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Berapakah volume dari 88 gram gas karbondioksida CO2​?​ (Diketahui: Ar​C=12;O=16)

Pertanyaan

Berapakah volume dari 88 gram gas karbondioksida C O subscript 2?blank subscript blank

(Diketahui: A subscript r space C equals 12 semicolon space O equals 16)

Pembahasan Soal:

Volume 88 gram gas C O subscript 2 diukur pada kondisi standar (STP). Penentuan volume dapat dihitung berdasrkan hubungan mol (n) dan volume molar (V subscript m).

table attributes columnalign right center left columnspacing 0px end attributes row cell n space C O subscript 2 end cell equals cell m over M subscript r end cell row blank equals cell 88 over 44 end cell row blank equals cell 2 space mol end cell row blank blank blank row V equals cell n cross times V subscript m end cell row blank equals cell 2 space mol cross times 22 comma 4 space L space mol to the power of negative sign 1 end exponent end cell row blank equals cell 44 comma 8 space L end cell end table 

Jadi, volume dari 88 gram gas C O subscript 2 adalah 44,8 L.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. MT

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Hitunglah massa dari :  2,24 liter CO2​ pada STP  (Ar C=12, Ar O=16)

Pembahasan Soal:

Untuk menghitung massa begin mathsize 14px style C O subscript 2 end style, jika volume gas diketahui pada keadaan STP dengan menggunakan rumus konsep mol.

Hubungan mol, massa zat dengan volume gas (STP)

begin mathsize 14px style mol equals fraction numerator massa space zat over denominator Mr end fraction equals fraction numerator V subscript open parentheses STP close parentheses end subscript over denominator 22 comma 4 end fraction Mr space C O subscript 2 equals 1 cross times Ar space C plus 2 cross times Ar space O space space space space space space space space space space space space equals 1 cross times 12 plus 2 cross times 16 equals 44 fraction numerator massa space C O subscript 2 over denominator 44 end fraction equals fraction numerator 2 comma 24 over denominator 22 comma 4 end fraction massa space C O subscript 2 equals 4 comma 4 space gram end style 

Maka, massa karbondioksida adalah 4,4 gram. 

0

Roboguru

Diketahui: Ar​:C=12gmol−1,N=14gmol−1,O=16gmol−1,S=32gmol−1,H=1gmol−1,Fe=56gmol−1 Tentukan massa dari 4,48 liter gas dinitrogen pentaoksida (STP)!

Pembahasan Soal:

Menentukan mol N subscript bold 2 O subscript bold 5

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space N subscript 2 O subscript 5 end cell equals cell fraction numerator V space N subscript 2 O subscript 5 over denominator V subscript STP end fraction end cell row cell mol space N subscript 2 O subscript 5 end cell equals cell fraction numerator 4 comma 48 space L over denominator 22 comma 4 space L end fraction end cell row cell mol space N subscript 2 O subscript 5 end cell equals cell 0 comma 2 space mol end cell row blank blank blank end table 
 

Menentukan massa N subscript bold 2 O subscript bold 5 

mol space N subscript 2 O subscript 5 equals fraction numerator italic g space N subscript 2 O subscript 5 over denominator M subscript r space N subscript 2 O subscript 5 end fraction italic g space N subscript 2 O subscript 5 double bond mol space N subscript 2 O subscript 5 cross times M subscript r space N subscript 2 O subscript 5 italic g space N subscript 2 O subscript 5 double bond mol space N subscript 2 O subscript 5 cross times left curly bracket left parenthesis 2 cross times A subscript r space N right parenthesis plus left parenthesis 5 cross times A subscript r space O right parenthesis right curly bracket italic g space N subscript 2 O subscript 5 equals 0 comma 2 space mol cross times left curly bracket left parenthesis 2 cross times 14 space g space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 5 cross times 16 space g space mol to the power of negative sign 1 end exponent right parenthesis right curly bracket italic g space N subscript 2 O subscript 5 equals 21 comma 6 space g 

Jadi massa N subscript bold 2 O subscript bold 5 adalah 21,6 gram.space 

1

Roboguru

Diketahui massa atom relatif (Ar​) dari beberapa tom di bawah ini: H=1, O=16, N=14, C=12, S=32, Na=23, K=39, Ca=40, Cl=35,5  Tentukan berapa volume dari 22 gram CO2​ jika diukur pada keadaan standar...

Pembahasan Soal:

Menentukan mol C O subscript bold 2

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space C O subscript 2 end cell equals cell fraction numerator massa space C O subscript 2 over denominator M subscript r space C O subscript 2 end fraction end cell row cell mol space C O subscript 2 end cell equals cell fraction numerator massa space C O subscript 2 over denominator left curly bracket left parenthesis 1 cross times A subscript r space C right parenthesis plus left parenthesis 2 cross times A subscript r space O subscript 2 right parenthesis right curly bracket end fraction end cell row cell mol space C O subscript 2 end cell equals cell fraction numerator 22 space gram over denominator left curly bracket left parenthesis 1 cross times 12 space gram space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 2 cross times A subscript r space 16 space gram space mol to the power of negative sign 1 end exponent right parenthesis right curly bracket end fraction end cell row cell mol space C O subscript 2 end cell equals cell fraction numerator 22 space gram over denominator left parenthesis 12 space gram space mol to the power of negative sign 1 end exponent plus 32 space gram space mol to the power of negative sign 1 end exponent right parenthesis end fraction end cell row cell mol space C O subscript 2 end cell equals cell fraction numerator 22 space gram over denominator 44 space gram space mol to the power of negative sign 1 end exponent end fraction end cell row cell mol space C O subscript 2 end cell equals cell 0 comma 5 space mol end cell end table 
 

Menentukan Volume C O subscript bold 2

table attributes columnalign right center left columnspacing 0px end attributes row blank blank V end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell C O end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript 2 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank mol end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell C O end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript 2 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank V end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript STP end cell end table V table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell C O end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript 2 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 0 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank mol end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 22 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank L end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank mol end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank to the power of negative sign 1 end exponent end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank V end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell C O end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank subscript 2 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 11 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank L end table  

Jadi volume dari 22 gram C O subscript bold 2 jika diukur pada keadaan standar adalah 11,2 L.space 

0

Roboguru

Berapakah massa 28 liter gas NH3​ jika diukur pada suhu 0∘C dan tekanan 1 atm? (Ar​H=1;N=14)

Pembahasan Soal:

Penentuan massa gas N H subscript 3 dapat dihitung berdasarkan hubungan mol (n) dengan volume molar (V subscript m) dan  massa molar (M subscript m) gas.

  • Hitung mol gas N H subscript 3.

table attributes columnalign right center left columnspacing 0px end attributes row V equals cell n cross times V subscript m end cell row cell 28 space L end cell equals cell n cross times 22 comma 4 space L space mol to the power of negative sign 1 end exponent end cell row n equals cell fraction numerator 28 space L over denominator 22 comma 4 space L space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 1 comma 25 space mol end cell end table 

  • Hitung massa gas N H subscript 3 dengan massa molar, M subscript m equals space 17 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 m space N H subscript 3 end cell equals cell n cross times M subscript m end cell row blank equals cell 1 comma 25 space mol cross times 17 space g space mol to the power of negative sign 1 end exponent end cell row blank equals cell 21 comma 25 space g end cell end table 

Jadi, massa gas N H subscript 3 adalah 21,25 g.

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

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