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Aruzhan07
@Aruzhan07
August 2022
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определите массовую долю едкого натра при растворении 4,6 г натрия в 150г воды.
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savinovvitek
2Na + 2H2O = 2NaOH + H2(стрелка вверх)
n(Na) = 4,6/23 = 0,2моль
n(NaOH) = n(Na)
m(NaOH) = 0,2*40 = 8г
m(H2O прореаг.) = 0,2*18 = 3,6г
m(H2O осталось) = 150-3,6 = 146,4г
m(к. смеси) = 146,4+8 = 154,4г
w(NaOH) = 8*100%/154,4 = 5,18%
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АкеркеАбраимова1
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Answers & Comments
n(Na) = 4,6/23 = 0,2моль
n(NaOH) = n(Na)
m(NaOH) = 0,2*40 = 8г
m(H2O прореаг.) = 0,2*18 = 3,6г
m(H2O осталось) = 150-3,6 = 146,4г
m(к. смеси) = 146,4+8 = 154,4г
w(NaOH) = 8*100%/154,4 = 5,18%