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Jika A r ​ : H = 1 g mol − 1 , O = 16 g mol − 1 , Ca = 40 g mol − 1 , Mg = 24 g mol − 1 , S = 32 g mol − 1 , Cl = 35 , 5 g mol − 1 , senyawa H 2 ​ O 2 ​ , Ca ( ClO ) 2 ​ , dan MgSO 4 ​ ⋅ 2 H 2 ​ O memiliki massa molekul berturut-turut ....

Jika , ,, , ,, senyawa , , dan  memiliki massa molekul berturut-turut ....space 

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A. Chandra

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

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yang tepat berturut-turut adalah , , dan .

 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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Menentukan nilai : Jadi, yang tepat berturut-turut adalah , , dan .

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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