Destroyer Help: Gen. Chem # 17

[busupshot83]

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I have a question regarding #17 in the General Chemistry section of the Destroyer.

On pg. 159, which contains the solution for the problem, focus on the following part:

...

(8 x 10^-7) = X^2
= 80 x 10^-8 = X^2
9 x 10^-4 = X = OH-

...

I undestand up until the = 80 x 10^-8 = X^2 part. Can someone explain how you go from (8 x 10^-7) = X^2 to = 80 x 10^-8 = X^2, and eventually 9 x 10^-4 = X = OH-?

 

[901preDent]

its the same thing! 80x10-8 and 8x10-7 are the same number and 9x10 -4 is the square root of that! hope this helps!

 

[901preDent]

it is done in the first place because the numbers are easier to work in your head is why they are changed in the first place.hope this helps

 

[busupshot83]

it is done in the first place because the numbers are easier to work in your head is why they are changed in the first place.hope this helps

Lol, I didn't even realize that they were the same thing. I owe you man, thank you for your prompt reply. I am now returning to studying after a couple of weeks off due to a personal emergency, and I am still missing a few marbles up in my head.

 

[DentalKitty]

I have a question regarding #17 in the General Chemistry section of the Destroyer.

On pg. 159, which contains the solution for the problem, focus on the following part:

...

(8 x 10^-7) = X^2
= 80 x 10^-8 = X^2
9 x 10^-4 = X = OH-

...

I undestand up until the = 80 x 10^-8 = X^2 part. Can someone explain how you go from (8 x 10^-7) = X^2 to = 80 x 10^-8 = X^2, and eventually 9 x 10^-4 = X = OH-?

Ok, so what you do is rearrange the scientific notation so you get closer to a number that's easier to take the square root of. 8 X 10^-7 is also equal to 80 X 10^-8 (make the front part bigger by a factor of ten, and then subtract 1 from the power of ten to balance it out). 80 is just about 81 which is 9^2. So you estimate the square root of 80 as 9 and the square root of 10^-8 as 10^-4 (divide the power by 2). This leaves you with x = 9 X 10^-4. Does that make sense?

 

[busupshot83]

Ok, so what you do is rearrange the scientific notation so you get closer to a number that's easier to take the square root of. 8 X 10^-7 is also equal to 80 X 10^-8 (make the front part bigger by a factor of ten, and then subtract 1 from the power of ten to balance it out). 80 is just about 81 which is 9^2. So you estimate the square root of 80 as 9 and the square root of 10^-8 as 10^-4 (divide the power by 2). This leaves you with x = 9 X 10^-4. Does that make sense?

Yep, thanks Kitty.

 

[busupshot83]

One more question on this problem:

How are you supposed to know that -log 9 x 10^-4 = 3 without a calculator? I asked two of my friends who are math tutors, and they were stumped.

 

[901preDent]

Well you arent exactly but i learned from an awesome indian chemistry professor ( shout out Dr. DAss) who could do logs in his head that if you are taking the log of a number like that raised to a power the log is always in close prox. tho the powere which it is raised! so like there the log of that number raised to -4 is close to 4 or in this case 3. hope this helps!

 

[901preDent]

i think this is the case. I am no math major at ALL so dont take my word for it!! later😎

 

[busupshot83]

Well you arent exactly but i learned from an awesome indian chemistry professor ( shout out Dr. DAss) who could do logs in his head that if you are taking the log of a number like that raised to a power the log is always in close prox. tho the powere which it is raised! so like there the log of that number raised to -4 is close to 4 or in this case 3. hope this helps!

Dr. Dass... I love you man.

 

[joonkimdds]

Well you arent exactly but i learned from an awesome indian chemistry professor ( shout out Dr. DAss) who could do logs in his head that if you are taking the log of a number like that raised to a power the log is always in close prox. tho the powere which it is raised! so like there the log of that number raised to -4 is close to 4 or in this case 3. hope this helps!

The Kaplan has different method than Dr. Dass.

every log bigger than 3 is bigger than 0.5
so...4-(0.5 or bigger number) is closer to 3 not 4.

but since it had negative in front of log, i think the correct answer should be -3, not +3.

 

[joefosho315]

Dr. Dass... I love you man.

There are 2 ways that you can think about this log problem. The first way is basically Dr. Dass's way, but I'll give a more detailed explanation (from Kaplan's blue book actually).

-log(n x 10^-m) = m - log👎

and since n will be between 1 and 10 (because it's using scientific notation), log👎 should be between 0 and 1. Thus, m - log👎 will always be between (m-1) and m. So applied to this problem, we have n = 9 and m = 4. So those values, once plugged in, will equal 4 - log(9) which approximately equals 3, since log(9) is really close to 1 (remember, log(10) = 1 while log(1) =0).

So that's the Dr. Dass/Kaplan way to do it.

Another way you can think of this log problem is that (9 x 10^-4) is approximately equal to (10 x 10^-4). This scientific notation rearranged then equals (1 x 10^-3). Thus, it should be easy to see that log(1 x 10^-3) equals 3.


Hope that helped!

P.S. - Just to make sure, we're not allowed to bring any calculators to the test right?

 

[joonkimdds]

There are 2 ways that you can think about this log problem. The first way is basically Dr. Dass's way, but I'll give a more detailed explanation (from Kaplan's blue book actually).

-log(n x 10^-m) = m - log👎

and since n will be between 1 and 10 (because it's using scientific notation), log👎 should be between 0 and 1. Thus, m - log👎 will always be between (m-1) and m. So applied to this problem, we have n = 9 and m = 4. So those values, once plugged in, will equal 4 - log(9) which approximately equals 3, since log(9) is really close to 1 (remember, log(10) = 1 while log(1) =0).

So that's the Dr. Dass/Kaplan way to do it.

Another way you can think of this log problem is that (9 x 10^-4) is approximately equal to (10 x 10^-4). This scientific notation rearranged then equals (1 x 10^-3). Thus, it should be easy to see that log(1 x 10^-3) equals 3.


Hope that helped!

P.S. - Just to make sure, we're not allowed to bring any calculators to the test right?

could you restate that it's Dr.Dass/JoonKimD.D.S. way to do it?
Kaplan taught me what the log is, but I came up with the idea of log(bigger than 3 = bigger than 0.5 logic. I deserve a credit

by the way, there was negative sign in front of log, so i belive the answer is -3, not +3 ?

 

[joefosho315]

could you restate that it's Dr.Dass/JoonKimD.D.S. way to do it?
Kaplan taught me what the log is, but I came up with the idea of log(bigger than 3 = bigger than 0.5 logic. I deserve a credit

by the way, there was negative sign in front of log, so i belive the answer is -3, not +3 ?

Hey Joon,

Actually, if you look at the Kaplan equation closely, it'll give the following:

-log(n x 10^-m) = m - log👎

So if you plug in m and n as I did above, you should get m - log👎 = 4 - log(9) which approximately equals 3. The negative sign is already included as part of the equation. Let me know if this doesn't make sense.

 

[DentalKitty]

could you restate that it's Dr.Dass/JoonKimD.D.S. way to do it?
Kaplan taught me what the log is, but I came up with the idea of log(bigger than 3 = bigger than 0.5 logic. I deserve a credit

by the way, there was negative sign in front of log, so i belive the answer is -3, not +3 ?

Nope, log of 10^-3 = -3 (10 is raised to the -3) so -log 10^-3 = +3. You have to remember to factor in the sign of the power...

 

[levyusupov]

The Kaplan has different method than Dr. Dass.

every log bigger than 3 is bigger than 0.5
so...4-(0.5 or bigger number) is closer to 3 not 4.

but since it had negative in front of log, i think the correct answer should be -3, not +3.

no, you got it in the reverse way. just because there is a negative sign in front of the log, the answer is positive. not the reverse !!!!