Chem questions help (1 Viewer)

photastic

Well-Known Member
Joined
Feb 11, 2013
Messages
1,848
Gender
Male
HSC
2014
Well for question 9, CO2 + H2O -> H2CO3 deltaH<0. Because temperature is increasing, equilibrium will shift to the left to absorb the heat, hence producing more CO2, eliminating A and B. Since CO2 is soluble at room temperature, eliminate D
 

integral95

Well-Known Member
Joined
Dec 16, 2012
Messages
779
Gender
Male
HSC
2013
ahhhhh if i remember, it's in the 2013 CSSA paper?
 

faker

New Member
Joined
Aug 1, 2014
Messages
11
Gender
Male
HSC
2014
Firstly writing the equation for a diprotic acid is an important skill:
H2A+Na(OH)---> Na2A+H2O
Thus the moles ration of acid to base is 1:2
Therefore since n=cv for both and the moles are in the ration of 1:2 we get this
therefore in fact 0.5(Concentration Na(OH) x volume Na(OH))=(Concentration of Oxalic Acid x Volume of Oxalic Acid)

First finding the moles of the standard acid: n=m/M= (12.6)/(126.068)=0.09992 moles
Then finding the concentration in mol/L: c=n/v = (0.09992)/(500 x 10^-3)=0.1984
Then from equation above 0.5(Concentration Na(OH) x 18.2x10^-3)=(0.1984x25x10^-3)
(Concentration Na(OH) x 18.2x10^-3)=(0.1984x25x10^-3)/(0.5)
(Concentration Na(OH))=(0.1984x25x10^-3)/(0.5)(18.2x10^-3)
= 0.545 mol/L
Note: I used very rough values thus the variation by 0.004 nonetheless you get the right idea :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top