Chem questions help (1 Viewer)
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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
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ahhhhh if i remember, it's in the 2013 CSSA paper?
Yep
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
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