M(FeCO3)=56+12+(16*3)=116г/моль
ω(э/в-ве)=Ar(э)*k/M(вещ-ва)*100%
ω(Fe/в-ве)=56/116*100%=48,2%
ω(C/в-ве)=12/116*100%=10,3%
ω(O3/в-ве)=100%-48,2%-10,3%=41,5%
2C4H10 + 5O2 ----соли Mn и Co---> 4CH3COOH + 2H2O
CH3COOH + CH3OH ----H+---> CH3COOCH3 + H2O
CH3COOCH3 + NaOH ----H2O----> CH3COONa + CH3OH
CH3COONa + HCl ----> CH3COOH + NaCl
<span>m (Cu) = w (Cu) * m (руды) = 0,02 * 10 = 0,2 г</span>
Ответ:
Объяснение:
СH4 + Br2 -> CH3-Br + HBr (hy)
2CH3-CH3+7O2 -> 4CO2 + 6H2O
CH2=CH2 + H2O-> CH3-CH2-OH (kat)