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because the concentration of the limiting reactant is not changing
Its B now am asking why B not D
View attachment 44033
HELP! Answer is C. Does it have something to do with the proportions of the acid and base?
Ok thanks a lotbecause the concentration of the limiting reactant is not changing
Thank you so muchActually, since the limiting reactant here is zinc (solid), it so would be more correct to use "mass or moles" being the same between the two experiments rather than "concentration"
The is no change in the
1) particle size of solid reactant (zinc)
2) concentration of the aqueous reactant (acid)
3 )initial temperature of reaction
Thanks a lot!Heat of neutralization for strong acids and strong alkali is estimated to be -57kJ PER MOLE of H2O formed.
For both experiments. (Involving strong acid and bases) , it reflects twos moles of water formed .
So heat of reaction is -57 x 2 for both equations.
CAn you explain of X againW: reacts with Br2 and uv to form intermediate CH3CHBrCH3 , which then reacts with CN-
X: reacts with Br2 and uv to form intermediate (CH3)2C(OH)Br ,which then reacts with CN-
Y: reacts with HBr to form intermediate CH3CHBrCH3 , which then reacts with aq. NaOH
Z: undergoes hydrolysis to form intermediate CH3CH2CH2OH , which then reacts with PBr3
Notice that the H on the centre Carbon is ultimately replaced by CN?CAn you explain of X again
okNotice that the H on the centre Carbon is ultimately replaced by CN?
This would require us to change the H with a halogen (either Br or Cl), one way to do it would be free radical substitution.
Then the halogen alkane intermediate will undergo a nucleophilic substitution by adding CN- (using NaCN or KCN)
Daaaang. Thanks again mateW: reacts with Br2 and uv to form intermediate CH3CHBrCH3 , which then reacts with CN-
X: reacts with Br2 and uv to form intermediate (CH3)2C(OH)Br ,which then reacts with CN-
Y: reacts with HBr to form intermediate CH3CHBrCH3 , which then reacts with aq. NaOH
Z: undergoes hydrolysis to form intermediate CH3CH2CH2OH , which then reacts with PBr3
No problem, I'm actually a tutor based in Singapore, so I guess we are neighbors.Daaaang. Thanks again mate
No problem, I'm actually a tutor based in Singapore, so I guess we are neighbors.
Would like to get an understanding of students exams schedule in other countries, so in Malaysia, both the A and O levels (?) are done in May and not November?
C40H82 ---> C16H34 + 2C12H24
How would you calculate the bond energy of this reaction? (bond energy C-C 350, C=C 610, C-H 410)
Is this the actual question?
We'll need to make some reasonable assumptions here that the C12H24 is an alkene rather than a cycloalkane. The C12H24 would be having one C=C bond as it is "missing" 2 hydrogen
Energy taken in to break bonds of C40H82
82 (C-H) and 39 (C-C) bonds
= 82 (410) + 39 (350)
=47270 kJ
Energy given out in forming bonds of C16H34 and 2 C12H24
34 (C-H) and 15 (C-C) bonds and 2 (C=C) and 20 (C-C bonds) and 48 (C-H)
= 34 (410) + 15 (350) + 2 (610) + 20 (350) + 48(410)
=46680 kJ
Enthalpy change = +47270 - 46680 = +590kJ/mol
Quite a bit of numbers at this late hour, hope I didn't make a mistake. There is a slightly shorter way, but I think this way should be more understandable.
Hello ! can someone plz explain the following : Q 4 , 5, 14,39 in
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf
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