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CHEM1101 Exam HELL (1 Viewer)
- Thread starter fush
- Start date
Ok, for J-13:
Kp=1.16.
You need to work out the amount of moles of CO2 and CaCO3.
For CO2, you need to use pV=nRT
n=pV/RT
n=(1.16*10)/(0.08206*1073)
n=0.13174...
For CaCO3:
n= 20/(40.08+16*3+12.01)
n=0.19982...
The difference between these two values is the amount of CaCO3 that remains at equilibrium:
0.19982-0.13174 = 0.06808
The percentage is (0.06808/0.19982)*100 = 34.1%
The reason why you only take into account CO2 is because Kp=[CO2]
Kp=1.16.
You need to work out the amount of moles of CO2 and CaCO3.
For CO2, you need to use pV=nRT
n=pV/RT
n=(1.16*10)/(0.08206*1073)
n=0.13174...
For CaCO3:
n= 20/(40.08+16*3+12.01)
n=0.19982...
The difference between these two values is the amount of CaCO3 that remains at equilibrium:
0.19982-0.13174 = 0.06808
The percentage is (0.06808/0.19982)*100 = 34.1%
The reason why you only take into account CO2 is because Kp=[CO2]
Last edited:
allstarr69
Member
- Joined
- Feb 7, 2004
- Messages
- 225
hmm i cant get the right answer... heres what ive done
grams of I3- at equilibrium = 12.5 - 0.330 = 12.17
moles of I3- at equilibrium = 12.12 / 380.7 = 0.032 moles per litre
moles of I2 at equilibrium = 0.33 / 253.8 = 0.0013 moles per litre
moles of I- at equilibrium = 0.1 - 12.17 / 380.7 = 0.068
Therefore eq. constant = 362
Hmm nit can you explain what Ive done wrong or more realistically what ive done right, if anything.
You said some moles of I- would have been removed and you have to take that into account. How do you do that?
Cheers/
grams of I3- at equilibrium = 12.5 - 0.330 = 12.17
moles of I3- at equilibrium = 12.12 / 380.7 = 0.032 moles per litre
moles of I2 at equilibrium = 0.33 / 253.8 = 0.0013 moles per litre
moles of I- at equilibrium = 0.1 - 12.17 / 380.7 = 0.068
Therefore eq. constant = 362
Hmm nit can you explain what Ive done wrong or more realistically what ive done right, if anything.
You said some moles of I- would have been removed and you have to take that into account. How do you do that?
Cheers/
hey can any1 help me wit these questions
Pb(s) + Sn2+(aq) <------->Pb2+(aq) + Sn(s)
Use the Nernst equation to calculate the ratio of cation concentrations at 298 K for which the cell potential, E = 0. (See the data sheet for a table of standard reduction potentials.)
and
Calculate the molar activity of 3H (in Curie), given its half-life of 12.26 years.
thanks
Pb(s) + Sn2+(aq) <------->Pb2+(aq) + Sn(s)
Use the Nernst equation to calculate the ratio of cation concentrations at 298 K for which the cell potential, E = 0. (See the data sheet for a table of standard reduction potentials.)
and
Calculate the molar activity of 3H (in Curie), given its half-life of 12.26 years.
thanks
nit
Member
The first one is just plugging the data into the nernst equation. Write out the nernst equation, put Ecell=0 and you get that E(standard)=RT/nF*ln(K), and K, the equilibrium constant, is the ratio of cations.
As for the next part, molar activity is the activity displayed by a mole of atoms and is independent of concentration of radiocative substance. A_m=lambda*N_A where where lambda= ln2/(T_1/2). Convert T_1/2 into seconds to get A in terms of disintegrations/second and then you use the fact that 1curie* = 3.7x10^10 disintegrations/second
*(no. disintegrations/second of 1 gram of Radon-226).
Allstarrr: One of your problems is that you can't say that m(I3-)=m(I2)initial - m(I2)used. I2 and I3- are obviously different species and so you'd have to take recourse to using moles to solve that.
As for equilibrium constants and their units, strictly speaking we shouldn't be using concentrations but rather activities which are a dimensionless measure of concentrations as such. So it's strictly not correct to use units in equilibrium expressions though we probably have to every time we calculate K or Q.
As for the next part, molar activity is the activity displayed by a mole of atoms and is independent of concentration of radiocative substance. A_m=lambda*N_A where where lambda= ln2/(T_1/2). Convert T_1/2 into seconds to get A in terms of disintegrations/second and then you use the fact that 1curie* = 3.7x10^10 disintegrations/second
*(no. disintegrations/second of 1 gram of Radon-226).
Allstarrr: One of your problems is that you can't say that m(I3-)=m(I2)initial - m(I2)used. I2 and I3- are obviously different species and so you'd have to take recourse to using moles to solve that.
As for equilibrium constants and their units, strictly speaking we shouldn't be using concentrations but rather activities which are a dimensionless measure of concentrations as such. So it's strictly not correct to use units in equilibrium expressions though we probably have to every time we calculate K or Q.
For the Curie one, I thnk the answer that is stated is wrong. I ended up with 2.92*10^4, and I asked someone and they said it should be right.
For the other question:
E= E0 - RT/nF*ln(k) (I prefer using this equation than to the log one, but it's up to you)
E=0,
E0=RT/nF*ln(k)
E0= 0.13+(-0.14) = -0.01V
-0.01=(8.314*298)/(2*9.65*10^4)*ln(k) (n=2 because there is a transfer of 2 electrons)
ln(k)=-0.778988...
e^ln(k)=e^-0.778988...
k= 0.45887...
k=0.46 (i think the answer is 0.45- just rounding errors)
For the other question:
E= E0 - RT/nF*ln(k) (I prefer using this equation than to the log one, but it's up to you)
E=0,
E0=RT/nF*ln(k)
E0= 0.13+(-0.14) = -0.01V
-0.01=(8.314*298)/(2*9.65*10^4)*ln(k) (n=2 because there is a transfer of 2 electrons)
ln(k)=-0.778988...
e^ln(k)=e^-0.778988...
k= 0.45887...
k=0.46 (i think the answer is 0.45- just rounding errors)
allstarr69
Member
- Joined
- Feb 7, 2004
- Messages
- 225
yeh i think i finally get it... unfortunately takes me quite a while before i get it ... time i wont have in the exam hall lol.
Well at least i can get most of the rest of the questions right.
Nit, you are WAY too smart!!!!!!! Can we swap brains just for monday?
Oh why thank you very much.
Well at least i can get most of the rest of the questions right.
Nit, you are WAY too smart!!!!!!! Can we swap brains just for monday?
Oh why thank you very much.
allstarr69
Member
- Joined
- Feb 7, 2004
- Messages
- 225
Tenille you arent someone i know r u........ youd know if you were by my name
thanks for the help
sorry but can some1 tell me how to do this one too...its kind of the same i still cant do it.
The half life of I-131 is 8.06 days. Calculate the activity, in Bq, of 12.0 g of pure I-131. Calculate the specific activity of I-131 in Ci /mol
sorry but can some1 tell me how to do this one too...its kind of the same i still cant do it.
The half life of I-131 is 8.06 days. Calculate the activity, in Bq, of 12.0 g of pure I-131. Calculate the specific activity of I-131 in Ci /mol
As for your question bobbie212, you need to work out the time in seconds, then work out the decay rate, which is equal to ln2/T(1/2). Once you've done that, you multiply your answer by avagadro's number. Your answer is then in Bq/mol. If you want to find out the answer in Bq, you multiply it by the amount of moles of I there are.
To find out the answer in Ci/mol, you just grab your answer that was in Bq/mol and divide it by 3.7*10^10.
QUOTE=Allstarrr]Tenille you arent someone i know r u........ youd know if you were by my name[/QUOTE]
I'm not sure. I've met so many people at uni I can only remember like 2 or 3 names. Lol.
To find out the answer in Ci/mol, you just grab your answer that was in Bq/mol and divide it by 3.7*10^10.
QUOTE=Allstarrr]Tenille you arent someone i know r u........ youd know if you were by my name[/QUOTE]
I'm not sure. I've met so many people at uni I can only remember like 2 or 3 names. Lol.
this is a simple question
like NAOH(100ML)+HCL(100ML) and temp change is 13.4(exothermic)
so enthalpy in kJ would be 4.18*13.4*0.2*-.1
the answer from webct took the mass value to be 100g instead of 200. and got answer twice as small. but why?
like NAOH(100ML)+HCL(100ML) and temp change is 13.4(exothermic)
so enthalpy in kJ would be 4.18*13.4*0.2*-.1
the answer from webct took the mass value to be 100g instead of 200. and got answer twice as small. but why?
Enthalpy is in kJ/mol. So you work out m*C*(change in temp) for each substance, then divide each by the amount of moles, then add them together.Steven12 said:
And for the november exam, are you talking about advanced or normal chem? Because if it's normal, you can do those exams since they do have CHEM1011 during semester 2.
yea,i kind of know what you are saying
sorry didnt put up the whole question
the question was 100ml of NAOH is mixed with 100ml of HCL , both 100ml and 2M. the temp went from 24.6-38 deg.
assuming a perfect calorimeter, determine the standard enthalpy change for neutralisation reaction.
so
4.18*100*-13.4=5601(j) (same for both solutions)
5601/2+5601/2=5601(j/M)
right?
and so 5.6 kJ/M but the answer was 56kj per mol
i dont know what i am missing?
sorry didnt put up the whole question
the question was 100ml of NAOH is mixed with 100ml of HCL , both 100ml and 2M. the temp went from 24.6-38 deg.
assuming a perfect calorimeter, determine the standard enthalpy change for neutralisation reaction.
so
4.18*100*-13.4=5601(j) (same for both solutions)
5601/2+5601/2=5601(j/M)
right?
and so 5.6 kJ/M but the answer was 56kj per mol
i dont know what i am missing?