the dropper will have markings of 1-3 cm3.that would come aroung like 20 drops
in case it does not... 15-20 drops then.
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
the dropper will have markings of 1-3 cm3.that would come aroung like 20 drops
owkaythe dropper will have markings of 1-3 cm3.
in case it does not... 15-20 drops then.
Hello, why are propene and methane .25% respectively? How do you know that they are equally .25%? It could be .15 and .35 also too, right?oh goog... look.. we have 3 products.. ethene methane and propene...
if i mole is produced by cracking, then o.5% will be of ethene(as mentioned in the q) then 0.25 of propene and .25 of methane is produced... .25+.25+.5=1...
to convert .25 into proper fraction, we need to multiply it by 4 to bring it to a whole number... by doing this.. we will get 1 mole of propene, 1 mole of methane and 2 moles of ethene(.5*4).. no of Carbons now we have are 8 all together... so B is the answer...
4)http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf
Q38, Q4,Q5? Doubt can anyone, i will be thankfull to him
Well 39 it is B coz
Catalytic converter converts oxides of nitrogen to nitrogen gas
and CO and unburnt hydrocarbons to CO2 which means it is oxidation.
No idea why 24 is wrong :')
View attachment 57326
Why part iii no change?
Wouldn't the equlibrium shift to the left so the potential will decrease so wouldn't the Ecell increase?
View attachment 57327
mark scheme says concentration of fe3+ is 0.2 but question is showing 0.1... how is it 0.2??
Also for Ka don't you multiply Ka by the concentration of the solution and not just Fe3+ ions??
4)
View attachment 57351
5)
Shape of BF3 is trigonal planner so option B and D are ruled off.
Due to presence of lone pair at nitrogen atom will repel B - F bond pairs hence its shape will not be trigonal planner its bond angle will decrease so A cannot be answer hence its C.
38)
Only 1 is correct.
First step will be Nucleophilic substitution reaction. Cl will be replaced by CN, on hydrolysis of CN with NaOH(aq) it will give COO-Na+
2 is incorrect, I guess there wont be any such reaction.
3 is incorrect as Cl will be replaced by OH.
2) Use of N from NPK will be 15/60i got it , thx for taking ur time for solving.
by the way which variant of papper 1 are u giving?
i also have doubt in q2 and q15 same papper
To balance the number of electron, you need to multiply second equation with 2.
- The following half reactions occur when potassium iodate(V), KIO3, in hydrochloric acid solution oxidises iodine to ICl –.
IO– +2Cl– +6H+ +4e– →ICl– +3HO 322
I +4Cl– →2ICl– +2e– 22
What is the ratio of IO – to I in the balanced chemical equation for the overall reaction? 32
A1:1 B1:2 C1:4 D2:1
Initial Moles = IM and eq mole = EM9 Nitrogen reacts with hydrogen to produce ammonia.
N2(g) + 3H2(g) 2NH3(g)
A mixture of 2.00mol of nitrogen, 6.00mol of hydrogen, and 2.40mol of ammonia is allowed to reach equilibrium in a sealed vessel of volume 1 dm3 under certain conditions. It was found that 2.32 mol of nitrogen were present in the equilibrium mixture.
What is the value of Kc under these conditions?
(1.76)2 (2.32)(6.96)3
(1.76)2 (2.32)(6.32)3
(2.08)2 (2.32)(6.32)3
(2.40)2 (2.32)(6.00)3
Na reacts with both alcohols and acids as well. So 4 moles of Na is there.26 How many moles of hydrogen, H2, are evolved when an excess of sodium metal is added to one mole of citric acid?
CO2H
HOCCH2CO2H
CH2CO2H citric acid
A1B2C3D4
Q15 how u find total mass?2) Use of N from NPK will be 15/60
Moles of N = (15/60)/14
Concentration of N in fertilizer : (15/60)/(14*5) = 0.03
15) CaCO3 --------> CaO + CO2
Mr of CaCO3 = 100.1
Total mass of CaCO3 = 1200g
n of CaCO3 used in total = 1200/100.1 = 11.9
1 : 1 ratio of CaCO3 and CO2
Mr of CO2 = 44
m of CO2 = 44 * 11.9 = round up to 527g
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now