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1)При взаимодействии меди массой 6,4г с хлором обьемом 33,6л получили соль. Вычислите ее массу.
Дано:
m(Cu)= 6,4 г
V(Cl)=33,6л
m(CuCl2)-?
6,4г 33,6л
1)Cu + Cl2 = CuCl2
1моль 1 моль 1 моль
64г 22,4л
2) .n(Cl)=V/Vм
n(Cl)= 33,6/22,4л=1,5моль
n=m/M => m= M*n
M(Cl)=35г/моль
m(Cl)= 35г/моль * 1,5 моль= 52,5 г
m(CuCl2)= m(Cu)+ m(Cl)
m(CuCl2)= 6,4г+ 52,5г= 58, 9 г
ответ: m(CuCl2)=58,9 г
Объяснения:
22,4л(объем газа при нормальных условиях )
Vм – Молярный объем (22,4л)
M- Молярная масса(в таблице Менделеева)
n-количество вещества
Общая формула двухатомных спиртов CnH2n(OH)2
при сгорании
CnH2n(OH)2 будет образовываться nCO2 + (n+1)H2O.
нас интересует только H2O
итак:
n*12+2n*1+2*(16+1) это молекулярный вес двухатомного спирта, где n - число атомов углерода
(n+1)*(18) это вес молей воды, образующейся при сжигании 1 моль двухатомного спирта с числом атомов углерода n.
n*12+2n*1+2*(16+1) = 14n + 34 - молекулярная масса спирта
(n+1)*(18) = 18n + 18 - молекулярная масса полученной воды
пропорция
(14n + 34) / (18n + 18) = 6,2 г / 5,4 г
5,4 (14n + 34) = 6,2 (18n+18)
n = 2
Нашли, что в молекуле 2 атома углерода
Двухатомный спирт с 2 атомами углерода в молекуле - это этиленгликоль C2H4(OH)2
или этандиол-1,2
HO-CH2-CH2-OH