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Pertanyaan

Jika hasil kali kelarutan MgCO 3 ​ sebesar 3 , 5 x 1 0 − 8 , massa garam yang terlarut dalam 250 mL air sebanyak ... ( A r ​ Mg = 24 g mol − 1 , C = 12 g mol − 1 , dan O = 16 g mol − 1 )

Jika hasil kali kelarutan  sebesar  , massa garam yang terlarut dalam 250 mL air sebanyak ...

  

  1. 5,55 mgspace 

  2. 4,67 mgspace 

  3. 3,92 mgspace 

  4. 1,57 mgspace 

  5. 1,87 mgspace 

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Y. Rochmawatie

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

jawaban yang benar adalah C.

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Pembahasan

Pembahasan
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Jadi, jawaban yang benar adalah C.

 begin mathsize 14px style Mg C O subscript 3 open parentheses italic s close parentheses equilibrium Mg to the power of 2 plus sign left parenthesis italic a italic q right parenthesis space plus space 2 space C O subscript 3 to the power of 2 minus sign end exponent left parenthesis italic a italic q right parenthesis italic space italic space italic space italic s italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic s italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic 2 italic s K subscript italic s italic p end subscript double bond open square brackets Mg to the power of 2 plus sign close square brackets space left square bracket C O subscript 3 to the power of 2 minus sign end exponent right square bracket space K subscript italic s italic p end subscript double bond open parentheses s close parentheses open parentheses s close parentheses K subscript italic s italic p end subscript equals s squared s squared space equals space 3 comma 5 space cross times space 10 to the power of negative sign 8 end exponent space mol forward slash L space space space s equals square root of 3 comma 5 cross times space 10 to the power of negative sign 8 end exponent space mol forward slash L end root space space space space space italic s bold equals bold 1 bold comma bold 87 bold cross times bold space bold 10 to the power of bold minus sign bold 4 end exponent bold space bold mol bold forward slash italic L bold space  Kelarutan space Mg C O subscript 3 space adalah space 1 comma 87 cross times 10 to the power of negative sign 4 end exponent space mol forward slash L italic M subscript italic r italic space Mg C O subscript 3 equals 84 space g forward slash mol M space equals fraction numerator n space open parentheses mol close parentheses over denominator V space open parentheses L close parentheses end fraction n space equals space M cross times V n equals space 1 comma 87 cross times 10 to the power of negative sign 4 end exponent space mol forward slash L space cross times space 2 comma 5 cross times 10 to the power of negative sign 1 end exponent space L n equals space 4 comma 675 cross times 10 to the power of negative sign 5 end exponent space mol  massa equals space n cross times Mr massa equals space 4 comma 675 cross times 10 to the power of negative sign 5 end exponent space mol space cross times space 84 space g space mol to the power of negative sign 1 end exponent massa equals space 3 comma 927 cross times 10 to the power of negative sign 3 end exponent g space massa equals 3 comma 92 space mg end style

 

Jadi, jawaban yang benar adalah C.

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