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Hasil kali kelarutan Ag 2 ​ SO 4 ​ yang memiliki kelarutan sebesar 5 , 4 g L − 1 adalah ... ( A r ​ Ag = 108 g mol − 1 , S = 32 g mol − 1 , dan O = 16 g mol − 1 )

Hasil kali kelarutan   yang memiliki kelarutan sebesar   adalah ...

 

  1. begin mathsize 14px style 5 comma 78 space x space 10 to the power of negative sign 3 end exponent end style 

  2. begin mathsize 14px style 1 comma 97 space x space 10 to the power of negative sign 5 end exponent end style 

  3. begin mathsize 14px style 3 comma 34 space x space 10 to the power of negative sign 5 end exponent end style 

  4. begin mathsize 14px style 9 comma 82 space x space 10 to the power of negative sign 6 end exponent end style 

  5. begin mathsize 14px style 4 comma 91 space x space 10 to the power of negative sign 6 end exponent end style 

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Y. Rochmawatie

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Jadi, dan karena pilihan jawaban tidak ada maka pilih yang paling mendekati yaitu B.

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell kelarutan space Ag subscript 2 S O subscript 4 space end cell equals cell space 5 comma 4 space g forward slash L end cell row cell italic M subscript italic r italic space Ag subscript 2 S O subscript 4 space end cell equals cell space 312 space g forward slash mol 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 blank blank cell bold Persamaan bold space bold reaksinya end cell row cell Ag subscript bold 2 S O subscript bold 4 bold italic open parentheses italic s bold italic close parentheses end cell bold rightwards harpoon over leftwards harpoon cell bold 2 Ag to the power of bold plus sign bold italic left parenthesis italic a italic q bold italic right parenthesis bold plus bold space S O subscript bold 4 to the power of bold 2 bold minus sign end exponent bold italic left parenthesis italic a italic q bold italic right parenthesis end cell row blank blank cell 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 space italic space italic space italic space italic space italic space italic space italic 2 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 space italic space italic s end cell row cell K subscript italic s italic p end subscript end cell equals cell left parenthesis 2 s right parenthesis squared open parentheses s close parentheses space end cell row cell K subscript italic s italic p end subscript end cell equals cell 4 s cubed end cell row cell K subscript italic s italic p end subscript end cell equals cell 4 left parenthesis 0 comma 0173 space mol forward slash L bold right parenthesis cubed space end cell row cell K subscript bold sp end cell bold equals cell bold 2 bold comma bold 07 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent end cell end table end style

 

Jadi, begin mathsize 14px style K subscript bold sp Ag subscript bold 2 S O subscript bold 4 bold equals bold 2 bold comma bold 07 bold cross times bold 10 to the power of bold minus sign bold 5 end exponent end style dan karena pilihan jawaban tidak ada maka pilih yang paling mendekati yaitu B.

 

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