more Que's in chem.
may02, P1 (CIE)
Q6,21,11
Q6) P1V1+P2V2= P3V3
he (1X2) + ne (2X1)= x X 3
x= 4/3
A
What you do is take a V3= 3dm3 (1dm3 of he and 2 dm3 of ne..since they are connected)
Q11) C
It will keep increasing as more products are formed...later, reactants act as a limiting agent...
If you are wondering why the curve can't plateau out, it is because no more substrate has been added to keep the rate constant....as the conc of substrate falls, rate of reaction falls
Q21) You already had 6 as soon as you figure the formula...right?
now when the haloalkane is either one of these C2H4Cl2 or C2H2Cl4 or C2H3Cl3, they could have isomers...
in C2H4Cl2 the Cl could be on each carbon or only on 1 carbon...2 possible structures
C2H2Cl4 the Cl could be 2 on each carbon or 3 on 1 and 1 on the other
C2H3Cl6 the Cl could be all 3 on one, or divide as 2 on 1 and 1 on the other
6+ the 3 possibilities = 9
C