj0iNjMuMDc3NjE5MzUzMzMzM2ODQiIHkyPSI5Ni40NjcwODg2NDQ4NzI0MiI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+ PC9zdG9wPjxzdG9wIHN0b3AtY29sb3I9ImN1cnJlbnRDb2xvciIgb2Zmc2V0PSIxMDAlIj48L3N0b3A+ PC9saW5lYXJHcmFkaWVudD48bGluZWFyR3JhZGllbnQgaWQ9ImxpbmUtbWkxbXpmeGZxZ2VmdXNoOXhxLTE5IiBncmFkaWVudFVuaXRzPSJ1c2VyU3BhY2VPblVzZSIgeDE9Ijg4LjU2MTY0NDc3MTEzNTgzIiB5MT0iNTIuMzI3NDM4OTY1MjI0MzI0IiB4Mj0iMTE3LjMzODMyMzgxMTc5NjYxIiB5Mj0iMzkuNTE1MjI4MjUwODAwNTc1Ij48c3RvcCBzdG9wLWNvbG9yPSJjdXJyZW50Q29sb3IiIG9mZnNldD0iMjAlIj48L3N0b3A + PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjEwMCUiPjwvc3RvcD48L2xpbmVhckdyYWRpZW50PjxsaW5lYXJHcmFkaWVudCBpZD0ibGluZS1taTFtemZ4ZnFnZWZ1c2g5eHEtMjEiIGdyYWRpZW50VW5pdHM9InVzZXJTcGFjZU9uVXNlIiB4MT0iODUuMjY4OTkxMTQ4Mjc0MDciIHkxPSIyMC45OTk5OTk5OTk5OTk5OTMiIHgyPSI4OC41NjE2NDQ3NzExMzU4MyIgeTI9IjUyLjMyNzQzODk2NTIyNDMyNCI +PHN0b3Agc3RvcC1jb2xvcj0iY3VycmVudENvbG9yIiBvZmZzZXQ9IjIwJSI+. 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+PC90ZXh0PjwvZz48L3N2Zz4="/> Dalam diagram, awan elektron saling tumpang tindih "bahu-membahu" untuk membentuk ikatan $\pi$; B. Jari-jari atom karbon lebih besar daripada jari-jari atom oksigen; C. $\mathrm { H } ^ { - }$ memiliki dua elektron pada lapisan terluarnya, membentuk struktur stabil dua elektron; D. Pada $\mathrm { BF } _ { 3 }$, seharusnya terdapat 6 elektron di lapisan terluar atom B, dan tidak ada pasangan elektron tunggal pada B. Rumus elektronik $\mathrm { BF } _ { 3 }$ seharusnya $\mathrm { BF } _ { 3 }$. B tidak memiliki pasangan elektron tunggal; Jawabannya adalah C.
Kembali ke Topik

Chemical Nomenclature and Equations

化学用语与方程式

37 Soal Latihan

Berlatih dengan soal berbahasa Mandarin untuk mempersiapkan ujian CSCA. Anda dapat mengaktifkan/menonaktifkan terjemahan saat berlatih.

Ringkasan Topik

Istilah dan persamaan kimia adalah bahasa dasar kimia, terutama mencakup simbol unsur, rumus kimia, valensi, simbol ionik dan penulisan dan penyetaraan persamaan kimia. Dalam ujian, bagian materi ini sering muncul dalam bentuk soal pilihan ganda, dengan fokus pada penguasaan aturan dasar dan kemampuan menilai detailnya, seperti perhitungan valensi, rumus kimia analisis yang benar dan salah, pencocokan persamaan dan sebagainya.

Jumlah Soal:37

Poin Penting

  • 1Aturan perhitungan untuk simbol dan valensi elemen umum
  • 2Penulisan dan penamaan rumus kimia yang benar (molekuler, struktural, elektronik)
  • 3Pengenalan Simbol Ionik dan Skema Struktur Atom
  • 4Menyeimbangkan persamaan kimia dan menentukan jenis reaksi

Tips Belajar

Disarankan untuk menghafal senyawa unsur umum dengan mengkategorikannya dan berlatih menuliskannya dalam kaitannya dengan senyawa tertentu, dengan memperhatikan aturan penamaan untuk rumus kimia yang serupa.

Bisa soal satuan ≠ Lulus ujian

Ujian simulasi lengkap sesuai silabus resmi, gabungan topik seperti ujian asli

Dapatkan Ujian Simulasi →

Sumber Belajar Terkait

Panduan Kimia
Materi persiapan ujian lengkap
Glosarium Kimia
Tinjau istilah dan konsep terkait

Jelajahi latihan mata pelajaran lainnya:

Latihan MatematikaLatihan Fisika