дано
m(ppa H2SO4) = 200 g
W(H2SO4) = 20%
η(Na2SO4) = 80%
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m прак (Na2SO4)-?
m(H2SO4) = 200 * 20% / 100% = 40 g
M(H2SO4) = 98 g/mol
n(H2SO4) = m/M = 98 / 40 = 2.45 mol
H2SO4+Na2O-->Na2SO4+H2O
n(H2SO4) = n(Na2SO4) = 2.45 mol
M(Na2SO4) = 142 g/mol
m теор (Na2SO4) = n*M = 2.45 * 142 = 347.9 g
m практ (Na2SO4) = 347.9 * 80% / 100% = 278.32 g
ответ 278.32 г
1. Г если за 10 сек 0.03, то за 1 сек 0.003моль/(л*с)
A
1)2С3H7COOH+K2O-->2С3H7COOK+H2O
2)2С3H7COOH+MgO-->(С3H7COO)2Mg+H2O
3)3С3H7COOH+Al(OH)3-->(С3H7COO)3Al+3H2O
4)С3H7COOH+NaOH-->С3H7COONa+H2O
Б
1)2<span>СН3-CH(СН3)-СООН+K2O-->2<span>СН3-CH(СН3)-СООK+H2O</span></span>
<span><span>2)2<span>СН3-CH(СН3)-СООН+MgO-->(<span>СН3-CH(СН3)-СОО)2Mg+H2O</span></span></span></span>
<span><span><span><span>3)3<span>СН3-CH(СН3)-СООН+Al(OH)3-->(<span>СН3-CH(СН3)-СОО)3Al+3H2O</span></span></span></span></span></span>
<span><span><span><span><span><span>4)<span>СН3-CH(СН3)-СООН+NaOH--><span>СН3-CH(СН3)-СООNa+H2O</span></span></span></span></span></span></span></span>
BaCl2 + 2AgNO3 = Ba(NO3)2 + 2AgCl осадок белого цвета
Pb(NO3)2 + K2S = 2KNO3 + PbS осадок серого цвета