-{-(CH2-CH=CH-CH2-)x-[CH(C6H5)-CH2-]y-}n-
M =( 54x+104y )
nC = 4x + 8y = nCO2= 161,28/22,4 = 7,2 moli
n monomer = 95,6/(54x+ 104y) = 23,9/(13,5x+ 26y)
23,9/(13,5x+26y) moli cauciuc.............................7,2 moli at. C
1mol..........................................................(4x+8y)
23,9(4x+8y) = 7,2(13,5x+26y) 95,6x + 191,2y = 97,2x + 187,2y
1,6x= 4y x/y = 5/2
A) : A
B) : F M = 5·54 + 2·104 = 478 g/mol n cauciuc = 95,6/478 = 0,2moli
nBr2 = 0,2·x = 0,2·5 = 1 mBr2 = 160·0,2 = 32g
C) : A nC tertiar = 2·5 + 6·2 = 22
D): A
E) : F M = 478g/mol mC = (4·5 + 8·2)12 = 432g
%C = 432·100/478 = 90,376%