STPM化学专区

SCT

原帖由 末日审判 于 7-7-2009 06:34 PM 发表
各位高手，帮帮下！
1.Selenium,Se,occurs in nature as a mixture of six isotopes having the relative abundance given below:
   nucleon(mass) number           %abundance
                 74            ...

(1) ans : 40
不明白(2)和(3)在问什么 orbital A,B,C,D,E? 只知道orbital s,p,d,f... 

四月一日的小皮

原帖由 星星之玲 于 14-2-2009 09:05 PM 发表
这是physical chemistry 的chapter12 phase equilibrium问题谁能帮我算一下啊?谢谢!

The partition coefficient of a substance X between ether and water is 8.0 .
An  aqueous solution containing 5.0g of  ...

[X]in ether / [X] in water = 8.0
let x is the amount of X extracted
then  (x/100)/(5-x/100) = 8
                         x = 8(5-x)
                        9x = 40
                         x =4.44

四月一日的小皮

原帖由 星星之玲 于 14-2-2009 09:05 PM 发表
这是physical chemistry 的chapter12 phase equilibrium问题谁能帮我算一下啊?谢谢!

The partition coefficient of a substance X between ether and water is 8.0 .
An  aqueous solution containing 5.0g of  ...

[X]in ether / [X] in water = 8.0
let x is the amount of X extracted
then  (x/100)/(5-x/100) = 8
                         x = 8(5-x)
                        9x = 40
                         x =4.44

四月一日的小皮

原帖由 cgcy 于 18-4-2009 12:50 AM 发表
Mechanism :
NO + F2 -----> NOF + F*
F* + NO -----> NOF
Overall equation:
2NO + F2 -----> 2NOF
Question: Please justify the mechanism in term of reaction kinetics.

Rate = k1[F2][NO]=k-1[NOF][F*]
Rate = k2[F*][NO]=k-2[NOF]
[F*]= k1[F2][NO]/k-1[NOF]
THEN, k2[NO] (k1[F2][NO]/k-1[NOF]) = k-2[NOF]
(k2k1/k-1)[NO]^2[F2]= k-2[NOF]^2
THUS,  Rate=K[NO]^2[F2]=K-1[NOF]^2,given that  K=(k2k1/k-1) and (K x K-1)=1
correct me if i'm wrong

bell_25

我这里也有个问题。。
可不可以帮我看我的答案有没有错。。
题目是
SO2 reacts with O2 to produce SO3 which is then used to manufacture H2SO4. A mixture containing SO2 and O2 with a mole of ratio of 2:1, total pressure at beginning 3 atm is passed over V2O5 at 450 degree celcius. The equilibrium partial pressure of SO3 was found to be 1.9atm.

2 SO2 (g) + O2 (g)   -----> 2 SO3 (g)
Calculate the equilibrium partial pressure of SO2 and O2.
Determine the equilibrium constant, Kp at 450 degree celcius.

我的答案是 0.1atm 和 0.05atm
Kp 是7220 atm^-1

如果有任何地方做错了。。麻烦指点。。谢谢。。

idontwant2b

想知道

cooridnate bond
ｄｉｍｅｒ
ａｄｄｕｃｔ
ｌｉｇａｎｄ


是什么分别？

四月一日的小皮

原帖由 bell_25 于 2-8-2009 12:15 AM 发表
我这里也有个问题。。
可不可以帮我看我的答案有没有错。。
题目是
SO2 reacts with O2 to produce SO3 which is then used to manufacture H2SO4. A mixture containing SO2 and O2 with a mole of ratio of 2:1 ...

               2 SO2 (g) + O2 (g)   -----> 2 SO3 (g)
initial            2               1                        0
final         2(1-0.95)    1-0.95             2(0.95)
                = 0.1           =0.05
Kp =  (1.90^2)/(0.1^2)(0.05)  = 7220 atm^-1

appletlv

我觉得是increase咯
across period, nuclear charge increase, but inner electron no increase, so attraction outer electron to nucleus increase, screening effect decrease, so more compact,so density increase

HaPpy_GirL_91

有谁有STPM Chemistry Experiment 3 的答案。。。
帮帮我。。

@影子@

各位，
想问下，inorganic chemistry那本书最好？

Dicardo

如果题目只写： 'when aqueous NaCl undergo electrolysis under standard condition with carbon eletrode, which will be produce at anode ?"

是 Chlorine 还是 Oxygen gas 啊？

jenyeh90

恩，到這里看看吧
http://chemed.chem.purdue.edu/ge ... ch20/faraday.php#aq

Dicardo

原帖由 jenyeh90 于 24-8-2009 01:09 PM 发表
恩，到這里看看吧
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch20/faraday.php#aq

我怎么知道“aqueous"是指”dilute“ 还是 "concentrated" ？

Allison

aqueous 就是有水，所以是dilute的

Dicardo

原帖由 Allison 于 24-8-2009 03:57 PM 发表
aqueous 就是有水，所以是dilute的

concentrated 也是有水，不是吗？

Allison

erm。。。
如果是concentrated NaCl，那代表Na & Cl ion 比 H2O多
所以Cl ion 会discharged
aqueous 的话， H2O 比较多所以OH ion discharged
你是lower的？

Dicardo

原帖由 Allison 于 24-8-2009 04:10 PM 发表
erm。。。
如果是concentrated NaCl，那代表Na & Cl ion 比 H2O多
所以Cl ion 会discharged
aqueous 的话， H2O 比较多所以OH ion discharged
你是lower的？

upper 的。。。嘻嘻。。
下星期预考了。。现在才来赶。。。

Allison

我也一样。。。现在还在上网
完蛋了

darksider

原帖由 Allison 于 2009/8/24 04:10 PM 发表
erm。。。
如果是concentrated NaCl，那代表Na & Cl ion 比 H2O多
所以Cl ion 会discharged
aqueous 的话， H2O 比较多所以OH ion discharged
你是lower的？

Correction
我想这样的解释会好一点：

In Anode, two possible ions can be oxidised , ie , cl- and h2o
Cl2(g) + 2 e− ⇄ 2 Cl−    E°red= +1.36V  - (1)
O2(g) + 4 H+ + 4 e− ⇄ 2 H2O E°red= +1.23V - (2)

Overvoltage needed for the formation of oxygen is higher than that of chlorine gas because activation energy needed for transfer of electrons at the electrode-solution surface is high which corresponds to slow rate (transfer of electron from anode to cathode). Thus, extra voltage must be applied so that the reaction can proceed in a satisfactory rate. Since rate of formation of chlorine gas is faster than that of oxygen gas due to lower activation energy,Cl- is oxidised and chlorine gas is produced. (Reaction which requires lesser overvoltage will occur instead of the one requiring higher overvoltage)

Hopefully I didn't confuse any of you.

Additional notes  (not related)
----------------------
Let say the electrolyes used are all aqueous

i) In anode , possible half reactions :
O2(g) + 4 H+ + 4 e− ⇄ 2 H2O  E°red= +1.23V --- (1)
O2(g) + 2 H2O + 4 e− ⇄ 4 OH−(aq) E°red= +0.40V --- (2)
Cl2(g) + 2 e− ⇄ 2 Cl−    E°red=  +1.36V  - (3) ----->>> just an example, replace it with any other anion's half eq

According to MPM,
if the electrolye used is neutral, equation (2) can be eliminated from the possible half equations and we consider only (1) and (3) instead. This is because in neutral solution, most of the water molecules remain undissociated and [OH−]is very small , ie , 1x10−7mol dm-3. So we compare only equation (1) and (3) and the one with lower E°red will be discharged.

If the electrolyte used is alkaline, then equation (1) should be ignored because the [OH−] cannot be ignored now. Hence, we compare equation (2) and equation (3)


ii) In cathode, possible half reactions :
2 H2O + 2 e− ⇄ H2(g) + 2 OH− E°red= -0.8277V -----(1)
2 H+ + 2 e− ⇄ H2(g) E°red= 0.00V ----(2)
Cu2+ + 2 e− ⇄ Cu(s) E°red= +0.34V ----(3) ---- just an example, replace it with any other cations will do


If the electrolye used is neutral, equation (2) should be ignored because large amount of h2o remains undissociated and [H+] is relatively small , ie , 1x10−7mol dm-3. Hence, we compare only equations (1) and (3) if the electrolye is neutral.


If the electrolye used is acidic, equation (1) can be ignored since  [H+] is relatively large now. Hence, we compare only equations (2) and (3).


According to MPM, most candidates do not realise the presence of h2o molecules in aqueous electrolye.


Again, I hope I didn't confuse you. Good luck!


Btw, for most textbooks I've used, I found out Chemistry by John McMurry and Robert C.Fay is the best. None of the malaysian text books could be compared to it. However it is expensive but you know in internet we have such thing called download, so

[ 本帖最后由 darksider 于 27-8-2009 07:41 PM 编辑 ]

darksider

If a solution is aqueous, it has water molecules.

If the solution is dilute, it means that the water molecules are present in large amount.

If the solution is concentrated, it means that the water molecules are present in relatively small amount.