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Pertanyaan

Jika diketahui begin mathsize 14px style K subscript italic s italic p end subscript space Hg subscript 2 Cl subscript 2 space equals space 1 space x space 10 to the power of negative sign 18 end exponent end style , kelarutan begin mathsize 14px style Hg subscript 2 Cl subscript 2 end style dalam larutan begin mathsize 14px style Ca Cl subscript 2 end style 0,01 M adalah ...

  1. begin mathsize 14px style 2 comma 5 space x space 10 to the power of negative sign 15 end exponent end style  

  2. begin mathsize 14px style 5 comma 0 space x space 10 to the power of negative sign 15 end exponent end style  

  3. begin mathsize 14px style 1 comma 0 space x space 10 to the power of negative sign 14 end exponent end style  

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

  5. begin mathsize 14px style 5 comma 0 space x space 10 to the power of negative sign 14 end exponent end style 

Y. Rochmawatie

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah opsi A.

Pembahasan

begin mathsize 14px style Ca Cl subscript 2 left parenthesis italic a italic q right parenthesis rightwards harpoon over leftwards harpoon space Ca to the power of 2 plus sign left parenthesis italic a italic q right parenthesis space plus space space 2 Cl to the power of minus sign left parenthesis italic a italic q right parenthesis space space space space 0 comma 01 space space space space space space space space space space space space space 0 comma 01 space space space space space space space space space space space space space 0 comma 02 space space Hg subscript 2 Cl subscript 2 left parenthesis italic a italic q right parenthesis rightwards harpoon over leftwards harpoon Hg subscript 2 to the power of 2 plus end exponent left parenthesis italic a italic q right parenthesis space plus space 2 Cl to the power of minus sign left parenthesis italic a italic q right parenthesis space space space space s space space space space space space space space space space space space space space space space space space s space space space space space space space space space space space space space space space space space space space 2 s end style . 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell Ksp space Hg subscript 2 Cl subscript 2 end cell equals cell Qsp space open square brackets Hg to the power of 2 plus sign close square brackets open square brackets Cl to the power of minus sign close square brackets squared end cell row cell 1 space x space 10 to the power of negative sign 18 end exponent end cell equals cell open parentheses s close parentheses left parenthesis 0 comma 02 right parenthesis squared space end cell row cell 1 space x space 10 to the power of negative sign 18 end exponent end cell equals cell open parentheses s close parentheses left parenthesis 0 comma 0004 right parenthesis end cell row s equals cell fraction numerator 1 cross times 10 to the power of negative sign 18 end exponent over denominator 4 cross times 10 to the power of negative sign 4 end exponent end fraction end cell row s equals cell 2 comma 5 cross times 10 to the power of negative sign 15 end exponent end cell row blank blank blank end table end style.  

Jadi kelarutan begin mathsize 14px style Hg subscript 2 Cl subscript 2 end style dalam larutan begin mathsize 14px style Ca Cl subscript 2 space 0 comma 1 space M end style adalah Error converting from MathML to accessible text.

Jadi, jawaban yang tepat adalah opsi A.

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