Chemistry 2006/ 2007 Club***

romansten9 · Oct 3, 2006

I don't know the answer to the density question, we aren't to that point in our text book. But I do know that temperature is going to change the density. I would suggest if we are going to tutor each other, that we don't "guess" This stuff can be confusing enough! So make sure you know what you're doing if you attempt to help someone else, or you'll confuse them worse! thanks! : )

bon jovi · Oct 4, 2006

ok...Hello all... I have little problem...I have to use heat of fusion, SH of water, and/or heat of vaporization...

a) cal need to melt 50.0g of ice at 0C and to warm liquid to 65C (need 2 steps for this)

this is what i thought i should do...

I need for 1g H2O-80 cal.

80cal/1g h2o * 50.0g = 4000 cal

then : (65C*50g)*1.00cal/gC=3250cal

4000+3250=7250 cal...

however, resolt in the book gives me 7300 cal...What I am doing wrong?:uhoh21:

also

kJ given off when 15g of steam condenses at 100C and the liquid cools to 0C

g*540cal/g=15*540=8100cal = 33.91 kJ

result in the book is 40kJ...again what I am doing wrong? Please HELP.... :confused:

thanks

Bon Jovi:Melody:

abrainerd · Oct 4, 2006

bon jovi, you ARE doing these correctly, you just are not remembering to round off to the significant figure. 7250 rounds off to 7300 and 33.91 would round off to 40kj. Your answers are correct.

bon jovi · Oct 4, 2006

bon jovi, you ARE doing these correctly, you just are not remembering to round off to the significant figure. 7250 rounds off to 7300 and 33.91 would round off to 40kj. Your answers are correct.

:bowingpur

You just made my day! I thought that I was totaly clooless on this....

Thanks a lot

miranda819, BSN · Oct 4, 2006

Can anyone help me with this question?

How many grams of Cl2 are needed to produce 1.50 moles of chloroform (CHCl3)?

CH1 + 3Cl2 -> CHCl3 + 3HCl

I've got the 1.50 moles of CHCl3 and the molecular weight of CHCl3 is 119.37g, but I don't know where to plug these in to get the grams for Cl2. Help please!

Thank you!

catzy5 · Oct 4, 2006

Can anyone help me with this question?
How many grams of Cl2 are needed to produce 1.50 moles of chloroform (CHCl3)?
CH1 + 3Cl2 -> CHCl3 + 3HCl
I've got the 1.50 moles of CHCl3 and the molecular weight of CHCl3 is 119.37g, but I don't know where to plug these in to get the grams for Cl2. Help please!
Thank you!

I believe what you need is the molecular weight of Cl2.

this is a Mole to Mole relationship. You want to know how many grams of Cl2 you will need to produce 1.50 moles of CHCl3

so you would set it up like this.

1.50 mol CHCl3 X mole/mole ratio (3MolesCl2/1Mol Chcl3)

next take that answer of moles of Cl2 X molecular mass of Cl2

I hope that helps. Your answer should end up in Grams of Cl2

miranda819, BSN · Oct 4, 2006

Thanks for your help! I think I finally got it!

1.50 mol CHCl3 x 3 mol Cl3/1 mol CHCl3 = 4.50 mol Cl2

then I take that answer 4.50 mol Cl2 x 70.9 Cl2/1 mol Cl2 = 319 g of Cl2

Does that seem right?

Thanks again for the help!

miranda819, BSN · Oct 4, 2006

Now I'm completely lost on this one:

Given this reaction:

2Fe2O3 + 6C + 3O2 -> 4Fe + 6CO2

How many g of C are necessary to react completley with 0.58 g of Fe2O3?

Where do I even begin?

catzy5 · Oct 4, 2006

Thanks for your help! I think I finally got it!
1.50 mol CHCl3 x 3 mol Cl3/1 mol CHCl3 = 4.50 mol Cl2
then I take that answer 4.50 mol Cl2 x 70.9 Cl2/1 mol Cl2 = 319 g of Cl2
Does that seem right?
Thanks again for the help!

yup thats what I got.

catzy5 · Oct 4, 2006

Now I'm completely lost on this one:
Given this reaction:
2Fe2O3 + 6C + 3O2 -> 4Fe + 6CO2
How many g of C are necessary to react completley with 0.58 g of Fe2O3?
Where do I even begin?

its the same exact thing its a mole to mole relationship the only thing different is you are going from grams to moles.

so if you change 0.58g Fe2O3 to moles then do your mole/ratio just lke before.

use that answer of moles of Carbon and convert moles to grams.

ok

it always looks overwhelming but break it down step by step its the exact same problems just take a deep breath and work out what you know.

miranda819, BSN · Oct 4, 2006

Thank you again! I took a step back and redid the problem and came up with 0.13 g of C.

I think I need to stay slow and steady, I'm starting to get it! Yeah!!

sim2002 · Oct 6, 2006

Hello all,

I am in the same boat with everyone. Chem is Ok for now. It is really nice to know that we all are going thru the same things.

Keep positive.