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It solely depends on what reaction happens, just record your values and plot them.
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thankyou bt umm how do we knw k where we hav to draw the odr 2nd line to gt a intersection point :OI guess a picture will say it all, just look:
http://img.ctrlv.in.s3.amazonaws.com/img/519d426bc59ba.jpg
Ufffff, please listen carefully once so no questions will arise (beat lol).thankyou bt umm how do we knw k where we hav to draw the odr 2nd line to gt a intersection point :O
It is an orange-brown gas.When N02 from nitrate is formed how do we test that?
For the NH3 we add the NAoh and al foil and when we place the litmus at the mouth of the tube?
How to we make thw litmus damp? xP like sprinkle it with water?
oh :O okay the horizontal nd the rising points dey make intersecting lines ........thankyouuu so much sorrry fr being such a dumb :/Ufffff, please listen carefully once so no questions will arise (beat lol).
The results you get show and increase in temperature (line goes towards up), and then suddenly they become constant (remains horizontal) or it will go down.
If you look at the points, all the points horizontal at the end are one line, the ones increasing from initial reading are the other.
r u sure this will come can u giv the values btw i didnt get where we hsould plot these points
Oxidation of Mn in MnO4- is +7 (x+(4*-2)=-1)Could anyone from here possibly tell me how to solve http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s09_qp_32.pdf Question 1, part c? The part where we have to balance the equation.
thankyouuu so much martin flakes my paper was awesum thanks fr explaining me that graph partOxidation of Mn in MnO4- is +7 (x+(4*-2)=-1)
And its oxidation number changes to Mn2+ Change is from +7 to +2 (change: 5)
--
Second half equation, x+(4*-2)=-2 C is +6 in left side
And (x+(2*-2)=0) so x is +4 on right side Change is from +6 to +4 (change: 2)
--
Then cross multiply the changes with half equations so you will get
2 MNO4- + 16H+ + 10e- --> 2 Mn2+ + 8H2O
5 C2O4^2- --> 10 CO2 + 10 e-
=========
Once you get number of moles of C2O42- by cross multiplying with moles of KMnO4, you can calculate the concentration of FB3.
Then for finding the number of moles of initial undiluted, you cross multiply the moles of sample taken for titration with 250 cm3 like
1 mol --> 25 cm3
x mol --> 250 cm3
then it will give u a new mol value.
Now since in volumetric flask you just added water, it contains the same number of moles as in undiluted.
So just use c = n / v and use V as volume first step added of undiluted by burette.
Good luck!
Atta Boy, That's Pure Desire!It took mr forever in the first question, adding excess reagent and leaving the b****te open while throwing away the wrong exp.... Gonna lose about 7/6 marks... But I fuhckhing swear imma rape paper 1 so hard! X(
+/- 2Nevermind answers! I just want to know What's the maximum difference we can have between our value and examiners value? (To be in the range of accpetable)
Perfect! Thanks. I got my titration 25.40 while our Examiner got 23.50+/- 2
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