<span><span>2<span>Fe</span></span> + <span>3<span>Cl2</span></span> → <span>2<span>FeCl<span>3
вроде так
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Вoзьмём m(NaOH,10%) за x, тогда m(NaOH,100%) в нём будет 0,1х. m(NaOH,100%) в 40%-ом р-ре =m(р-ра)*40%/100%=300*40/100=120г. m(NaOH,100%) в конечном р-ре= 0,1x+120=m(всего р-ра)*20%/100%=(300+х)*20%/100%, решаем ур-е 0,1х+120=60+0,2х, 0,1х=60, х=600г. Ответ: m(NaOH,10%)=600г.
1, 4)
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2 а 3
б 5
в 1
г 2
C2H6-C2H4+H2
C2H4+H2O-C2H5OH
C2H5OH+HCl-C2H5Cl+H2O
2C2H5Cl+2Na-C4H10+2NaCl
C4H10+O2-2CH3COOH+H2
CH3COOH+NaOH-CH3COONa+H2O
CH3COONa+NaOH-CH4+Na2CO3
CH4+Cl2-CH3Cl+HCl
2CH3Cl+Na-C2H6+2NaCl
2C2H6+7O2-4CO2+6H2O
6CO2+6H2O-C6H12O6+6O2
C6H12O6-C2H5OH+2CO2
2C2H5OH+2O2-CH3COOH+2H2O
CH3COOH+C2H5OH-CH3COOC2H5