Chemia ćwiczenia nowej ERY 3 od str 22do 30 :) dzieki z gory ;/

32. a) mCxH2x-2 = 82u 86u = 12x + 1 * 2x - 2 86u = 12x + 2x - 2 864u = 14x x = 6 wzór sumaryczny : C6H10 Nazwa : heksyn b) m CxH2x-2 = m C6H10 86u = (12*6) : (1 * 10 - 2) = 72 : 10 = 36 : 5 33. a) C4H8 mC4H8 = 12u * 4 + 8 * 1u = 56u m C4 = 12u * 4 = 48u m H8 = 8u % C4 100% - 56u x - 48u x = 85,71% % H8 100% - 56u x - 8u x = 14,29% b) C4H6 m C4H6 = 12u * 4 + 4u = 54u m C4 = 48u m H6 = 6u % H6 100% - 54u x - 6u x = 11,11% % C4 100% - 54u x - 48u x = 88,89% 34. a) C2H4, ETYLEN b) mC2H4 = 12*2 + 1*4 = 28u m C2 = 24u m H4 = 4u % C 100% - 28u x - 24u x = 85,71% % H 100% - 28u x - 4u x = 14,29% c) m C2H4 = 2*12 + 1*4 = 28u 28u = 12u*2 : 1u*4 = 24u : 4u = 6 :1 37. C2H4 + 2O2 -> 2CO + 2 H20 C2H4 + O2 -> 2C + 2 H20 C2H4 + 3O2 -> 2CO2 + 2 H20 C2H4 + Cl2 -> c2h4cL2 C2H4 + Br2 -> ... n C3H6 -> CH2 = CH - CH3 CH2 = CH2 + HCl -> C2H5Cl ch2 = ch2 + h2 -> ch3 - ch3 39. HC ≡ CH 1, 3, 1 40. m C2H2 = 2 * 12 = 26u m C2 = 24u m H2 = 2u 100% - 93,31% = 7,69% 26u - 100% 24u - c % x = 92, 31% 42. 2 C2H2 + 5O2 -> 4CO2 + 2H20 2C2H2 + O2 -> 4C + 2H20 2C2H2 + 302 -> 4CO + 2H20