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Pertanyaan

Hasil kali kelarutan Mg ( OH ) 2 ​ dalam air pada suhu tertentu adalah 2 × 1 0 − 11 . Tentukan kelarutan dalam: Larutan yang mengandung 0,5 M MgCl 2 ​

Hasil kali kelarutan  dalam air pada suhu tertentu adalah . Tentukan kelarutan Mg open parentheses O H close parentheses subscript 2 dalam:

Larutan yang mengandung 0,5 M  

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J. Siregar

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

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Jawaban

kelarutan dalam larutan 0,5 M adalah .

kelarutan Mg open parentheses O H close parentheses subscript 2 dalam larutan Mg Cl subscript 2 0,5 M  adalah Error converting from MathML to accessible text..

Pembahasan

Penentuan kelarutan dalam larutan 0,5 M dapat dihitung berdasarkan pengaruh ion senama. Jadi, kelarutan dalam larutan 0,5 M adalah .

Penentuan kelarutan Mg open parentheses O H close parentheses subscript 2 dalam larutan Mg Cl subscript 2 0,5 M dapat dihitung berdasarkan pengaruh ion senama.  

table attributes columnalign right center left columnspacing 0px end attributes row cell Mg Cl subscript 2 end cell rightwards arrow cell Mg to the power of 2 plus sign and 2 Cl to the power of minus sign end cell row blank blank cell 0 comma 5 space M space space space space 0 comma 5 space M end cell row blank blank blank row cell Mg open parentheses O H close parentheses subscript 2 end cell rightwards harpoon over leftwards harpoon cell Mg to the power of 2 plus sign and 2 O H to the power of minus sign end cell row blank blank cell space space space space s 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 2 s end cell row cell K subscript sp end cell equals cell open square brackets Mg to the power of 2 plus sign close square brackets open square brackets O H to the power of minus sign close square brackets squared end cell row cell 2 cross times 10 to the power of negative sign 11 end exponent end cell equals cell left parenthesis 0 comma 5 right parenthesis open square brackets O H to the power of minus sign close square brackets squared end cell row cell 2 cross times 10 to the power of negative sign 11 end exponent end cell equals cell left parenthesis 0 comma 5 right parenthesis open square brackets O H to the power of minus sign close square brackets squared end cell row cell open square brackets O H to the power of minus sign close square brackets end cell equals cell square root of fraction numerator 2 cross times 10 to the power of negative sign 11 end exponent over denominator 0 comma 5 end fraction end root end cell row blank equals cell square root of 4 cross times 10 to the power of negative sign 11 end exponent end root end cell row blank equals cell square root of 40 cross times 10 to the power of negative sign 12 end exponent end root end cell row blank equals cell 6 comma 3 cross times 10 to the power of negative sign 6 end exponent end cell row blank blank blank row s equals cell 1 half cross times open square brackets O H to the power of minus sign close square brackets end cell row blank equals cell 1 half cross times 6 comma 3 cross times 10 to the power of negative sign 6 end exponent end cell row blank equals cell 3 comma 15 cross times 10 to the power of negative sign 6 end exponent end cell end table   

Jadi, kelarutan Mg open parentheses O H close parentheses subscript 2 dalam larutan Mg Cl subscript 2 0,5 M  adalah Error converting from MathML to accessible text..

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