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Q9- only bonds are forming here. One X-X bond and 6 X-H bonds. so 6*395=2370http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Chemistry%20(9701)/9701_s11_qp_13.pdf
anyone pls tell me how to do ques. 9 and 12
Jazak allah khair!
Q9- only bonds are forming here. One X-X bond and 6 X-H bonds. so 6*395=2370
energy to form X-X + energy to form X-H = -2775
X-X - 2370=-2775
X-X=-405
Instead of - we put a + sign because they asked for bond energy and bond energy is always +ve according to its definition. so bond energy= +405
Q12- 4Al + 3O2 --> 2Al2O3
0.02 moles of Al were used so find moles of Al2O3. its in 4:2 ratio so 0.01 moles are formed of Al2O3.
Al2O3 + 6HCl --> 2AlCl3 + 3H2O
0.01 moles of Al2O3 need 0.06 moles of HCl.
moles = V*C
0.06/2 = 0.03dm^3 = 30cm^3
heey i hate chemistry and i dont wanna fail i am doing varient 11 any help ??
thnxx in advance
in HIO I has oxidation number of +1I thought the question would be easy but couldnt solve this please help!
View attachment 28416
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w12_qp_11.pdf
Can anyone solve Q:11,15,16,17,20
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
can anyone solve:Q:7,24,37
plz anyone
Thanks you made it easy!in HIO I has oxidation number of +1
in HIO3 I has oxidation number of +5
in I2 I has oxidation number of 0.
look the oxidation number of element in L.H.S=R.H.S
so multiply HIO by 5 to get +5 oxidation number thus balancing oxidation number in LHS and RHS.
as there are 5 iodine ions in left so there should be 5 iodine ions in right.
n=2 p=1 get it now?
Thanks37...: the voltmeter is connected across both resistors R and Q hence it shows a reading of 4 V even if the slider position is changed....
my pleasure mateThanks you made it easy!
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