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CAN SOMEONE HELP WITH QUESTION 5 PLEASE, it's driving me crazy, I cant figure out what to do :hair:

http://www-pvhs.stjohns.k12.fl.us/teachers/veatchd/12F0080F-0118C716.19/Stat%2BEq%2BWS.pdf

 

This is how far I get

 

Horizontally: T1 Cos30 = T2 Cos50 = 0 resultant force

Vertically: T1 Sin30 + T2 Sin50 = 1000N

 

But then I dont know how to work out the rest :tears:

 

Is that right so far?

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  • 1 month later...
Can someone translate: "My older sister wants to know if you know who MIKA is. He's the guy who sings "Elle Me Dit." into french for me please??

 

Ma soeur aînée veut savoir si tu sais qui est Mika. C'est le gars qui chante "Elle me dit."

 

That's how I would translate it. :thumb_yello:

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When doing enthalpy calculations with given enthalpies of combustion or formation, how do you work out whether to take products from reactants or reactants from products? My teacher really confused me about this!:blink:

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Did you make it with the "reaction triangle" or only with formulas? I do it with the reaction triangle because I get confused with the formulas all the time, too, and I think it's much easier with the reaction triangle.

 

With the reaction triangle it's only necessary that your arrows are in the right direction, it must be:

 

enthalpy1.jpg

 

And than you only need to add up the small arrows and you have your enthalpy of the reaction with the big arrow.

 

A = B + C

 

And if one arrow is in the wrong direction, you only have to switch the sign

 

enthalpy2.jpg

 

A = (-B) + C

 

For example if you want Hf (C5H10O5), and given are Hc (C5H10O5) = -2130 kJ/mol and Hf (CO2) = -394 KJ/mol and Hf (H20) = -286 kJ/mol

 

enthalpy3.jpg

 

H = (-3400) + (- - 2130) = -3400 + 2130 = - 1270 kJ/mol

 

It's "- - 2130" because the arrow is in the wrong direction.

 

And it's totally unimportant how you write your triangle, you don't need to worry to think about which reaction must be the "big-arrow-reaction", you just need to look that your arrows are in the right direction in the triangle, depending on what is given.

 

enthalpy4.jpg

 

(- H) + (-3400) = -2130

- H = 1270

H = - 1270 kJ/mol

 

Again "- H" because now this arrow is in the wrong direction.

 

I hope that was what you wanted to know and it was understandable in English :aah:

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Hi Astor, I'm doing AQA. What about you?

 

Thanks so much for your thorough answer Dermoment, it is the kind of thing I'm stuck on. I'll look again in the morning because it's making me freak out now.

 

I actually find organic chemistry much easier than Hess' cycle chemistry, but that might be because I have a different teacher for it.

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When doing enthalpy calculations with given enthalpies of combustion or formation, how do you work out whether to take products from reactants or reactants from products? My teacher really confused me about this!:blink:

 

You can always use:

SUM (Hproducts) - SUM (Hreactants)

it's pretty easy:

for example:find the standard enthalpy of reaction:4NH3+3O2= 2N2+6H2O

delta H for H2O:-242

delta H for NH3:-46

solve:4NH3+3O2=4X(-46)+(3X0)=-184

2N2+6H2O=(2X0)+6X(-242)=-1452

delta H total:SUM (Hproducts) - SUM (Hreactants)=(-1452)-(-184)=-1268 KJ

It can be confusing sometimes,actually,it was the only reason I couldn't get world gold medal for chemistry 2 years ago!

Edited by Yasi
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You can always use:

SUM (Hproducts) - SUM (Hreactants)

it's pretty easy:

for example:find the standard enthalpy of reaction:4NH3+3O2= 2N2+6H2O

delta H for H2O:-242

delta H for NH3:-46

solve:4NH3+3O2=4X(-46)+(3X0)=-184

2N2+6H2O=(2X0)+6X(-242)=-1452

delta H total:SUM (Hproducts) - SUM (Hreactants)=(-1452)-(-184)=-1268 KJ

It can be confusing sometimes,actually,it was the only reason I couldn't get world gold medal for chemistry 2 years ago!

 

The problem is that you sometimes have to do reactants-products! I think if you're asked enthalpy of combustion you do reactants-products and you do products-reactants for enthalpy of formation. I hope that's the right way round!

 

The other problem for me is when they ask you to calculate one of the values in the triangle!

 

Glad I'm not the only one who finds it confusing! Thanks for your help. :thumb_yello:

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The problem is that you sometimes have to do reactants-products! I think if you're asked enthalpy of combustion you do reactants-products and you do products-reactants for enthalpy of formation. I hope that's the right way round!

 

The other problem for me is when they ask you to calculate one of the values in the triangle!

 

Glad I'm not the only one who finds it confusing! Thanks for your help. :thumb_yello:

 

If the data is entalpies of formation, it's products minus reactants.

If the data is enthalpies of combustion, it's reactants minus products.

Sorry,I don't remember anything about triangle,actually,we never solved problems that way!

your welcome,hope it helped:thumb_yello:

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Hi Astor, I'm doing AQA. What about you?

 

Thanks so much for your thorough answer Dermoment, it is the kind of thing I'm stuck on. I'll look again in the morning because it's making me freak out now.

 

I actually find organic chemistry much easier than Hess' cycle chemistry, but that might be because I have a different teacher for it.

I'm on OCR - dayum, I wondered if we could on be the same one :teehee:

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