You are Here: Home

# Super quick chemistry question. watch

1. [CHEM4 AQA Equilibria] but i guess other people can do it too.

Kc = [A]^2[B]/[C][D]^3

FiND the units of Kc.

So, if you do it this way (adding the powers and subtracting), you get -1. (2+1) - (3+1). So (mol dm^-3)^-1. So, are the units mol-1 dm^3

Or if you just cancel out the powers, don't you get 1/mol dm^-3

Or am I just plain confused?
2. lool you are just plain confused

btw im clueless so don't quote me
3. (Original post by Ayostunner)
lool you are just plain confused

btw im clueless so don't quote me
I know, my fellow T.I. lover, I know.
4. (Original post by Get Cape.Wear Cape.Fly.)
[CHEM4 AQA Equilibria] but i guess other people can do it too.

Kc = [A]^2[B]/[C][D]^3

FiND the units of Kc.

So, if you do it this way (adding the powers and subtracting), you get -1. (2+1) - (3+1). So (mol dm^-3)^-1. So, are the units mol-1 dm^3

Or if you just cancel out the powers, don't you get 1/mol dm^-3

Or am I just plain confused?
each concentration is measured in mol dm-3

so you can go cancelling down the powers at the top and the bottom remembering that [A] means mol dm[<sup]-3[/sup] raised to the power of 1.

[A]^2[B]/[C][D]^3

is raised to the power of 3 at the top and 4 at the bottom so overall it is raised to the power of -1

Hence the units are mol-1 dm3
5. (Original post by Get Cape.Wear Cape.Fly.)
[CHEM4 AQA Equilibria] but i guess other people can do it too.

Kc = [A]^2[B]/[C][D]^3

FiND the units of Kc.

So, if you do it this way (adding the powers and subtracting), you get -1. (2+1) - (3+1). So (mol dm^-3)^-1. So, are the units mol-1 dm^3

Or if you just cancel out the powers, don't you get 1/mol dm^-3

Or am I just plain confused?
Yeah you are right, just look in terms of powers, that power of 1's will cancel. The 2 will cancel the 2 of 3 from the [d]^3 leaving with 1/mol dm^-3 = dm^3 mol^-1
6. I get the same from both way, mol^-1 dm^3.
7. (Original post by charco)
each concentration is measured in mol dm-3

so you can go cancelling down the powers at the top and the bottom remembering that [A] means mol dm[<sup]-3[/sup] raised to the power of 1.

[A]^2[B]/[C][D]^3

is raised to the power of 3 at the top and 4 at the bottom so overall it is raised to the power of -1

Hence the units are mol-1 dm3
That's what I thought, but the answer says mol dm^-3. A book mistake?
8. (Original post by Preasure)
I get the same from both way, mol^-1 dm^3.
Ok goood, maybe the book is wrong ...again
9. (Original post by Get Cape.Wear Cape.Fly.)
That's what I thought, but the answer says mol dm^-3. A book mistake?
yup, book typo
10. (Original post by Get Cape.Wear Cape.Fly.)
That's what I thought, but the answer says mol dm^-3. A book mistake?
Yeah, the book has a made a mistake which isn't surprising to me
11. i don't know why we even bother with that book tbh...

The damn thing seems to think that N2H2 is Ammonia :\

(p94 if you're sad enough to want to check )
12. (Original post by Bslforever)
i don't know why we even bother with that book tbh...

The damn thing seems to think that N2H2 is Ammonia :\

(p94 if you're sad enough to want to check )
haha! I noticed that too! I actually sat there thinking about it for about 3 minutes
13. (Original post by Get Cape.Wear Cape.Fly.)
haha! I noticed that too! I actually sat there thinking about it for about 3 minutes
I just love the part at the back of the book above the blurb where it says

Written by the experts
Checked by the examiners

I honestly hope the examiners show a little more care when checking my paper tbh...
14. (Original post by Bslforever)
I just love the part at the back of the book above the blurb where it says

Written by the experts
Checked by the examiners

I honestly hope the examiners show a little more care when checking my paper tbh...
Agreed
15. That VERY same mistake was made in the OFFICIAL MARK SCHEME.

I spotted it a while back, lemme find the thread.

EDIT: Here
16. mol-1dm3
17. Treat it as indices. The way I do it is the following:

Cancel down as approprate, then convert to or

Of course you coould just multiply in the 1st step and do

Hope that helps!

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 17, 2010
Today on TSR

### He lied about his age

Thought he was 19... really he's 14

### University open days

Wed, 25 Jul '18
2. University of Buckingham
Wed, 25 Jul '18
3. Bournemouth University