Pertanyaan

Tentukan jumlah molekul dari 10 gram CaCO 3 ! ( A r C = 12 g mol − 1 , O = 16 g mol − 1 , Ca = 40 g mol − 1 )

Tentukan jumlah molekul dari 10 gram $CaCO_{3}$ !

$(A_{r} C = 12 g mol^{- 1}, O = 16 g mol^{- 1}, Ca = 40 g mol^{- 1})$ $\;$

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L. Avicenna

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

Jawaban terverifikasi

Jawaban

jumlah molekulnya adalah .

jumlah molekulnya adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold 6 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold comma end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold 02 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold cross times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold 10 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank to the power of bold 23 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank bold partikel end table end style

Pembahasan

Jadi jumlah molekulnya adalah .

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell mol space Ca C O subscript 3 end cell equals cell fraction numerator massa space Ca C O subscript 3 over denominator M subscript r space Ca C O subscript 3 end fraction end cell row cell mol space Ca C O subscript 3 end cell equals cell fraction numerator massa space Ca C O subscript 3 over denominator left parenthesis 1 cross times A subscript r space Ca right parenthesis plus left parenthesis 1 cross times A subscript r space C right parenthesis plus left parenthesis 3 cross times A subscript r space O right parenthesis end fraction end cell row cell mol space Ca C O subscript 3 end cell equals cell fraction numerator 10 space g over denominator left parenthesis 1 cross times 40 space g space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 1 cross times 12 space g space mol to the power of negative sign 1 end exponent right parenthesis plus left parenthesis 3 cross times 16 space g space mol to the power of negative sign 1 end exponent right parenthesis end fraction end cell row cell mol space Ca C O subscript 3 end cell equals cell fraction numerator 10 space g over denominator 40 space g space mol to the power of negative sign 1 end exponent plus 12 space g space mol to the power of negative sign 1 end exponent plus 48 space g space mol to the power of negative sign 1 end exponent end fraction end cell row cell mol space Ca C O subscript 3 end cell equals cell fraction numerator 10 space g over denominator 100 space g space mol to the power of negative sign 1 end exponent end fraction end cell row cell mol space Ca C O subscript 3 end cell equals cell 0 comma 1 space mol end cell row blank blank blank row cell Jumlah space partikel space Ca C O subscript 3 end cell equals cell mol space Ca C O subscript 3 cross times 6 comma 02 cross times 10 to the power of 23 end cell row cell Jumlah space partikel space Ca C O subscript 3 end cell equals cell 0 comma 1 space mol cross times 6 comma 02 cross times 10 to the power of 23 end cell row cell Jumlah space partikel space Ca C O subscript 3 end cell equals cell 6 comma 02 cross times 10 to the power of 22 space partikel end cell end table end style$

Jadi jumlah molekulnya adalah . space