Roboguru

Volume 0,02 mol gas oksigen yang benar pada kondisi RTP (25*C, 1 atm) adalah ....

Pertanyaan

Volume 0,02 mol gas oksigen yang benar pada kondisi RTP (25*C, 1 atm) adalah ....

  1. 0,448 Lspace 

  2. 4,48 L space 

  3. 0,48 L space 

  4. 0,048 L space 

  5. 4,8 Lspace 

Pembahasan Soal:

V equals fraction numerator n cross times R cross times T over denominator P end fraction V equals fraction numerator 0 comma 02 cross times 0 comma 082 cross times 298 over denominator 1 end fraction equals 0 comma 48872 space L

Dengan demikian, maka jawaban yang tepat adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Budi

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 06 Juni 2021

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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

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. 

Roboguru

Berapakah volume dari 2,5 mol gas nitrogen dioksida pada keadaan kamar (RTP)?

Pembahasan Soal:

Volume 2,5 mol gas nitrogen dioksida pada keadaan RTP:

V space N O subscript 2 double bond n cross times 24 comma 4 space L space space space space space space space space space space equals 2 comma 5 cross times 24 comma 4 space L space space space space space space space space space space equals 61 space L 

Jadi, volume gas nitrogen dioksida tersebut adalah 61 L.

Roboguru

Alkohol () terbakar menurut persamaan reaksi berikut.    Hitung volume oksigen (RTP) yang diperlukan untuk membakar sempurna 9,2 gram alkohol. ( H = 1, C = 12, dan O = 16)

Pembahasan Soal:

Menentukan nilai mol O subscript 2:

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript C subscript 2 H subscript 5 O H end subscript end cell equals cell m over Mr end cell row blank equals cell fraction numerator 9 comma 2 over denominator left parenthesis left parenthesis 2 cross times 12 right parenthesis plus left parenthesis 6 cross times 1 right parenthesis plus left parenthesis 1 cross times 16 right parenthesis right parenthesis end fraction end cell row blank equals cell fraction numerator 9 comma 2 over denominator 46 end fraction end cell row blank equals cell 0 comma 2 space mol end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell n subscript O subscript 2 end subscript end cell equals cell fraction numerator Koef space O subscript 2 over denominator Koef space C subscript 2 H subscript 5 O H end fraction cross times n subscript C subscript 2 H subscript 5 O H end subscript end cell row blank equals cell 3 over 1 cross times 0 comma 2 end cell row blank equals cell 0 comma 6 space mol end cell end table 

Sehingga volume oksigen (RTP) yang diperlukan:

table attributes columnalign right center left columnspacing 0px end attributes row cell V subscript O 2 end subscript end cell equals cell n cross times 24 comma 4 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 oksigen yang diperlukan adalah 14,64 liter.

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.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

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