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Đáp án:
$[Na^+] = [Cl^-] = 2M$
Giải thích các bước giải:
$NaCl \to Na^+ + Cl^-$
Ta có :
$[Na^+] = [Cl^-] = C_{M_{NaCl}} = 2M$
$C_{M_{NaCl}}=2M$
Bảo toàn nguyên tố:
$[Na^+]=[Cl^-]=2M$
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