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Jika dalam 1 gram oksigen terdapat 2a buah atom oksigen ,maka massa dari 10a buah atom belerang adalah ..... (Ar O = 16, S = 32)

Pertanyaan

Jika dalam 1 gram oksigen terdapat 2a buah atom oksigen ,maka massa dari 10a buah atom belerang adalah ..... (Ar O = 16, S = 32)space 

  1. ...space 

  2. ...space 

Pembahasan Soal:

fraction numerator Massa space O over denominator Massa space S end fraction equals fraction numerator italic n cross times italic M italic r over denominator italic n cross times italic M italic r end fraction fraction numerator Massa space O over denominator Massa space S end fraction equals fraction numerator begin display style fraction numerator jumlah space atom over denominator bil point space avogadro end fraction end style cross times italic M italic r over denominator begin display style fraction numerator jumlah space atom over denominator bil point space avogadro end fraction end style cross times italic M italic r end fraction fraction numerator 1 space g over denominator Massa space S end fraction equals fraction numerator begin display style fraction numerator 2 a over denominator 6 comma 02 cross times 10 to the power of 23 end fraction end style cross times 16 over denominator begin display style fraction numerator 10 a over denominator 6 comma 02 cross times 10 to the power of 23 end fraction end style cross times 32 end fraction fraction numerator 1 space g over denominator Massa space S end fraction equals fraction numerator begin display style 2 cross times 16 end style over denominator begin display style 10 cross times 32 end style end fraction fraction numerator 1 space g over denominator Massa space S end fraction equals fraction numerator begin display style 32 end style over denominator begin display style 320 end style end fraction space space Massa space S equals 320 over 32 equals 10 space g 

Jadi, massa belerangnya adalah 10 g.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Anisa

Mahasiswa/Alumni Universitas Negeri Semarang

Terakhir diupdate 02 Mei 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

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 partikel, mol, massa zat, dan   dari senyawa 2 mol !

Pembahasan Soal:

A. Mencari Jumlah Partikel 2 mol undefined:

          begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic n space H Cl end cell equals cell fraction numerator Jumlah space partikel space H Cl over denominator italic L end fraction end cell row cell Jumlah space partikel space H Cl end cell equals cell italic n space H Cl cross times italic L end cell row cell Jumlah space partikel space H Cl end cell equals cell 2 space mol cross times 6 comma 02 x 10 to the power of 23 space partikel space mol to the power of negative sign 1 end exponent end cell row cell Jumlah space partikel space H Cl end cell equals cell 1 comma 204 cross times 10 to the power of 24 space partikel end cell end table end style  

Jadi jumlah partikel HCl adalah Error converting from MathML to accessible text..
 

B. Mol undefined sudah diketahui dari soal yaitu sebesar 2 mol.
 

C. Mencari begin mathsize 14px style italic M subscript italic r bold space H Cl end style:

     begin mathsize 14px style M subscript r space H Cl equals left parenthesis 1 cross times A subscript r space H right parenthesis plus left parenthesis 1 cross times A subscript r space Cl right parenthesis M subscript r space H Cl equals left parenthesis 1 cross times 1 right parenthesis plus left parenthesis 1 cross times 35 comma 5 right parenthesis M subscript r space H Cl equals 1 plus 35 comma 5 M subscript r space H Cl equals 36 comma 5 end style 

     Jadi begin mathsize 14px style italic M subscript italic r bold space H Cl end style adalah 36,5.
 

D. Massa undefined 

    begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic n space H Cl end cell equals cell fraction numerator italic g over denominator M subscript r space H Cl end fraction end cell row italic g equals cell italic n space H Cl cross times M subscript r space H Cl end cell row italic g equals cell 2 space mol cross times 36 comma 5 space gram space mol to the power of negative sign 1 end exponent end cell row italic g equals cell 73 space gram end cell row blank blank blank end table end style 

   Jadi massa 2 mol undefined adalah 73 gram.undefined 

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

Berapa mol dan berapa atom masing-masing unsur yang terdapat dalam 10,3 g ?

Pembahasan Soal:

- Menghitung jumlah mol masing-masing unsur

mol space Na subscript 2 C O subscript 3 equals fraction numerator massa over denominator M subscript r space Na subscript 2 C O subscript 3 end fraction mol space Na subscript 2 C O subscript 3 equals fraction numerator 10 comma 3 space g over denominator 106 end fraction mol space Na subscript 2 C O subscript 3 equals 0 comma 097 space mol

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Na end cell equals cell 2 cross times mol space Na subscript 2 C O subscript 3 end cell row cell mol space Na end cell equals cell 2 cross times 0 comma 097 end cell row cell bold mol bold space Na end cell bold equals cell bold 0 bold comma bold 194 bold space bold mol end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space C end cell equals cell mol space Na subscript 2 C O subscript 3 end cell row cell bold mol bold space C end cell bold equals cell bold 0 bold comma bold 097 bold space bold mol end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell mol space O end cell equals cell 3 cross times mol space Na subscript 2 C O subscript 3 end cell row cell mol space O end cell equals cell 3 cross times 0 comma 097 end cell row cell bold mol bold space O end cell bold equals cell bold 0 bold comma bold 291 bold space bold mol end cell end table
 

- Menghitung jumlah atom masing-masing unsur

begin inline style sum with blank below atom space Na double bond mol cross times 6 comma 02 cross times 10 to the power of 23 sum with blank below atom space Na equals 0 comma 194 cross times 6 comma 02 cross times 10 to the power of 23 sum with blank below atom space Na equals bold 1 bold comma bold 17 begin bold style cross times end style begin bold style 10 to the power of 23 space atom end style end style

table attributes columnalign right center left columnspacing 0px end attributes row cell sum with blank below atom space C end cell equals cell mol cross times 6 comma 02 cross times 10 to the power of 23 end cell row cell begin inline style sum with blank below end style atom space C end cell equals cell begin inline style 0 comma 097 end style begin inline style cross times end style begin inline style 6 end style begin inline style comma end style begin inline style 02 end style begin inline style cross times end style begin inline style 10 to the power of 23 end style end cell row cell begin inline style sum with blank below end style atom space C end cell equals cell bold 5 bold comma bold 84 bold cross times bold 10 to the power of bold 22 bold space bold atom end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell begin inline style sum with blank below end style atom space O end cell equals cell mol cross times 6 comma 02 cross times 10 to the power of 23 end cell row cell begin inline style sum with blank below end style atom space O end cell equals cell 0 comma 291 cross times 6 comma 02 cross times 10 to the power of 23 end cell row cell begin inline style sum with blank below end style atom space O end cell equals cell bold 1 bold comma bold 75 bold cross times bold 10 to the power of bold 23 bold space bold atom end cell end table


Jadi, jumlah mol dan atom masing-masing unsur dalam 10,3 g Na subscript bold 2 C O subscript bold 3 seperti pada hasil di atas.space

0

Roboguru

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