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Hitung jumlah yang tertera diminta berikut : jumlah ion  dalam 1,51 kg

Pertanyaan

Hitung jumlah yang tertera diminta berikut :

jumlah ion begin mathsize 14px style Li to the power of plus sign end style dalam 1,51 kg begin mathsize 14px style Li subscript 2 S end style undefined

Pembahasan Soal:

Penentuan jumlah partikel ion begin mathsize 14px style Li to the power of plus sign end style dapat dihitung dengan menggunakan rumus hubungan mol (n), massa molar (begin mathsize 14px style M subscript m end style) dan bilangan Avogadro (L).

  • Hitung massa molar senyawa begin mathsize 14px style Li subscript 2 S end style. Massa molar begin mathsize 14px style Li subscript 2 S end style sama dengan massa atom relatif dalam satuan begin mathsize 14px style g space mol to the power of negative sign 1 end exponent end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M subscript m space Li subscript 2 S end cell equals cell M subscript r space Li subscript 2 S end cell row cell M subscript m space Li subscript 2 S end cell equals cell 2 cross times A subscript r space Li plus 1 cross times A subscript r space S end cell row cell M subscript m space Li subscript 2 S end cell equals cell 2 cross times 7 plus 1 cross times 32 end cell row cell M subscript m space Li subscript 2 S end cell equals cell 14 plus 32 end cell row cell M subscript m space Li subscript 2 S end cell equals cell 46 space g space mol to the power of negative sign 1 end exponent end cell end table end style     

  • Hitung mol begin mathsize 14px style Li subscript 2 S end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell massa space Li subscript 2 S end cell equals cell n cross times M subscript m end cell row cell italic n italic space Li subscript 2 S end cell equals cell fraction numerator massa space Li subscript 2 S over denominator M subscript m end fraction end cell row cell italic n italic space Li subscript 2 S end cell equals cell fraction numerator 1 comma 51 space kg cross times 1000 space g forward slash kg over denominator 46 space g space mol to the power of negative sign 1 end exponent end fraction end cell row cell italic n italic space Li subscript 2 S end cell equals cell 32 comma 83 space mol end cell end table end style   

  • Hitung mol ion begin mathsize 14px style Li to the power of plus sign end style berdasarkan reaksi ionisasi begin mathsize 14px style Li subscript 2 S end style.

begin mathsize 14px style Li subscript 2 S yields 2 Li to the power of plus sign and S to the power of 2 minus sign  italic n space Li to the power of plus sign equals fraction numerator koefisien space Li to the power of plus sign over denominator koefisien space Li subscript 2 S end fraction cross times italic n italic space Li subscript italic 2 S italic n space Li to the power of plus sign equals 2 over 1 cross times 32 comma 83 space mol italic n space Li to the power of plus sign equals 65 comma 66 space mol end style 

  • Hitung jumlah partikel begin mathsize 14px style Li to the power of plus sign end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Jumlah space Partikel space Li to the power of plus sign end cell equals cell n cross times L end cell row cell Jumlah space Partikel space Li blank to the power of plus end cell equals cell 65 comma 66 space mol cross times 6 comma 02 cross times 10 to the power of 23 space ion space mol to the power of negative sign 1 end exponent end cell row cell Jumlah space Partikel space Li blank to the power of plus end cell equals cell 3 comma 95 cross times 10 to the power of 25 space ion end cell end table end style  

Jadi, jumlah partikel begin mathsize 14px style Li to the power of plus sign end style adalah begin mathsize 14px style 3 comma 95 cross times 10 to the power of 25 space ion end style.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 14 Maret 2021

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