STPM 化学问题 马拉松！

lewisham

A.....................

huhuxboy

对了。
请出题。。。
多谢

morphv02c01

Which of the following suggests the presence of free radicals in the chlorination of methane?

1. Hydrogen chloride is present in the product.
2. The reaction proceeds most quickly in sunlight or ultraviolet light.
3. Ethane is present in small quantities in the product.

waiyee926

我的天。。。 我全都忘掉了。。 赫赫。。。 亏你们还可以去翻回书

hvaly

哈哈，我也是忘到一干二净了，不过没有错的话答案应该是2和3。

再见是为了明天

原帖由 huhuxboy 于 1-6-2007 09:42 PM 发表
C6H5COONa dissociate completely in water to form C6H5COO- and Na+ ions.

C6H5COO- + H2O <==> C6H5COOH + OH-
Let the base dissociation constant of this reaction be Kb

C6H5COO- is almos ...

i don understand why [OH-]^2/0.5 ??? not [C6H5COOH][OH-]/0.5 ???
and why the equilibrium shifts to the left ???how to see?

再见是为了明天

N2O4(g) <===> 2NO2(g)
The degree of dissociation of N2O4 at 100KPa and 150celcius is 15% .Calculate the Kp for the reaction at 150celcius .
A. 0.106KPa
B. 0.353KPa
C. 2.650KPa
D  9.220KPa

morphv02c01

D. 9.220 Kpa

morphv02c01

对了..........  

morphv02c01

If the rate of decay of a radioactive isotope decreases from 200counts per minute to 25 counts per minute after 24 hours, what is the half life?
A. 3h
B. 4h
C. 6h
D. 8h

再见是为了明天

原帖由 morphv02c01 于 3-6-2007 07:28 PM 发表
D. 9.220 Kpa

solution pls...

hamilan911

原帖由 morphv02c01 于 3-6-2007 07:36 PM 发表
If the rate of decay of a radioactive isotope decreases from 200counts per minute to 25 counts per minute after 24 hours, what is the half life?
A. 3h
B. 4h
C. 6h
D. 8h

25/200 = 1/8 = 3halr life
3 half life = 24 hours
so half life = 8 hours
D.

sylvia_r

原帖由 再见是为了明天 于 3-6-2007 05:41 PM 发表
N2O4(g) <===> 2NO2(g)
The degree of dissociation of N2O4 at 100KPa and 150celcius is 15% .Calculate the Kp for the reaction at 150celcius .
A. 0.106KPa
B. 0.353KPa
C. 2.650KPa
D  9.220KPa

Before equilibrium,
let no. of mol of N2O4 be 1 mol.

After equilibrium,
no of mol of N2O4 is 0.85mol (dissociate 15%) and
no. of mol of NO2 is 0.30mol

Partial pressure of N2O4 is (0.85/1.15) * 100kPa
Partial pressure of NO2 is (0.30/1.15) * 100kPa

Kp= (P NO2)^2 / (P N2O4)

再见是为了明天

100cm^3 aqueous silver nitrate is added to 100cm^3 aqueous sodium chloride of concentration 0.15mol dm^-3 .What is the minimum concentration of silver nitrate for a precipitate to occur?
[Ksp AgCl =1.0x10^-10 mol^2 dm^-6]

A. 1.33x10^-10 moldm^-3
B. 3.34x10^-10 moldm^-3
C. 6.67x10^-10 moldm^-3
D. 2.67x10^-9  moldm^-3

sylvia_r

原帖由 再见是为了明天 于 5-6-2007 09:00 PM 发表
100cm^3 aqueous silver nitrate is added to 100cm^3 aqueous sodium chloride of concentration 0.15mol dm^-3 .What is the minimum concentration of silver nitrate for a precipitate to occur?


A.  ...

concentration of NaCl= 0.15 * (100/200) (dillute effect)
                     =0.075mol dm^-3

concentration of AgNO3 to form percipitate = 1^-10/ 0.075
                                           = 1.33^-9 mol dm^-3
but brcause of dillute effect, this concentration is for 200cm^3 one, so the concentration of AgNO3 in 100cm^3 = 1.33^-9 * 200/100
                 =2.67^-9 mol dm^-3

Is it correct?

lewisham

c.............

morphv02c01

let [ Ag ] = A
[ NaCl ] = 0.15moldm^3
therefore [ Cl- ] = 0.15moldm^3
For precipitation to occur, Ksp must be equal to Q (ionic pdt)
therefore Ksp of AgCl= [ Ag ] . [ Cl- ]
          1.0x10^-10= A x 0.15moldm^3
          A= 6.67x10^-10




真paiseh。。 错到很离谱。。

[ 本帖最后由 morphv02c01 于 6-6-2007 08:41 PM 编辑 ]

再见是为了明天

原帖由 sylvia_r 于 6-6-2007 08:40 AM 发表


concentration of NaCl= 0.15 * (100/200) (dillute effect)
                     =0.075mol dm^-3

concentration of AgNO3 to form percipitate = 1^-10/ 0.075
                                ...

对了！ 。。。你介意给我你的msn address 吗？？
因为我们可以互相讨论功课。。。

morphv02c01

A saturated solution of calcium hydroxide is prepared at 18degree celcius. The mixture is then filtered to remove the undissolved solid. When 25cm^3 of the filtrate is titrated against 0.1moldm^-3 HCl, 13.06cm^3 of the acid are required for neutralisation. What s the Ksp for calcium hydroxide in mol^3dm^-9 at 18degree celcius?

A. 7.12 x 10^-5
B. 3.56 x 10^-5
C. 1.78 x 10^-5
D. 2.84 x 10^-4

再见是为了明天

原帖由 morphv02c01 于 6-6-2007 08:51 PM 发表
A saturated solution of calcium hydroxide is prepared at 18degree celcius. The mixture is then filtered to remove the undissolved solid. When 25cm^3 of the filtrate is titrated against 0.1moldm^- ...

Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O

0.1x13.06    2
--------- = ---
M x 25.00    1

concentration of Ca(OH)2 = 0.02612 moldm^-3

Ca(OH)2 <===> Ca2+ + 2OH-
1 mol of Calcium hydroxide produces 1mol of calcium ion and 2 mol of hydroxide ion

Ksp of calcium hydroxide = [Ca2+][OH-]^2
                         = (0.02612)(2x0.026120)^2
                         = 7.13x10^-5