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Lengkapi tabel berikut! (Ar N=14, H=1, S=32, O=16, C=12)

Pertanyaan

Lengkapi tabel berikut! (Ar N=14, H=1, S=32, O=16, C=12)

Pembahasan Soal:

  • Molekul CH4 

menentukan Mr CH4 :

M subscript r space space C H subscript 4 space equals space A subscript r space C plus left parenthesis 4 cross times A subscript r space H right parenthesis space space space space space space space space space space space equals space 12 plus left parenthesis 4 cross times 1 right parenthesis space space space space space space space space space space space equals space 12 plus 4 space space space space space space space space space space space equals space 16 

menentukan mol CH4 :

n space equals space fraction numerator V over denominator 22 comma 4 end fraction n space equals space fraction numerator 5 comma 6 over denominator 22 comma 4 end fraction n space equals 0 comma 25 space mol 

menentukan massa CH4 :

massa space equals space n space cross times space M subscript r massa space equals space 0 comma 25 cross times 16 massa space equals 4 space gram 

menentukan JP CH4 :

JP space equals space n space cross times bilangan space avogadro JP space equals space 0 comma 25 space cross times space left parenthesis 6 comma 02 cross times 10 to the power of 23 right parenthesis JP space equals 1 comma 505 cross times 10 to the power of 23 

  • Molekul C4H10 

menentukan Mr C4H10 :

M subscript r space space C subscript 4 H subscript 10 space equals left parenthesis 4 cross times space A subscript r space C right parenthesis plus left parenthesis 10 cross times A subscript r space H right parenthesis space space space space space space space space space space space equals space left parenthesis 4 cross times 12 right parenthesis plus left parenthesis 10 cross times 1 right parenthesis space space space space space space space space space space space equals space 48 plus 10 space space space space space space space space space space space equals 58 

menentukan mol C4H10 :

n space equals space massa over M subscript r n space equals space fraction numerator 1 comma 16 over denominator 58 end fraction n space equals 0 comma 02 space mol  

menentukan JP C4H10 :

JP space equals space n space cross times bilangan space avogadro JP space equals space 0 comma 02 space cross times space left parenthesis 6 comma 02 cross times 10 to the power of 23 right parenthesis JP space equals 1 comma 204 cross times 10 to the power of 22 

menentukan volume C4H10 :

V space equals space n cross times 22 comma 4 space space space equals 0 comma 02 cross times 22 comma 4 space space space space equals 0 comma 448 space L 

Jadi, tabel lengkap sebagai berikut :

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Nurul

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Terakhir diupdate 05 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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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

0

Roboguru

HITUNGLAH jumlah molekul dari: 11 gram

Pembahasan Soal:

Untuk mengetahui jumlah molekul dari massa, dapat menggunakan konsep hubungan mol dengan massa dan massa atom relatif, selanjutnya konsep hubungan mol dengan bilangan avogadro.

Penentuan mol 11 g undefined dengan konsep hubungan mol dengan massa dan Mr:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Ar space C end cell equals cell 12 space g space mol to the power of negative sign 1 end exponent end cell row cell Ar space O end cell equals cell 16 space g space mol to the power of negative sign 1 end exponent end cell row cell Mr space C O subscript 2 end cell equals cell Ar space C plus 2 left parenthesis Ar space O right parenthesis end cell row blank equals cell 12 plus 2 left parenthesis 16 right parenthesis end cell row blank equals cell 44 space g space mol to the power of negative sign 1 end exponent end cell row n equals cell fraction numerator massa space over denominator Mr end fraction end cell row blank equals cell fraction numerator 11 space g over denominator 44 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 0 comma 25 space mol end cell row blank blank blank end table end style


Penentuan jumlah partikel 0,25 mol undefined dengan konsep hubungan mol dengan bilangan avogadro:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell bil point space Avogadro end cell equals cell 6 comma 02 cross times 10 to the power of 23 space molekul end cell row n equals cell fraction numerator jumlah space partikel over denominator bil point space Avogadro end fraction end cell row cell jumlah space partikel end cell equals cell n cross times bil point space Avogadro end cell row blank equals cell 0 comma 25 space mol cross times left parenthesis 6 comma 02 cross times 10 to the power of 23 space molekul right parenthesis end cell row blank equals cell 1 comma 505 cross times 10 to the power of 23 space molekul end cell end table end style 


Jadi, jumlah partikel 11 g undefined adalah begin mathsize 14px style 1 comma 505 cross times 10 to the power of 23 end style molekul.

0

Roboguru

Lengkapi tabel berikut! (Ar N=14, H=1, S=32, O=16, C=12)

Pembahasan Soal:

  • molekul NH3 

menentukan Mr NH3 :

M subscript r space N H subscript 3 equals space A subscript r space N plus left parenthesis 3 cross times A subscript r space H right parenthesis space space space space space space space space space equals space 14 plus left parenthesis 3 cross times 1 right parenthesis space space space space space space space space space equals space 17  

menentukan mol NH3 :

n space equals massa over M subscript r n space equals fraction numerator 3 comma 4 over denominator 17 end fraction n space equals space 0 comma 2 space mol

menentukan jumlah partikel NH3:

JP space equals space n space cross times space bilangan space avogadro space space space space equals space 0 comma 2 space cross times left parenthesis 6 comma 02 cross times 10 to the power of 23 right parenthesis space space space space equals 1 comma 204 space cross times space 10 to the power of 23

menentukan volume NH3:

V space equals space n space cross times space 22 comma 4 space space space equals 0 comma 2 cross times 22 comma 4 space space space equals 4 comma 48 space L 

  • molekul CO2

menentukan MCO2 :

M subscript r space C O subscript 2 equals space A subscript r space C plus left parenthesis 2 cross times A subscript r space O right parenthesis space space space space space space space space space equals space 12 plus left parenthesis 2 cross times 16 right parenthesis space space space space space space space space space equals space 12 plus 32 space space space space space space space space space equals 44  

menentukan mol CO2 :

n space equals fraction numerator V over denominator 22 comma 4 end fraction n space equals fraction numerator 11 comma 2 over denominator 22 comma 4 end fraction n space equals space 0 comma 5 space mol 

menentukan massa CO2:

massa space equals space n space cross times M subscript r massa space equals space 0 comma 5 space cross times 44 massa space equals 22 space gram 

menentukan jumlah partikel CO2:

JP space equals space n space cross times space bilangan space avogadro space space space space equals 0 comma 5 space cross times left parenthesis 6 comma 02 cross times 10 to the power of 23 right parenthesis space space space space equals 3 comma 01 cross times 10 to the power of 23  

  •  molekul H2SO4 :

menentukan Mr H2SO4 :

M subscript r space H subscript 2 S O subscript 4 equals left parenthesis 2 cross times A subscript r space H right parenthesis plus A subscript r space S plus left parenthesis 4 cross times A subscript r space O right parenthesis space space space space space space space space space space space space space equals left parenthesis 2 cross times 1 right parenthesis plus 32 plus left parenthesis 4 cross times 16 right parenthesis space space space space space space space space space space space space space equals 2 plus 32 plus 64 space space space space space space space space space space space space space equals 98   

menentukan mol H2SO4:

n space equals fraction numerator JP over denominator bilangan space avogadro end fraction n space equals 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 n space equals space 0 comma 5 space mol 

menentukan massa H2SO4 :

massa space equals n space cross times space M subscript r space massa space equals 0 comma 5 space cross times space 98 massa space equals 49 space gram space 

menentukan volume H2SO4 :

V space equals space n space cross times space 22 comma 4 space space space equals space 0 comma 5 space cross times space 22 comma 4 space space space equals 11 comma 2 space L 

Jadi, tabel lengkap seperti berikut :

0

Roboguru

Hitung jumlah yang tertera diminta berikut : massa dalam kg, dari  atom Zn

Pembahasan Soal:

Penentuan massa (m) dapat dihitung dengan menggunakan rumus hubungan mol (n), jumlah partikel dan massa molar (begin mathsize 14px style M subscript m end style).

  • Hitung mol Zn berdasarkan data jumlah partikel.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space Partikel space Zn end cell equals cell n cross times L end cell row cell italic n space Zn end cell equals cell fraction numerator Jumlah space Partikel space Zn over denominator L end fraction end cell row cell italic n space Zn end cell equals cell fraction numerator 6 comma 15 cross times 10 to the power of 27 space atom over denominator 6 comma 02 cross times 10 to the power of 23 space atom space mol to the power of negative sign 1 end exponent end fraction end cell row cell italic n space Zn end cell equals cell 10216 space mol end cell end table end style 

  • Hitung massa molar Zn. Zn adalah atom, maka massa molarnya sama dengan massa atom relatif dan menggunakan satuan begin mathsize 14px style g space mol to the power of negative sign 1 end exponent end style. (begin mathsize 14px style A subscript r space Zn equals 65 end style)

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript m space Zn end cell equals cell A subscript r space Zn end cell row cell M subscript m space Zn end cell equals cell 65 space g space mol to the power of negative sign 1 end exponent end cell end table end style  

  • Hitung massa Zn.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell massa space Zn end cell equals cell n cross times M subscript m space Zn end cell row cell massa space Zn end cell equals cell 10216 space mol cross times 65 space g space mol to the power of negative sign 1 end exponent end cell row cell massa space Zn end cell equals cell 664.037 space g end cell row cell massa space Zn end cell equals cell 664 comma 037 space kg end cell end table end style     


Jadi, massa Zn adalah 664, 037 kg.

0

Roboguru

HITUNGLAH jumlah molekul dari 0,9 gram

Pembahasan Soal:

Untuk mengetahui jumlah molekul dari massa, dapat menggunakan konsep hubungan mol dengan massa dan massa atom relatif, selanjutnya konsep hubungan mol dengan bilangan avogadro.

Penentuan mol 0,9 g begin mathsize 14px style H subscript 2 O end style dengan konsep hubungan mol dengan massa dan Mr:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Ar space H end cell equals cell 1 space g space mol to the power of negative sign 1 end exponent end cell row cell Ar space O end cell equals cell 16 space g space mol to the power of negative sign 1 end exponent end cell row cell Mr space H subscript 2 O end cell equals cell 2 left parenthesis Ar space H right parenthesis plus Ar space O end cell row blank equals cell 2 left parenthesis 1 right parenthesis plus 18 end cell row blank equals cell 18 space g space mol to the power of negative sign 1 end exponent end cell row n equals cell massa over Mr end cell row blank equals cell fraction numerator 0 comma 9 space g over denominator 18 space g space mol to the power of negative sign 1 end exponent end fraction end cell row blank equals cell 0 comma 05 space mol end cell end table end style


Penentuan jumlah partikel 0,05 mol undefined dengan konsep hubungan mol dengan bilangan avogadro:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell bil space Avogadro end cell equals cell 6 comma 02 cross times 10 to the power of 23 space molekul end cell row n equals cell fraction numerator jumlah space partikel over denominator bil space avogadro end fraction end cell row cell jumlah space partikel end cell equals cell n cross times bilangan space avogadro end cell row blank equals cell 0 comma 05 space mol cross times left parenthesis 6 comma 02 cross times 10 to the power of 23 space molekul right parenthesis end cell row blank equals cell 3 comma 01 cross times 10 to the power of 22 space molekul end cell end table end style 


Jadi, jumlah partikel 0,9 gram undefined adalah begin mathsize 14px style 3 comma 01 cross times 10 to the power of 22 end style molekul. 

0

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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