Roboguru

Jumlah atom H yang terdapat dalam 16 gram hidrazin () adalah ... ( H = 1 dan N = 14)

Pertanyaan

Jumlah atom H yang terdapat dalam 16 gram hidrazin (N subscript 2 H subscript 4) adalah ... (italic A subscript r H = 1 dan N = 14)

  1. 16 space 

  2. 64 space 

  3. 3 comma 01 cross times 10 to the power of 23 

  4. 1 comma 204 cross times 10 to the power of 23 

  5. 1 comma 204 cross times 10 to the power of 24 

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript H end cell equals cell fraction numerator 4 cross times Ar space H over denominator italic M subscript r italic space N subscript 2 H subscript 4 end fraction cross times m subscript N subscript 2 H subscript 4 end subscript end cell row blank equals cell fraction numerator 4 cross times 1 over denominator left parenthesis left parenthesis 2 cross times 14 right parenthesis plus left parenthesis 4 cross times 1 right parenthesis right parenthesis end fraction cross times 16 end cell row blank equals cell 4 over 32 cross times 16 end cell row blank equals cell 2 space gram end cell end table  

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript H end cell equals cell m over italic A subscript r end cell row blank equals cell 2 over 1 end cell row blank equals cell 2 space mol end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space atom space H end cell equals cell n cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 2 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 12 comma 04 cross times 10 to the power of 23 end cell row blank equals cell 1 comma 204 cross times 10 to the power of 24 space atom end cell end table 

Jadi, jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Chandra

Mahasiswa/Alumni Universitas Negeri Jakarta

Terakhir diupdate 30 April 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Lengkapi tabel berikut ini!

Pembahasan Soal:

Langkah 1: Melengkapi tabel NO

1. Mol NO (n)

    table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript NO end cell equals cell fraction numerator jumlah space partikel subscript NO over denominator L end fraction end cell row blank equals cell fraction numerator 6 comma 02 cross times 10 to the power of 23 space over denominator 6 comma 022 cross times 10 to the power of 23 end fraction end cell row blank equals cell 1 space mol end cell end table


2. Mr NO 

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript NO end cell equals cell Ar subscript N and Ar subscript O end cell row blank equals cell 14 plus 16 end cell row blank equals cell 30 space begin inline style bevelled gram over mol end style end cell end table


3. Massa NO (m)

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript NO end cell equals cell n subscript NO cross times Mr subscript NO end cell row blank equals cell 1 space mol space cross times 30 space begin inline style bevelled gram over mol end style end cell row blank equals cell 30 space gram end cell end table


4. Volume NO (STP)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript NO end cell equals cell n subscript NO cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 1 space mol space cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 22 comma 4 space L end cell end table


5. Volume NO (27bold degree bold space C, 1 atm)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript NO end cell equals cell n subscript NO cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 1 space mol cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 24 space L end cell end table


Langkah 2: Melangkapi tabel H subscript bold 2 

1. Jumlah partikel H subscript bold 2 (x)

    table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times L end cell row blank equals cell 0 comma 1 space mol cross times 6 comma 022 cross times 10 to the power of 23 space partikel end cell row blank equals cell 6 comma 022 cross times 10 to the power of 22 space partikel end cell row blank blank blank end table

2. Mr H subscript bold 2

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript H subscript 2 end subscript end cell equals cell 2 cross times Ar subscript H end cell row blank equals cell 2 cross times 1 space begin inline style bevelled gram over mol end style end cell row blank equals cell 2 space begin inline style bevelled gram over mol end style end cell end table
 

3. Massa H subscript bold 2

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times Mr subscript H subscript 2 end subscript end cell row blank equals cell 0 comma 1 space mol space cross times 2 space begin inline style bevelled gram over mol end style end cell row blank equals cell 0 comma 2 space gram end cell end table


4. Volume H subscript bold 2 (STP)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 1 space mol space cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 2 comma 24 space L end cell end table


5. Volume H subscript bold 2 (27bold degree bold space C, 1 atm)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript H subscript 2 end subscript end cell equals cell n subscript H subscript 2 end subscript cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 1 space mol cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 2 comma 4 space L end cell end table


Langkah 3: Melangkapi tabel N H subscript bold 3

1. mol N H subscript bold 3 (n)

    table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript N H subscript 3 end subscript end cell equals cell fraction numerator V subscript STP subscript N H subscript 3 end subscript end subscript over denominator 22 comma 4 space bevelled L over mol end fraction end cell row blank equals cell fraction numerator 4 comma 48 space L over denominator 22 comma 4 space bevelled L over mol end fraction end cell row blank equals cell 0 comma 2 space mol end cell end table
 

2. Jumlah partikel N H subscript bold 3 (x)
     table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel subscript N H subscript 3 end subscript end cell equals cell n subscript N H subscript 3 end subscript cross times L end cell row blank equals cell 0 comma 2 space mol cross times 6 comma 022 cross times 10 to the power of 23 space partikel end cell row blank equals cell 1 comma 2 cross times 10 to the power of 23 space partikel end cell end table


3. Mr N H subscript bold 3

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript N H subscript 3 end subscript end cell equals cell Ar subscript N plus left parenthesis 3 cross times Ar subscript H right parenthesis end cell row blank equals cell 14 space begin inline style bevelled gram over mol end style plus left parenthesis 3 cross times 1 space begin inline style bevelled gram over mol right parenthesis end style end cell row blank equals cell 17 space begin inline style bevelled gram over mol end style end cell end table
 

4. Massa N H subscript bold 3

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript N H subscript 3 end subscript end cell equals cell n subscript N H subscript 3 end subscript cross times Mr subscript N H subscript 3 end subscript end cell row blank equals cell 0 comma 2 space mol space cross times 17 space begin inline style bevelled gram over mol end style end cell row blank equals cell 3 comma 4 space gram end cell end table


5. Volume N H subscript bold 3 (27bold degree bold space C, 1 atm)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript N H subscript 3 end subscript end cell equals cell n subscript N H subscript 3 end subscript cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 2 space mol cross times 24 space begin inline style bevelled L over mol end style end cell row blank equals cell 4 comma 8 space L end cell end table


Langkah 4: Melangkapi tabel C H subscript 4

1. mol C H subscript 4 (n)

    table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C H subscript 4 end subscript end cell equals cell fraction numerator V subscript RTP subscript C H subscript 4 end subscript end subscript over denominator 24 space bevelled L over mol end fraction end cell row blank equals cell fraction numerator 12 comma 3 space L over denominator 24 space bevelled L over mol end fraction end cell row blank equals cell 0 comma 5125 space mol end cell end table 
 

2. Jumlah partikel C H subscript 4 (x)
     table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel subscript C H subscript 4 end subscript end cell equals cell n subscript C H subscript 4 end subscript cross times L end cell row blank equals cell 0 comma 512 space mol cross times 6 comma 022 cross times 10 to the power of 23 space partikel end cell row blank equals cell 3 comma 08 cross times 10 to the power of 23 space partikel end cell end table


3. Mr C H subscript 4

    table attributes columnalign right center left columnspacing 0px end attributes row cell Mr subscript C H subscript 4 end subscript end cell equals cell Ar subscript C plus left parenthesis 4 cross times Ar subscript H right parenthesis end cell row blank equals cell 12 space begin inline style bevelled gram over mol end style plus left parenthesis 4 cross times 1 space begin inline style bevelled gram over mol right parenthesis end style end cell row blank equals cell 16 space begin inline style bevelled gram over mol end style end cell end table
 

4. Massa C H subscript 4

    table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript C H subscript 4 end subscript end cell equals cell n subscript C H subscript 4 end subscript cross times Mr subscript C H subscript 4 end subscript end cell row blank equals cell 0 comma 512 space mol space cross times 16 space begin inline style bevelled gram over mol end style end cell row blank equals cell 8 comma 2 space gram end cell end table 


5. Volume C H subscript 4 (STP)

    table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript C H subscript 4 end subscript end cell equals cell n subscript C H subscript 4 end subscript cross times 22 comma 4 space begin inline style bevelled L over mol end style end cell row blank equals cell 0 comma 512 space mol cross times 22.4 space begin inline style bevelled L over mol end style end cell row blank equals cell 11 comma 48 space L end cell end table  


Dengan demikian, tabel lengkapnya adalah

Roboguru

Massa atom relatif () H = 1 dan O = 16, tetapan Avogadro = , jumlah molekul air yang terdapat dalam 180 gram  adalah ...

Pembahasan Soal:

Menentukan nilai begin mathsize 14px style italic M subscript r space H subscript 2 O end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space H subscript 2 O end cell equals cell left parenthesis 2 cross times italic A subscript r space H right parenthesis plus left parenthesis 1 cross times italic A subscript r space O right parenthesis end cell row blank equals cell left parenthesis 2 cross times 1 right parenthesis plus left parenthesis 1 cross times 16 right parenthesis end cell row blank equals cell 18 space g space mol to the power of negative sign 1 end exponent end cell end table end style 

Menentukan nilai mol begin mathsize 14px style H subscript 2 O end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript H subscript 2 O end subscript end cell equals cell m over italic M subscript r end cell row blank equals cell 180 over 18 end cell row blank equals cell 10 space mol end cell end table end style 

Menentukan jumlah molekul begin mathsize 14px style H subscript 2 O end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space molekul space H subscript 2 O end cell equals cell n cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 10 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 6 comma 02 cross times 10 to the power of 24 space molekul end cell end table end style 

Jadi, jawaban yang benar adalah C.

Roboguru

Kerjakan soal konsep mol berikut:

Pembahasan Soal:

Diketahui:

italic M subscript r space C O subscript 2 equals 44 space g space mol to the power of negative sign 1 end exponent 

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C O subscript 2 end subscript end cell equals cell fraction numerator jumlah space partikel over denominator 6 comma 02 cross times 10 to the power of 23 end fraction end cell row blank equals cell fraction numerator 3 comma 01 cross times 10 to the power of 23 over denominator 6 comma 02 cross times 10 to the power of 23 end fraction end cell row blank equals cell 0 comma 5 space mol end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell m subscript C O subscript 2 end subscript end cell equals cell n cross times italic M subscript r end cell row blank equals cell 0 comma 5 cross times 44 end cell row blank equals cell 22 space gram end cell end table 

Jadi, jawaban yang benar adalah 22 gram.

Roboguru

Kalium superoksida () digunakan dalam sistem penyangga hidup karena menyerap  dan membebaskan oksigen menurut reaksi berikut.    (belum setara)  a. Berapa mol oksigen dibebaskan jika 110 gram   dire...

Pembahasan Soal:

Persamaan reaksi setara:

4 K O subscript 2 open parentheses italic s close parentheses space plus space 2 C O subscript 2 open parentheses italic g close parentheses space rightwards arrow space 2 K subscript 2 C O subscript 3 open parentheses italic s close parentheses space plus space 3 O subscript 2 open parentheses italic g close parentheses 

a. Mol oksigen dibebaskan

table attributes columnalign right center left columnspacing 0px end attributes row cell italic n subscript O subscript bold 2 end subscript end cell equals cell fraction numerator Koef space O subscript 2 over denominator Koef space C O subscript 2 end fraction cross times n subscript C O subscript 2 end subscript end cell row blank equals cell 3 over 2 cross times 110 over 44 end cell row blank equals cell bold 3 bold comma bold 75 bold space bold mol end cell end table 

b. Jumlah molekul oksigen

table attributes columnalign right center left columnspacing 0px end attributes row cell bold Jumlah bold space bold Molekul bold space O subscript bold 2 end cell equals cell n cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 3 comma 75 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell bold 22 bold comma bold 6 bold cross times bold 10 to the power of bold 23 bold space bold molekul end cell end table 

Jadi, jawaban yang benar sesuai penjelasan diatas.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved