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Chemistry: Post your doubts here!

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Sorry! Had to go out for something! I labelled X (mass of basic carbonate) and Y (mass of CuO)
alright see n= m/mr so for the basic carbonate nMr = m which is 1x (221+18x) and for the CuO m = 2 x 79.5 = 159

now mass of basic carbonate/ mass of CuO i = gradient..substitute the values and find x. did u get it ? make sure u make a big triangle on ur graph when finding your gradient to make the answer as accurate as u can
 
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alright see n= m/mr so for the basic carbonate nMr = m which is 1x (221+18x) and for the CuO m = 2 x 79.5 = 159

now mass of basic carbonate/ mass of CuO i = gradient..substitute the values and find x. did u get it ? make sure u make a big triangle on ur graph when finding your gradient to make the answer as accurate as u can

OHHHH GOT IT!!! Thanks! I just wasn't understanding how to use the gradient in it!!! Thank you!!!
 
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alright see n= m/mr so for the basic carbonate nMr = m which is 1x (221+18x) and for the CuO m = 2 x 79.5 = 159

now mass of basic carbonate/ mass of CuO i = gradient..substitute the values and find x. did u get it ? make sure u make a big triangle on ur graph when finding your gradient to make the answer as accurate as u can

Could you help me with ON 2007 Q1 f also?? I don't understand the equation!!
 
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how to know if a substance is a reducing or oxidising agent. For example: how is CO a reducing agent.
 
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how to know if a substance is a reducing or oxidising agent. For example: how is CO a reducing agent.
if a substance is oxidized then it is a reducing agent and if it is reduced then it is an oxidising agent
 
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Aww that's fine. Could you help me before Friday?
ofc!! because the next coming days i will be doing paper 5 mostly and i will need u and all the other paper 5 candidates to discuss stuff about it and doubts like how yesterday we kept discussing. i guess this will help us to score a good grade, right ?
 
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if you made your calculations above correctly then you will have 0.04 moles of NaOH that reacted with ethanoic acid which means at equilibrium we will have 0.1-0.04= 0.06 moles of ethanoic acid at equilibrium. acid to alcohol ratio is 1:1 so also 0.06 moles of alcohol left at equilibrium and 0+0.04 = 0.04 moles of water and sodium ethanoate al equil.

THIS IS THE LOGICAL WAY. i don't know why the mark scheme made the opposite anyone out here can explain ??
 
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if you made your calculations above correctly then you will have 0.04 moles of NaOH that reacted with ethanoic acid which means at equilibrium we will have 0.1-0.04= 0.06 moles of ethanoic acid at equilibrium. acid to alcohol ratio is 1:1 so also 0.06 moles of alcohol left at equilibrium and 0+0.04 = 0.04 moles of water and sodium ethanoate al equil.

THIS IS THE LOGICAL WAY. i don't know why the mark scheme made the opposite anyone out here can explain ??
you mean what i said was wrong?
 
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i mean the calculation to number i c(i) ?
anyways any help with organic chemitry?
 
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