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KabDan98
@KabDan98
July 2022
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чему будет равно произведение растворимости соли agI , если растворимость соли равна 1,2x10^-8 моль/л
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dec77cat
А=1,2*10⁻⁸ моль/л
AgCl = Ag⁺ + Cl⁻
ПР=[Ag⁺][Cl⁻]
[Ag⁺]=[Cl⁻]=а
ПР=a²
ПР=(1,2*10⁻⁸)²=
1,44*10⁻¹⁶
1 votes
Thanks 1
KabDan98
откуда вы Cl взяли и почему а в квадрате?
dec77cat
а*а=а^2
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Answers & Comments
AgCl = Ag⁺ + Cl⁻
ПР=[Ag⁺][Cl⁻]
[Ag⁺]=[Cl⁻]=а
ПР=a²
ПР=(1,2*10⁻⁸)²=1,44*10⁻¹⁶