Roboguru

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 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 :

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Nurul

Mahasiswa/Alumni UIN Syarif Hidayatullah Jakarta

Terakhir diupdate 05 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

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 MASSA DARI:  molekul

Pembahasan Soal:

Konsep yang digunakan adalah hubungan mol dengan jumlah partikel dan mol dengan massa dan massa atom relatif.

Penentuan mol undefined:

bil.Avogadron====6,022×1023molekulbil.Avogadrojumlahpartikel6,02×1023molekul6,02×1023molekul1mol

Penentuan massa undefined:

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

Oleh karena itu, massa 6,02×1023 molekul  H2O adalah 18 gram.

Jadi, massa H2O adalah 18 gram. 

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

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

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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