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vika0906
@vika0906
August 2022
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Натрий массой 26г. растворили в 72г. воды.Вычислите массовую долю(в %) полученного раствора.
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dec77cat
Verified answer
2Na + 2H₂O = 2NaOH + H₂
m(NaOH)=M(NaOH)m(Na)/M(Na)
m(H₂)=M(H₂)m(Na)/2M(Na)
m(p)=m(Na)+m(H₂O)-m(H₂)=m(Na)+m(H₂O)-M(H₂)m(Na)/2M(Na)
m(p)=m(Na){1-M(H₂)/2M(Na)}+m(H₂O)
w=m(NaOH)/m(p)
w=M(NaOH)m(Na) / M(Na){m(Na){1-M(H₂)/2M(Na)}+m(H₂O)}
w=40*26 / 23*{26{1-2/2*23}+72}=0,467 (46,7%)
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Answers & Comments
Verified answer
2Na + 2H₂O = 2NaOH + H₂m(NaOH)=M(NaOH)m(Na)/M(Na)
m(H₂)=M(H₂)m(Na)/2M(Na)
m(p)=m(Na)+m(H₂O)-m(H₂)=m(Na)+m(H₂O)-M(H₂)m(Na)/2M(Na)
m(p)=m(Na){1-M(H₂)/2M(Na)}+m(H₂O)
w=m(NaOH)/m(p)
w=M(NaOH)m(Na) / M(Na){m(Na){1-M(H₂)/2M(Na)}+m(H₂O)}
w=40*26 / 23*{26{1-2/2*23}+72}=0,467 (46,7%)