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Higher Chemistry Formulae - HSN forum

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Higher Chemistry Formulae


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#1 Ally

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Posted 19 May 2004 - 09:31 PM

So far I can remember the following equations:

no. of moles=concentration*volume

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delta.gifH=Hp-Hr

delta.gifH=c.m.delta.gift

Q=I*t

[H+] * [OH-] = 10^-14mol^2l^-2



If you're having any difficulties feel free to post in the forum, PM me or email me.

I've put the above equations into a pdf file.

Attached Files


Edited by ally, 28 May 2005 - 10:18 PM.


#2 little_minx

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Posted 21 May 2004 - 04:26 PM

hey!

Any tips for calculations where reactants are in excess?

Laura xxx

#3 Ally

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Posted 21 May 2004 - 05:10 PM

Right usually for this - there are many methods of tackling Chemistry calculations so you may find that this method doesn't suit you - you need the balanced equation and then you find the moles of one of the reactancts (using the formula n=c*v). From this you find out the no. of moles of the other reactant/product and then from all this info you can deduce whether the reactant is in excess or not.

If you're having any specific problems with this (like from past papers) just tell me the question and I'll be happy to help you.

#4 Ally

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Posted 21 May 2004 - 07:41 PM

If you prefer doing it the other way this may be of use to you:

moles ---> particles (Refering to Avogrado's here)
particles ---> moles
volume ---> moles
moles ---> volume


#5 little_minx

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Posted 21 May 2004 - 09:21 PM

Hey!

I can't do the whole "Link" ad "Proportion" - I prefer doing it the old school way!

Thanks for your help, I might have to take you up on it!

Laura xxx

#6 mel

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Posted 24 May 2004 - 10:33 AM

Can anyone please help me with this queston??

Calculate the mass of substance required to make up the solution

50cm cubed of 4 mol l-1 KCl (aq)


I hope you can understand that, I couldnt find some of the keys for cubed, etc!



#7 Ally

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Posted 25 May 2004 - 10:53 AM

QUOTE (mel @ May 24 2004, 10:33 AM)
Can anyone please help me with this queston??

Calculate the mass of substance required to make up the solution

50cm cubed of 4 mol l-1 KCl (aq)


I hope you can understand that, I couldnt find some of the keys for cubed, etc!


Use the formula:

n=c*v (as specified as above) - remembering the volume is in litres so we have to convert it.

So the volume is 0.05 (50/1000) and the concentration is 4

Substituting this in the equation gives:

n=concentration*volume
=4*0.05
=0.2 (Number of moles)

Now use the formula n=Mass/GFM:

The number of moles is 0.2
The GFM mass is (39.1+35.5) = 74.6g - Using a data booklet

Now sub these in to find the mass:

n=Mass/GFM
Mass=n*GFM (Rearranging the equation)
=0.2*74.6
=14.92g

If anyone disagrees with this answer please tell me where I went wrong.

#8 james1

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Posted 25 May 2004 - 10:59 AM

Looks correct!

#9 eck231

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Posted 30 May 2004 - 10:54 AM

Help please...im actually useless at chemistry

CaCO3 + 2HNO3 --> Ca(NO3)2 + H2O + CO2

A 2.14g piece of calcite was added to 50 cm³ of 0.2 mol l- power1.gif nitric acid in a beaker

Calculate the mass of calcite in grams left unreacted?

#10 Ally

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Posted 30 May 2004 - 11:54 AM

QUOTE (eck231 @ May 30 2004, 10:54 AM)
Help please...im actually useless at chemistry

CaCO3  +  2HNO3  -->  Ca(NO3)2  +  H2O  +  CO2

A 2.14g piece of calcite was added to 50 cm³  of 0.2 mol l- power1.gif nitric acid in a beaker

Calculate the mass of calcite in grams left unreacted?

OK here is the answer:

1. Find the number of moles of Nitric Acid:
=>n=c*v
      =0.2*0.05
      =0.01moles

2. 2HNO3 <-> CaCO3
      2mol <-> 1mol
  0.01mol <-> 0.005mol are needed.

3. Find the number of moles of CaCO3 actually present:
=>n=mass/GFM
      =2.14/100
      =0.0214moles are present

4. So that means there are 0.0164mols (0.0214-0.005) are in excess.

5. Now calculate the mass in excess (the mass left unreacted):
=>n=mass/GFM
=>mass=n*GFM (Equation rearranged)
            =0.0164*100

            =1.64g

#11 Ally

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Posted 30 May 2004 - 11:59 AM

I hope the above explanaation helps...

If anyone wants help, ask me and I'll (hopefully) be able to give a detailed explanation. biggrin.gif

#12 eck231

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Posted 30 May 2004 - 12:10 PM

thank you! ill never be able to do it but thank you!

#13 Phoenix

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Posted 01 June 2004 - 10:05 PM

Eh, what do people find easier? Using PVC acid = PVC alkali rule for neutralisation or simply N=CV and N=CV then working from there?

I get confused between which one to use and often end up making a muck up of the question due to my panicking!

#14 Ally

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Posted 16 October 2004 - 09:43 PM

A pdf file of the equations can be found at the top of page 1

#15 Discogirl17

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Posted 03 November 2004 - 07:32 PM

Ok Ally so I see you r coping well. BUT if ANYONE at all needs please do not hesittae to PM me with your questions
Half ideas,half quality, half a million pound law suit!





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