Mock for CE guys 2 (ans)

[H]bunbunbunbun

你show一下點change in O.N.......
C本身唔係0咩?

16.
個樣似係CHCH form出黎
點睇都唔似係其他= =

piyopiyo

C點會係0[[tk_30]]
C係 +4...

[ 本帖最後由 piyopiyo 於 27-4-2008 10:21 編輯 ]

[H]bunbunbunbun

why+4?
咁你點知H同O係咩

piyopiyo

原帖由 [H]bunbunbunbun 於 27-4-2008 10:30 發表
why+4?
咁你點知H同O係咩

唔知wor[[tk_30]]  做果陣只是直覺覺得是redox
但而家唸唸下有好似唔係...
呢個係dehydration ...
除非 H是 -1...


http://en.wikipedia.org/wiki/Organic_oxidation

[ 本帖最後由 piyopiyo 於 27-4-2008 10:58 編輯 ]

piyopiyo

第2題 why not A[[tk_30]]

[H]bunbunbunbun

Volume高左
H+ concentration不變
所以pH上升

btw
14點解B= =

[ 本帖最後由 [H]bunbunbunbun 於 27-4-2008 11:18 編輯 ]

piyopiyo

原帖由 [H]bunbunbunbun 於 27-4-2008 11:16 發表
Volume高左
H+ concentration不變
所以pH上升

btw
14點解B= =

但[OH-]都係不變[[tk_34]]

[H]bunbunbunbun

pH=-log[Conc. of H+]

Puppet

1. (3) × Too close b.p.

2.
A. ×
Volume increases => [H+(aq)] (molarity) decreases => pH = - log [H+(aq)], pH increases
Note that the solution is still neutral (assume ethanol is not an acid)
D.
Neutral iff [H+(aq)] = [OH-(aq)]
At higher temperatures, in pure water [H+(aq)] = [OH-(aq)] > 10^-7
pH < 7 but the solution is of course neutral

10.
General rule: Free elements as either reactant or products => must be redox (except C(graphite) -> C(diamond))
By conventional methond, C + 12(+1) + 6(-2) = 0 => C = 0
In fact, referring to the structure of glucose (as an example),
O.N. of 4 of the C atoms in C6H12O6 = 0
O.N. of C in aldo group of C6H12O6 = -1
ON. of the other terminal C of C6H12O6 = +1
So, please follow the general rule.

14. should be C
I typed it wrongly

[ 本帖最後由 Puppet 於 27-4-2008 12:25 編輯 ]

777kong

seems quite difficult