Pertanyaan

Dalam 200 mL air dapat larut 0,233 mg BaSO 4 ​ . Hasil kali kelarutan adalah.... (Mr = 233)

Dalam 200 mL air dapat larut 0,233 mg . Hasil kali kelarutan begin mathsize 14px style Ba S O subscript 4 end style adalah.... (Mr begin mathsize 14px style Ba S O subscript 4 end style= 233)

  1. begin mathsize 14px style 1 cross times 10 to the power of negative sign 10 end exponent end style

  2. begin mathsize 14px style 3 comma 6 cross times 10 to the power of negative sign 9 end exponent end style

  3. begin mathsize 14px style 3 cross times 10 to the power of negative sign 7 end exponent end style 

  4. begin mathsize 14px style 2 comma 5 cross times 10 to the power of negative sign 11 end exponent end style

  5. begin mathsize 14px style 1 cross times 10 to the power of negative sign 3 end exponent end style

V. Seftari

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

maka jawaban yang tepat adalah D

maka jawaban yang tepat adalah D

Pembahasan

Persamaan reaksi kesetimbangan kelarutan adalah: Dengan demikian, maka jawaban yang tepat adalah D

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M space Ba S O subscript 4 end cell equals cell space m over Mr cross times 1000 over V end cell row blank equals cell fraction numerator 0 comma 233 space mg over denominator 233 end fraction cross times fraction numerator 1 over denominator 200 mL end fraction end cell row blank equals cell left parenthesis 1 cross times 10 to the power of negative sign 3 end exponent right parenthesis cross times left parenthesis 5 cross times 10 to the power of negative sign 3 end exponent right parenthesis end cell row blank equals cell 5 cross times 10 to the power of negative sign 6 end exponent end cell end table end style


Persamaan reaksi kesetimbangan kelarutan undefined adalah:

undefined    

 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell K subscript sp space Ba S O subscript 4 end cell equals cell open square brackets Ba to the power of 2 plus sign close square brackets left square bracket S O subscript 4 to the power of 2 minus sign end exponent right square bracket end cell row blank equals cell left parenthesis 5 cross times 10 to the power of negative sign 6 end exponent right parenthesis left parenthesis 5 cross times 10 to the power of negative sign 6 end exponent right parenthesis end cell row blank equals cell 25 cross times 10 to the power of negative sign 12 end exponent end cell row blank equals cell 2 comma 5 cross times 10 to the power of negative sign 11 end exponent end cell row blank blank blank end table end style 

Dengan demikian, maka jawaban yang tepat adalah D

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