Roboguru

Volume 6,4 gram gas oksigen  pada keadaan Kamar (RTP) adalah ....()

Pertanyaan

Volume 6,4 gram gas oksigen O subscript 2 pada keadaan Kamar (RTP) adalah ....(Ar space O equals 16 comma space V subscript m equals 24 space L forward slash mol)

  1. 2,4 L

  2. 4,48 Lspace 

  3. 4,8 Lspace 

  4. 22,4 Lspace 

  5. 44,8 Lspace 

Pembahasan Soal:

n space O subscript 2 equals massa over Mr space space space space space space space equals fraction numerator 6 comma 4 over denominator 32 end fraction space space space space space space space equals 0 comma 2 space mol  V space O subscript 2 space open parentheses RTP close parentheses double bond n cross times Vm space space space space space space space space space space space space space space space space space space equals 0 comma 2 space mol cross times 24 space L forward slash mol space space space space space space space space space space space space space space space space space space equals 4 comma 8 space L 

Jadi, jawaban yang benar adalah C. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Anisa

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 01 Mei 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

0

Roboguru

Jika diketahui massa gas nitrogen  sebanyak 2,8 gram , maka tentukan volume gas nitrogen pada RTP!

Pembahasan Soal:

Menentukan mol N subscript bold 2

mol space N subscript 2 equals fraction numerator massa space N subscript 2 over denominator M subscript r space N subscript 2 end fraction mol space N subscript 2 equals fraction numerator massa space N subscript 2 over denominator 2 cross times A subscript r space N end fraction mol space N subscript 2 equals fraction numerator 2 comma 8 space gram over denominator 2 cross times 14 space gram space mol to the power of negative sign 1 end exponent end fraction mol space N subscript 2 equals fraction numerator 2 comma 8 space gram over denominator 28 space gram space mol to the power of negative sign 1 end exponent end fraction mol space N subscript 2 equals 0 comma 1 space mol 
 

Menentukan volume gas nitrogen pada keadaan RTP

V space N subscript 2 double bond mol space N subscript 2 cross times V subscript RTP V space N subscript 2 equals 0 comma 1 space mol cross times 24 space L space mol to the power of negative sign 1 end exponent V space N subscript 2 equals 2 comma 4 space L 

Jadi volume gas nitrogen pada keadaan RTP adalah 2,4 L. 

0

Roboguru

Tentukan volume dari 0,6 mol gas hidrogen yang diukur pada: b. keadaan kamar (RTP)

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript H end cell equals cell n cross times RTP end cell row blank equals cell 0 comma 6 cross times 24 comma 4 end cell row blank equals cell 14 comma 64 space Liter end cell end table 

Jadi, volume hidrogen yang benar adalah 14,64 Liter.

0

Roboguru

Diketahui massa atom relatif . Berapakah volume dari 17,6 gram yang diukur pada suhu dan tekanan tertentu jika volume 0,2 mol gas  adalah 5 liter?

Pembahasan Soal:

Menghitung mol begin mathsize 14px style C O subscript 2 end style
 

begin mathsize 14px style mol space C O subscript 2 equals massa over M subscript r mol space C O subscript 2 equals fraction numerator 17 comma 6 space g over denominator 44 space g forward slash mol end fraction mol space C O subscript 2 equals 0 comma 4 space mol end style


Menghitung volume begin mathsize 14px style C O subscript 2 end style


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator V space C O subscript 2 over denominator mol space C O subscript 2 end fraction end cell equals cell fraction numerator V space C O over denominator mol space C O end fraction end cell row cell fraction numerator V space C O subscript 2 over denominator 0 comma 4 end fraction end cell equals cell fraction numerator 5 over denominator 0 comma 2 end fraction end cell row cell V space C O subscript 2 end cell equals cell fraction numerator 5 cross times 0 comma 4 over denominator 0 comma 2 end fraction end cell row cell V space C O subscript 2 end cell equals cell 10 space L end cell end table end style


Jadi, volume gas karbondioksida pada keadaan tersebut adalah 10 liter.space

0

Roboguru

Salinlah tabel di bawah ini, kemudian isilah titik-titik yang tersedia pada poin D.

Pembahasan Soal:

Penentuan tekanan dapat ditentukan dengan persamaan:
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell P V end cell equals nRT row cell p cross times 0 comma 8 space L end cell equals cell open parentheses fraction numerator 0 comma 8 space g over denominator 32 space g forward slash mol end fraction close parentheses cross times 0 comma 082 cross times 273 end cell row P equals cell 0 comma 7 space atm end cell end table end style 
 

jumlah partikel dapat ditentukan dengan persamaan:
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space partikel end cell equals cell n cross times 6 comma 02.10 to the power of 23 end cell row blank equals cell open parentheses fraction numerator 0 comma 8 space g over denominator 32 end fraction close parentheses cross times 6 comma 02.10 to the power of 23 end cell row blank equals cell 1 comma 505 cross times 10 to the power of 22 end cell end table end style 
 

Jadi dapat disimpulkan jawaban yang tepat adalah tekanan = 0,7 atm dan jumlah partikel = 1,505.1022.space

0

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