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Try to solve the question .ill ask my sir tomm too !!
Btw its from year 1997
pfft...and how do you supose we do that??
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Try to solve the question .ill ask my sir tomm too !!
Btw its from year 1997
I am not attempting this one right now ......ill ask sir tommo....u do it by molar formulas !pfft...and how do you supose we do that??
I am not attempting this one right now ......ill ask sir tommo....u do it by molar formulas !
Fine and if u have any difficult questions do share them !way too lazy to do that!!
Which past paper ..?It takes 45 cm^3 of H3PO4 to neutralize 25 cm^3 of 1.8 mol/dm^3 of KOH. What's the concentration of H3PO4? I know how to do it but the answer i was getting wasn't matching the ms... (can't post it here because it was an old past paper)
You seriously don't wanna knowWhich past paper ..?
It takes 45 cm^3 of H3PO4 to neutralize 25 cm^3 of 1.8 mol/dm^3 of KOH. What's the concentration of H3PO4? I know how to do it but the answer i was getting wasn't matching the ms... (can't post it here because it was an old past paper)
First get the number of moles of KOH by doing this:It takes 45 cm^3 of H3PO4 to neutralize 25 cm^3 of 1.8 mol/dm^3 of KOH. What's the concentration of H3PO4? I know how to do it but the answer i was getting wasn't matching the ms... (can't post it here because it was an old past paper)
0.3 (and i am getting 0.3333333333) and they said in the instructions that all numerical answers should be to 2 d.p.) =/What is the answer in the marking scheme?
What answer did u get and whts in da mark scheme ?You seriously don't wanna know
I got the same answer !0.3 (and i am getting 0.3333333333) and they said in the instructions that all numerical answers should be to 2 d.p.) =/
The answer that came out when i calculated is 0.33 mol/dm3It takes 45 cm^3 of H3PO4 to neutralize 25 cm^3 of 1.8 mol/dm^3 of KOH. What's the concentration of H3PO4? I know how to do it but the answer i was getting wasn't matching the ms... (can't post it here because it was an old past paper)
That is the correct answer, if you write 0.33 it wouldnt matter because 0.3 has been obtained by rounding off 0.33 according to me0.3 (and i am getting 0.3333333333) and they said in the instructions that all numerical answers should be to 2 d.p.) =/
When u divide 0.015/0.045.thn u get 0.33333333333but the answer in mark scheme is only 0.3 !The answer that came out when i calculated is 0.33 mol/dm3
First find the number of moles of KOH that comes out as 0.045 moles
According to the equation 1 mol of H3PO4 reacts with 3 moles of KOH so by the ratio method there are 0.015 moles of H3PO4 in 45cm3.
As there are 0.015 moles in 45cm3 so there will 0.33 moles in 1000 cm3 that is the concentration of the acid: 0.33mol/dm3
I see, this focuses on reducing agent, the higher the percentage of SO2 the quicker KMnO4 would be decolourize.Hmmm... well i guess that would be an error then =/
Ok here's another one:
You have 3 types of coal. All three of them have sulphur in different amounts which burn to produce SO2. This gas reacts with KMnO4 and turns it's colour from purple to colorless. Which one has the most sulphur in it? 6 mark question so you have to mention all the apparatus and all. I got 4 because of lack of time (and a teacher who seemed to have a phadda with her husband or something) although i still don't get what i missed So, help required. Thanks
Even eighth graders would 'see' that try elaborating pleaseI see, this focuses on reducing agent, the higher the percentage of SO2 the quicker KMnO4 would be decolourize.
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