DoK1509

1)m(р-ра KHCO3)=q*v=75*1.07=80.25(г.) 2)m(в-ва KHCO3)=w*m=80.25*0.1 3)ню общее(KHCO3)=m(KHCO3)/M(KHCO3)=8.025/100=0.08(моль) 4)пусть m(ch3-cooh)=x ,то m(c2h5-cooh)=5.64-x 5)ню(ch3-cooh)=x/60 ню(c2h5-cooh)=5.64-x 6)ню1(KHCO3)=y 7)ню2(KHCO3)=0.08-y 8 x/60=y 5.64-x/74=0.08-y Решаем эту систему. 5.64-60y/74=0.08025-y 5.64-60y=5.94-74y 14y=0.3 y=0.043 x=0.043*60 x=2.6 w(ch3-cooh)=2.6/5,64=46% w(c2h5-cooh)=100-w(ch3-cooh)=54% Ответ:w(ch3-cooh)=46%;w(c2h5-cooh)=54%