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Hasil kali kelarutan Mg2​(PO4​)2​ yang memiliki kelarutan 2,62 mg L−1 adalah ....  (Ar​: Mg = 24 g mol−1 , P = 31 g mol−1, dan O = 16 g mol−1)

Pertanyaan

Hasil kali kelarutan begin mathsize 14px style Mg subscript 2 open parentheses P O subscript 4 close parentheses subscript 2 end style yang memiliki kelarutan 2,62 mg begin mathsize 14px style L to the power of negative sign 1 end exponent end style adalah .... 

begin mathsize 14px style left parenthesis italic A subscript italic r colon space Mg space equals space 24 space g space mol to the power of negative sign 1 space end exponent comma space P space equals space 31 space g space mol to the power of negative sign 1 end exponent comma space dan space O space equals space 16 space g space mol to the power of negative sign 1 end exponent right parenthesis end style 

  1. begin mathsize 14px style 1 comma 08 space x space 10 to the power of negative sign 20 end exponent end style 

  2. begin mathsize 14px style 5 comma space 40 space x space 10 to the power of negative sign 23 end exponent end style 

  3. undefined 

  4. begin mathsize 14px style 5 comma 40 space x space 10 to the power of negative sign 25 end exponent end style 

  5. begin mathsize 14px style 1 comma 08 space x space 10 to the power of negative sign 23 end exponent end style 

E. Dwihermiati

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Kelarutan begin mathsize 14px style Mg subscript 3 open parentheses P O subscript 4 close parentheses subscript 2 space end subscript adalah space 2 comma 62 space mg forward slash L end style 

begin mathsize 14px style M subscript r space Mg subscript 3 open parentheses P O subscript 4 close parentheses subscript 2 space equals space 262 space g forward slash mol end style 

Kelarutan dalam satuan begin mathsize 14px style mol forward slash Liter end style (konsentrasi) adalah : 

Error converting from MathML to accessible text.

Nilai M = s

begin mathsize 12px style Mg subscript 3 open parentheses P O subscript 4 close parentheses subscript 2 italic left parenthesis italic s italic right parenthesis equilibrium 3 Mg to the power of 2 plus sign italic left parenthesis italic a italic q italic right parenthesis plus space 2 left parenthesis P O subscript 4 to the power of 3 minus sign end exponent right parenthesis italic left parenthesis italic a italic q italic right parenthesis s space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 3 s space space space space space space space space space space space space space space space space space space space space space 2 s K subscript italic s italic p end subscript equals left parenthesis 3 s right parenthesis cubed left parenthesis 2 s right parenthesis squared space K subscript italic s italic p end subscript equals 108 s to the power of 5 K subscript italic s italic p end subscript equals 108 left parenthesis 10 to the power of negative sign 5 end exponent right parenthesis to the power of 5 space K subscript bold sp bold equals bold 1 bold comma bold 08 bold space italic x bold 10 to the power of bold minus sign bold 23 end exponent end style 

Jadi, jawaban yang benar adalah C.

108

4.0 (3 rating)

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