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Jika begin mathsize 14px style italic A subscript r end stylebegin mathsize 14px style H space equals space 1 space g space mol to the power of negative sign 1 end exponent end style, begin mathsize 14px style O space equals space 16 space g space mol to the power of negative sign 1 end exponent end style,begin mathsize 14px style Ca space equals space 40 space g space mol to the power of negative sign 1 end exponent end style, begin mathsize 14px style Mg space equals space 24 space g space mol to the power of negative sign 1 end exponent end style, begin mathsize 14px style S space equals space 32 space g space mol to the power of negative sign 1 end exponent end style,begin mathsize 14px style Cl space equals space 35 comma 5 space g space mol to the power of negative sign 1 end exponent end style, senyawa begin mathsize 14px style H subscript 2 O subscript 2 end style, begin mathsize 14px style Ca open parentheses Cl O close parentheses subscript 2 end style, dan begin mathsize 14px style Mg S O subscript 4 middle dot 2 H subscript 2 O end style memiliki massa molekul berturut-turut ....space 

A. Chandra

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

 begin mathsize 14px style italic M subscript italic r end style yang tepat berturut-turut adalah begin mathsize 14px style bold 34 bold space italic g bold space bold mol to the power of bold minus sign bold 1 end exponent end stylebegin mathsize 14px style bold 143 bold space italic g bold space bold mol to the power of bold minus sign bold 1 end exponent end style, dan begin mathsize 14px style bold 156 bold space italic g bold space bold mol to the power of bold minus sign bold 1 end exponent end style.

Pembahasan

Menentukan nilai begin mathsize 14px style italic M subscript r end style:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space H subscript 2 O subscript 2 end cell equals cell left parenthesis 2 cross times italic A subscript r space H right parenthesis plus left parenthesis 2 cross times italic A subscript r space O right parenthesis end cell row blank equals cell left parenthesis 2 cross times 1 right parenthesis plus left parenthesis 2 cross times 16 right parenthesis end cell row blank equals cell 2 plus 32 end cell row blank equals cell 34 space g space mol to the power of negative sign 1 end exponent end cell end table end style 

Error converting from MathML to accessible text. 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell italic M subscript r space Mg S O subscript 4 middle dot 2 H subscript 2 O end cell equals cell left parenthesis 1 cross times italic A subscript r space Mg right parenthesis plus left parenthesis 1 cross times italic A subscript r space S right parenthesis plus left parenthesis 6 cross times italic A subscript r space O right parenthesis plus left parenthesis 4 cross times italic A subscript r space H right parenthesis end cell row blank equals cell left parenthesis 1 cross times 24 right parenthesis plus left parenthesis 1 cross times 32 right parenthesis plus left parenthesis 6 cross times 16 right parenthesis plus left parenthesis 4 cross times 1 right parenthesis end cell row blank equals cell 24 plus 32 plus 96 plus 4 end cell row blank equals cell 156 space g space mol to the power of negative sign 1 end exponent end cell end table end style 

Jadi, begin mathsize 14px style italic M subscript italic r end style yang tepat berturut-turut adalah begin mathsize 14px style bold 34 bold space italic g bold space bold mol to the power of bold minus sign bold 1 end exponent end stylebegin mathsize 14px style bold 143 bold space italic g bold space bold mol to the power of bold minus sign bold 1 end exponent end style, dan begin mathsize 14px style bold 156 bold space italic g bold space bold mol to the power of bold minus sign bold 1 end exponent end style.

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Hitunglah Mr​CaBr2​, bila Ar​Ca=40,078 dan Ar​Br=79,904!

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