Posted by Nancy on Tuesday, July 23, 2013 at 1:52am.
I took a look at this question to see if I copied it correctly, and it does NOT say how much of the solution was used. It does use mols as the unit.
Dr. Bob, I am not sure what you mean by "decays into Argon." I realize that the 38 CL decays to Argon, but how would that work into the solution? I am confused as it appears I need to use PV =NRT and then also half life, but I just don't get 300K as the answer, which is what the book has.
If you have any insight, let me know. If not I will find out later this week how to solve it.
The nuclide 38cl decays by beta emission with a half life of 40min. A sample of .40 mol of H38cl is placed in a 6.24 liter container. After 80 min the pressure is 1650 mmHg. What is temperature of container?
I have tried calculating the decay rate of the 38cl and then plugging into the formula pv =nrt but I do not get the answer, 300 k. Any help would be great.
I would think it would be like this:
.40 mol is .0258% Hydrogen
.40 mole is .974 CL38
.40 x .974=.3896 x half life .25=.0974 + mole of hydrogen .01032
13.54/.1077= 125k but this isn't correct
The correct answer is 300K
anybody know how it was obtained?
Chemistry - DrBob222, Monday, July 22, 2013 at 12:39pm
I'm responding just to let you know that I've looked at the problem more than once over the last day or two. Is the HCl a gas? I suppose no if it is a 0.40 m solution. Is that m and not M? How much of the 0.40m solution? is used? The problem doesn't say. And do you realize that Cl38 goes to Ar38 upon releasing a beta particle.
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- Chemistry re-post - bobpursley, Tuesday, July 23, 2013 at 10:43am
HCL is a gas. There is a trick her.
2Cl2 decays to Cl2+2Ar
so note that for each two moles of CL2 decaying, you get three moles of gas as products. That may solve your problem
- Chemistry re-post - DrBob222, Tuesday, July 23, 2013 at 2:30pm
Following Bob Pursley's comment, IF HCl is a gas (then what does the m dstand for?)
k = 0.693/t1/2 = about 0.017 but you can do it more accurately.
ln(No/N) = kt
ln(0.4/N) = 0.0173*80
N = about 0.1
So we started at 0.4
end up with 0.1
used = 0.3 mol
0.3 x 3/2 = 0.45 gas formed and we had 0.1 left over which makes 0.55 mols gas.
Then PV = nRT and T = 300 K.
- Chemistry re-post - Devron, Thursday, July 25, 2013 at 1:37am
Okay guys, this problem is really bugging me. I initially saw it and declined to try and solve it. However, after Nancy reposted the question, I thought that I would see if I could take a stab at it. I did back calculations and obtained 0.55 moles, but I could not figure out how in the world that a closed system, which the wording of the problem indicates that it is, undergoing beta decay could produce more moles then was initially present before the decay process. I calculated that 0.1 moles of H38Cl gas would remain in the system after 80 minutes, and I figured that H38Cl is a gas, but I have never seen a problem where a substance decays and produces more moles of gas then was initially present. From what I can tell, the law of conservation states that the end process of a decay or reaction can never exceed that starting material. However, in this case it did. I'm just trying to understand the initial 1:1.5 ratio; I understand what Dr. Bob222 did, but not the initial step. A clearer explanation would be appreciated.
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