For which from the given equilibriums an increase of pressure does not exert influence on the displacement of an equilibrium: a) 2HBr (g) H2 (g) + Br2 (g); b) СО2 (g) + Н2 (g)  Н2O (g) + СО (g) ; c) N2O4 (g)  2NO2 (g) ; d) 2CO (g) + O2 (g)  2CO2 (g) ; e) N2 (g) + O2 (g)  2NO (g)

1. C2H5NH2COOH + HCl = NH3ClC2H5COOH

2. 2NH2CH2COOH + 2Na = 2NH2CH2COONa + H2

3. NH2C3H7COOH + C2H5OH (t>140°,h2so4(k))= NH2C3H7COOC2H5 + H2O

4. NH2CH2COOH + NH2C2H5COOH = NH2CH2CONHC2H5COOH + H2O

5. NH2C2H5COOH + KOH = NH2C2H5COOK + H2O

6. 1) C6H6 + OHNO2(k) (h2so4(k))= C6H5NO2 + H2O

2) C6H5NO2 + 3H2 (kat)= C6H5NH2 + 2H2O

3) C6H5NH2 + 3Br2 = C6H2Br3NH2 + 3HBr

7. 1) C2H5OH + CuO (t)= CH3COH + Cu + H2O

2) CH3COH + Ag2O (t,NH3*H2O)= CH3COOH + 2Ag

3) CH3COOH + Cl2 ((CH3COO)3Co)= CH2ClCOOH + HCl

4) CH2ClCOOH + 2NH3 = CH2NH2COOH + NH4Cl

8. C6H5NH2 + OHNO2 (k)= C6H4NH2NO2 + H2O

9. C6H5NH2 + HCl = [C6H5NH3]Cl

ответ: ответ: ω(C)=88.9%ω(C)=88.9%.

Объяснение:

Дано: Решение m(каменного угля)=0.27гm(каменного угля)=0.27г m(BaCO3)=3.94гm(BaCO3)=3.94г C+O2=CO2C+O2=CO2 CO2+Ba(OH)2=BaCO3↓CO2+Ba(OH)2=BaCO3↓ n(BaCO3)=m(BaCO3)M(BaCO3)=3.94197=0.02 мольn(BaCO3)=m(BaCO3)M(BaCO3)=3.94197=0.02 моль n(C)=n(CO2)=n(BaCO3)=0.02 мольn(C)=n(CO2)=n(BaCO3)=0.02 моль m(C)=n(C)⋅M(C)=0.02⋅12=0.24гm(C)=n(C)⋅M(C)=0.02⋅12=0.24г ω(C)=100⋅m(C)m(каменного угля)=100⋅0.240.27=88.9%ω(C)=100⋅m(C)m(каменного угля)=100⋅0.240.27=88.9% ω(C)=?