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### test 2 #6 steps

Posted: Fri Mar 13, 2020 1:01 pm
Can someone please explain the math steps involved to get the correct answer for this problem? I understand the set up, just not the steps you need to take to get the right answer of 0.42M. Thank you!

### Re: test 2 #6 steps

Posted: Fri Mar 13, 2020 1:10 pm
I think the mistake you did was using a wrong value in the equation, instead of using 0.257/2, you should've used 0.0592/n, where n is the number of mole of electrons.

### Re: test 2 #6 steps

Posted: Fri Mar 13, 2020 1:23 pm
Angus Wu_4G wrote:I think the mistake you did was using a wrong value in the equation, instead of using 0.257/2, you should've used 0.0592/n, where n is the number of mole of electrons.

That change in value shouldn't matter because you should still receive the same answer. I'm just having trouble with the calculus part of determining the "x" from the ln x/2.2.

### Re: test 2 #6 steps

Posted: Fri Mar 13, 2020 1:27 pm
Christineg1G wrote:Can someone please explain the math steps involved to get the correct answer for this problem? I understand the set up, just not the steps you need to take to get the right answer of 0.42M. Thank you!

Well first you’d multiply both sides by -2/0.0257 ( the reciprocal of the constant on the left side of the ln) to cancel it out. Then you get ln (x/2.2)= 0.0213(-2/0.0257). Then you exponentiate the equation and you get e^(0.0213(-2/0.0257))= x/2.2 and then you multiply each side by 2.2 and you get the value for x. You can do this bc the rule is log(base a) k = x becomes a^x =k. And for ln it’s ln(base e) k= x, which becomes e^x =k. Hope that helps.

### Re: test 2 #6 steps

Posted: Fri Mar 13, 2020 1:48 pm
Diana A 2L wrote:
Christineg1G wrote:Can someone please explain the math steps involved to get the correct answer for this problem? I understand the set up, just not the steps you need to take to get the right answer of 0.42M. Thank you!

Well first you’d multiply both sides by -2/0.0257 ( the reciprocal of the constant on the left side of the ln) to cancel it out. Then you get ln (x/2.2)= 0.0213(-2/0.0257). Then you exponentiate the equation and you get e^(0.0213(-2/0.0257))= x/2.2 and then you multiply each side by 2.2 and you get the value for x. You can do this bc the rule is log(base a) k = x becomes a^x =k. And for ln it’s ln(base e) k= x, which becomes e^x =k. Hope that helps.

Thank you!