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Chemistry: Post your doubts here!

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have you any idea of what are you saying?
what do you mean by this:
ZaqZainab said:
Listen if it has HIGH ionisation energry it mean it DOESNOT give off electrons easily and there will be less electrons in the lattice and if less electrons the pull wont be that grea and so the melting point will be low easy to melt
thn u tell me
 
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you are such a liar. you always solve all doubts and pretend like you know nothing
excuse me i know stuffs i am not that dumb i have doubts too just cause i can solve people doubts doesnt mean i know everything as the person said i can solve everything no that is wrong
and even if i could All praise to God i don't have pride
 
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can you just say what is confusing you in each question to make it easy for us to answer?
oh okay
in Q37 why the first statement is correct? I mean the empirical formula has 1 atom of O, so there must be aldehyde :/

in Q26 idk wht im doin wrond I tried it many times, but my answer isn't in the options :$ :/

in Q9 I coul make out tht glucose and fructose are structural isomers so option A and C is ruled out but cant understand how osmotic pressure is affected

Q29 don't get it at all :/

Btw its http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf
Q37 (D)
Q26 (B)
Q9 (D)
Q29 (D)
Q35 (B)
Q40 (A)
 
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there are 0.44 mol of Ag+ so there are 0.44 mol of Fe2+
it means out of 1 mol o.44 remain, rest of them have been converted into Ag and Fe3+
so there are 1-0.44=0.56 mol of Fe3+ and 0.56 mol of Ag
Kc=0.56*0.56 / 0.44*0.44 = 1.62
thats precisely what i did
but ans is D, 2.89. :/
it is obtained if we solve
0.56/0.44*0.44
but thats incorrect
 
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Something I think might be worth mentioning, in case some of you don't already know:
In an organic question, if you see LiAlH4 being used, know that it's a reducing agent, just like NaBH4. This is another one of the additions to the syllabus.
Also, H2 reduces an aldehyde to alcohol.
 
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there are 0.44 mol of Ag+ so there are 0.44 mol of Fe2+
it means out of 1 mol o.44 remain, rest of them have been converted into Ag and Fe3+
so there are 1-0.44=0.56 mol of Fe3+ and 0.56 mol of Ag
Kc=0.56*0.56 / 0.44*0.44 = 1.62
hey i have a question
in an experiment b mol of HI were put into a sealed vessel under pressure p.
ateqilibrium ,x mol of HI had dissociated
2HI--- H2 +I2
now tell how its answer is x2/4(b-x)2
i dont get how the 4 came
 
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