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Chemistry Paper 5 question

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TJ-56:
Can someone please help me with q2 (f) May June 2008?
My x and y values are 5.70 and 4.00 respectively, but what should I do next??
Please reply asap, exam is tomorrow
thanx in advance

Sue T:
x and y are what - depends wat u plotted on the x and y axis (if u've plotted gases released then i really cant be of any help - lots of ppl asked about ths - u never really get wat th markin scheme says but- i myt be able 2 help a bit if u tell me wats wat...)
hurry!

TJ-56:

--- Quote from: Sue T on May 15, 2011, 06:43:50 pm ---x and y are what - depends wat u plotted on the x and y axis (if u've plotted gases released then i really cant be of any help - lots of ppl asked about ths - u never really get wat th markin scheme says but- i myt be able 2 help a bit if u tell me wats wat...)
hurry!

--- End quote ---

Ok sorry for not being clearer and thanx for helping out.
So, my y-axis is the mass of CuO (referring to the headings of the table on the opposite page of the graph: C - A ) and the x-axis is the mass of the basic carbonate ( B - A ). So the 2 required values as I stated are 4 (y axis) and 5.70 (x axis), what do you suggest I do next?
And thanx again, I really appreciate your help.

leebux101:
ok for this question , here is what i did >>
get the gradient of the line using any values on the graph.
gradient = 0.693 for me
gradient is mass of copper oxide deposited / mass of basic carbonate (this is what i plotted on the y and x axis respectively)
so 0.693 = mass of copper oxide deposited / mass of basic carbonate
now mass = number of moles * Mr
so 0.693 = 2*Mr of Cuo / 1*Mr of carbonate  <<<< i got the moles from the mole ratio in the equation.
u will get 0.693 = 159/(221+18x)
now solve for x , ur x value should be around 0.5
if u dont get it please tell me ok?

leebux101:
did u get the value of x?

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