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Jumlah molekul yang terdapat dalam 15 gr senyawa  (Ar C = 12, H = 1) adalah ....

Pertanyaan

Jumlah molekul yang terdapat dalam 15 gr senyawa begin mathsize 14px style C subscript 2 H subscript 6 end style (Ar C = 12, H = 1) adalah ....

  1. begin mathsize 14px style 2 cross times 10 to the power of 22 end style 

  2. begin mathsize 14px style 5 cross times 10 to the power of 22 end style 

  3. begin mathsize 14px style 3 comma 01 cross times 10 to the power of 23 end style 

  4. begin mathsize 14px style 6 comma 02 cross times 10 to the power of 23 end style 

  5. begin mathsize 14px style 1 comma 0 cross times 10 to the power of 24 end style 

Pembahasan Soal:

Untuk menghitung jumlah molekul dari senyawanya, terlebih dahulu mencari jumlah mol dengan menggunakan konsep hubungan mol (n) dengan massa dan massa molekul relatif begin mathsize 14px style open parentheses M subscript r close parentheses end style.

Penentuan mol:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row massa equals cell 15 space g end cell row cell Mr space C subscript 2 H subscript 6 end cell equals cell 2 left parenthesis Ar space C right parenthesis plus 6 left parenthesis Ar space H right parenthesis end cell row blank equals cell 2 left parenthesis 12 right parenthesis plus 6 left parenthesis 1 right parenthesis end cell row blank equals cell 24 plus 6 end cell row blank equals 30 row n equals cell massa over M subscript r end cell row blank equals cell 15 over 30 end cell row blank equals cell 0 comma 5 space mol end cell end table end style  


Penentuan jumlah molekul:


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell bilangan 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 molekul over denominator bilangan space Avogadro end fraction end cell row cell 0 comma 5 space mol end cell equals cell fraction numerator jumlah space molekul over denominator 6 comma 02 cross times 10 to the power of 23 space molekul end fraction end cell row cell jumlah space molekul end cell equals cell 0 comma 5 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 23 space molekul end cell end table end style


Jadi, jawaban yang benar adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Juliana

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 01 April 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

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

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.

0

Roboguru

Lengkapi tabel dibawah ini, bila diketahui Ar P = 31, Ca = 40, O = 16, Zn = 65, N = 14, Mg = 24, K = 39, Cr = 52, Mn = 55. Ditulis pembahasan perhitungannya.

Pembahasan Soal:

Tabel di atas dapat dilengkapi dengan konversi antar satuan dengan jumlah mol.

begin mathsize 14px style Ca subscript 3 open parentheses P O subscript 4 close parentheses subscript 2 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row Mr equals cell ΣAr space Ca subscript 3 open parentheses P O subscript 4 close parentheses subscript 2 equals 3 cross times Ca plus 2 cross times P plus 2 cross times 4 cross times O end cell row blank equals cell 3 cross times 40 plus 2 cross times 31 plus 2 cross times 4 cross times 16 end cell row blank equals cell 120 plus 62 plus 128 end cell row blank equals 310 end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row Massa equals cell n cross times Mr end cell row blank equals cell 0 comma 4 cross times 310 end cell row blank equals cell 124 space g end cell end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row X equals cell n cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 0 comma 4 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 2 comma 408 cross times 10 to the power of 23 end cell end table end style 

V tidak dapat dicari, karena keadaan (T dan p) tidak diketahui.


begin mathsize 14px style Zn open parentheses N O subscript 3 close parentheses subscript 2 end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row Mr equals cell ΣAr space Zn open parentheses N O subscript 3 close parentheses subscript 2 equals Zn plus 2 cross times N plus 2 cross times 3 cross times O end cell row blank equals cell 65 plus 2 cross times 14 plus 2 cross times 3 cross times 16 end cell row blank equals cell 65 plus 28 plus 96 end cell row blank equals 189 end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell massa over M subscript r end cell row blank equals cell fraction numerator 60 space g over denominator 189 space g forward slash mol end fraction end cell row blank equals cell 0 comma 32 space mol end cell end table end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row X equals cell n cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 0 comma 32 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 1 comma 9264 cross times 10 to the power of 23 end cell end table end style 

V tidak dapat dicari, karena keadaan (T dan p) tidak diketahui.


begin mathsize 14px style Mg subscript 3 open parentheses P O subscript 4 close parentheses subscript 2 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row Mr equals cell ΣAr space Mg subscript 3 open parentheses P O subscript 4 close parentheses subscript 2 equals 3 cross times Mg plus 2 cross times P plus 2 cross times 4 cross times O end cell row blank equals cell 3 cross times 24 plus 2 cross times 31 plus 2 cross times 4 cross times 16 end cell row blank equals cell 72 plus 62 plus 128 end cell row blank equals 262 end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row Massa equals cell n cross times Mr end cell row blank equals cell 0 comma 2 cross times 262 end cell row blank equals cell 52 comma 4 space g end cell end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row X equals cell n cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 0 comma 2 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 1 comma 204 cross times 10 to the power of 23 end cell end table end style 

V tidak dapat dicari, karena keadaan (T dan p) tidak diketahui.


begin mathsize 14px style K subscript 2 Cr subscript 2 O subscript 7 end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row Mr equals cell ΣAr space K subscript 2 Cr subscript 2 O subscript 7 end cell row blank equals cell 2 cross times K plus 2 cross times Cr plus 7 cross times O end cell row blank equals cell 3 cross times 39 plus 2 cross times 52 plus 7 cross times 16 end cell row blank equals cell 117 plus 104 plus 112 end cell row blank equals 333 end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row Massa equals cell n cross times Mr end cell row blank equals cell 0 comma 5 cross times 333 end cell row blank equals cell 166 comma 5 space g end cell end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row X equals cell n cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 0 comma 5 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 3 comma 01 cross times 10 to the power of 23 end cell end table end style 

V tidak dapat dicari, karena keadaan (T dan p) tidak diketahui.


begin mathsize 14px style K Mn O subscript 4 end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row Mr equals cell ΣAr space K Mn O subscript 4 end cell row blank equals cell K and Mn plus 4 cross times O end cell row blank equals cell 39 plus 55 plus 4 cross times 16 end cell row blank equals cell 39 plus 55 plus 64 end cell row blank equals 158 end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row n equals cell massa over M subscript r end cell row blank equals cell fraction numerator 25 space g over denominator 158 space g forward slash mol end fraction end cell row blank equals cell 0 comma 16 space mol end cell end table end style 

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row X equals cell n cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 0 comma 16 cross times 6 comma 02 cross times 10 to the power of 23 end cell row blank equals cell 0 comma 9632 cross times 10 to the power of 23 end cell row blank equals cell 9 comma 632 cross times 10 to the power of 23 end cell end table end style 

V tidak dapat dicari, karena keadaan (T dan p) tidak diketahui.

Maka tabel lengkapnya adalah sebagai berikut.


Jadi, data lengkap senyawa-senyawa tersebut adalah seperti pada tabel di atas.
 

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

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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