Найдите массовую долю элементов в веществах mgo , n2o , no , n2o5 , naoh , c6h12o6 , cu(no3)2
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Ответы на вопрос:
1) m(mgo) = 40 w(mg) = 24/40 = 0,6% w(o) = 16/40 = 0,4% 2) m(n2o) = 44 w(n) = 14/44 = 0,31% w(o) = 16/44 = 0,36% 3) m(no) = 30 w(n) = 14/30 = 0,46% w(o) = 16/30 = 0,53% 4) m(n2o5) = 108 w(n0 = 14/108 = 0,12% w(o) = 16/108 = 0.14% 5) m(naoh) = 40 w(na) = 23/40 = 0.5% w(o) = 16/40 = 0.4% w(h) = 1/40 = 0.025% 6) m(c6h12o6) = 180 w(c) = 12/180 = 0.06% w(h) = 1/180 = 0.005% w(o) = 16/180 = 0.08% 7) m(cu(no3)2) = 188 w(cu) = 64/188 = 0.34% w(n) = 14/188 = 0.07% w(o) = 16/188 = 0.08%
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