STPM化学专区

atnub_fuji

哦，谢谢=)
che 跟读sylalbus就可以？感觉上che范围蛮大的..
JianWen 发表于 10-6-2010 07:57 PM

没错，最好一步一步慢慢来，千万不要颠倒次序来读。现在你们最好打好你们的基础，不然的话以后你们读到 equilibrium 的部分时你就会哭的了。。。。。。

silent91

If 100cm^3 of o.5mol dm^-3 sulphuric acid is added to 100cm^3 of 0.5mol dm^-3 calcium chloride solution at room temperature, what is the mass of the precipitate, in gram, formed?
given solobility product for CaSO4 is 6 X 10^-5 mol^2 dm-6 at room temp & the RMM OF CaSO4 is 136.0


这课真的很不dim~

四月一日的小皮

If 100cm^3 of o.5mol dm^-3 sulphuric acid is added to 100cm^3 of 0.5mol dm^-3 calcium chloride solut ...
silent91 发表于 12-6-2010 08:29 PM

Just roughly do, let me know if i have wrong.

1=>let calculate [(SO4)2-] & [Ca2+].

Mole of H2SO4 = 0.5 x 100 / 1000 =0.05 mole
H2SO4 -> 2H+ + (SO4)2-
1mole of H2SO4 produced 1mole of (SO4)2-.
thus,there is 0.05mole of (SO4)2-.
[(SO4)2-] = 0.05 x 1000 / 200 = 0.25 mol dm-3

Mole of CaCl2 = 0.5 x 100 / 1000 =0.05 mole
CaCl2 -> Ca2+ + 2Cl-
1mole of CaCl2 produced 1mole of Ca2+.
thus,there is 0.05 mole of Ca2+.
[Ca2+] = 0.05 x 1000 / 200 = 0.25 mol dm-3

2=> find [(SO4)2-] & [Ca2+] using Ksp.
let x is solubility.
then, Ksp = [(SO4)2-][Ca2+]
      Ksp = x^2
      x^2 = 6 X 10^-5 mol^2 dm-6
      x   = 7.746 x 10^-3 mol dm-3

that mean,minimum [(SO4)2-] and [Ca2+] to form precipitate are 7.746 x 10^-3 mol dm-6.

3=> find excess [(SO4)2-] & [Ca2+] to form precipitate.

excess [(SO4)2-] = 0.25 - 7.746 x 10^-3 = 0.242
excess [Ca2+] = 0.25 - 7.746 x 10^-3 = 0.242

CaSO4 -> Ca2+ + (SO4)2-
so, [CaSO4] = 0.242 mol dm-3
    [CaSO4] = 32.912 g dm-3 x 200 /1000 dm3
            = 6.58 g

silent91

Just roughly do, let me know if i have wrong.

1=>let calculate [(SO4)2-] & [Ca2+].

Mole of ...
四月一日的小皮 发表于 13-6-2010 03:37 PM

万分感激，很详细一下～

nextdecade

Where to download STPM 2009 Past year question and answer?

silent91

A monovalent metal, M, forms a salt M2X.
If the solubility of the salt is s unit,
what is the solubility product, Ksp.

答案有点奇怪，所以希望有人指点～

endless_story

谢谢解答！！！
我想问这几题。。。

1)State the nucleon numbers that correspond to the peaks in the mass spectrum of oxygen which contains O-16 and O-18.

2)Calculate the number of electrons contained in 2.8g of nitrogen gas.

3)A compound containing carbon and hydrogen was analysed and found to contain 92.3% of carbon. When 0.195g of the compound was vapourised completely, the vapour was found to occupy 56.0 cm^3 at at s.t.p. What is the molecular formula of the compound?

四月一日的小皮

A monovalent metal, M, forms a salt M2X.
If the solubility of the salt is s unit,
what is the solu ...
silent91 发表于 16-6-2010 08:12 PM

M２X －＞ ２M+ + X－

［M２X］= s
［M+］= ２s
［X－］ = s

Ksp（M２X） = （２s）＾２ x （s）
            =   ４s＾３
不懂是不是呢？xD

谢谢解答！！！
我想问这几题。。。

1)State the nucleon numbers that correspond to the peaks in t ...
endless_story 发表于 16-6-2010 09:29 PM

1=>nucleon number of O-16 = 16
   nucleon number of O-18 = 18

2=> Mr of N = 14
    RMM of N2 = 28
    mole of N2 = 2.8 /28 = 0.1 mol
    In N2, no. of proton = no. of electron
    that is, 7+7=14 electons
   no. of e- in N2 = 0.1 x 6.02 x 10^23 x 14
                   = 8.428 x 10^23

3=> Using ideal gas law...
    PV =nRT
    P= 1 atm = 101000 Pa
    V= 56cm^3 = 5.6 x 10^-5 m^3
    n= m/ Mr = 0.195 / Mr
    R= 8.31 Pa m^3 K-1 mol^-1
    T= 273K
   
   101000(5.6 X 10^-5)(Mr) = 0.195(8.31)(273)
  Mr = 78 g mol-1
  No. of C = 78 x 92.3 / (100 x 12) = 6
  No. of H = 78 x 7.7 / (100 x 1 ) = 6
  Molecular formula = C6H6

silent91

M２X －＞ ２M+ + X－

［M２X］= s
［M+］= ２s
［X－］ = s

Ksp（M２X） = （２s）＾２ x （s ...
四月一日的小皮 发表于 16-6-2010 11:35 PM

我做的方法跟你一样，只是书所提供的答案有有点不同。
它给2s^3

四月一日的小皮

我做的方法跟你一样，只是书所提供的答案有有点不同。
它给2s^3
silent91 发表于 17-6-2010 10:02 PM

    哪一本的？我找找。xD

silent91

哪一本的？我找找。xD
四月一日的小皮 发表于 17-6-2010 10:45 PM

  federal physical chemistry chapter 7
question objective~

找不到就不用紧。
因为提供的答案只有一个有S＾３
这题不是很肯定一定是power 3???
所以，如果真的要选就这个。
但我认同你的做法是很standard的，我所吸收的也是酱～

silent91

The half reactions of MnO4- + H2O are given below.

MnO4- + 8H+ +5e ----> Mn2+ + 4H2O
H202 -------> 2H+ + O2 + 2e

If 100cm^3 of 0.2mol dm^-3 potassium mangantae(VII) reacts with an excess if acidified hydrogen peroxide, what is the volume of oxygen, in dm^3, evolved at s.t.p?
[ the molar volume of oxygen at room temp is 22.4dm^3 mol^-1]

四月一日的小皮

The half reactions of MnO4- + H2O are given below.

MnO4- + 8H+ +5e ----> Mn2+ + 4H2O
H202 ------ ...
silent91 发表于 19-6-2010 09:14 PM

he half reactions of MnO4- + H2O are given below.

MnO4- + 8H+ +5e ----> Mn2+ + 4H2O
H202 -------> 2H+ + O2 + 2e

If 100cm^3 of 0.2mol dm^-3 potassium mangantae(VII) reacts with anexcess if acidified hydrogen peroxide, what is the volume of oxygen, indm^3, evolved at s.t.p?
[ the molar volume of oxygen at room temp is 22.4dm^3 mol^-1]

let
MnO4- + 8H+ +5e ----> Mn2+ + 4H2O   -------->(1)
H202 -------> 2H+ + O2 + 2e              -------->(2)
(1)x2 : 2MnO4- + 16H+ +10e ----> 2Mn2+ + 8H2O
(2)x5 : 5H202 -------> 10H+ + 5O2 + 10e  
----------------------------------------------------
2MnO4- + 16H+ + 5H202 -> 2Mn2+ + 8H2O + 10H+ + 5O2
----------------------------------------------------
mole of potassium mangantae(VII)= 100 x0.2 / 1000 = 0.02 mole
mole of oxygen = 0.02 x 5 / 2 = 0.05 mole
volume of oxygen = 0.05 x 22.4 = 1.12dm^3

silent91

he half reactions of MnO4- + H2O are given below.

MnO4- + 8H+ +5e ----> Mn2+ + 4H2O
H202 --- ...
四月一日的小皮 发表于 20-6-2010 03:47 AM

你怎么知道要balance e 而不是 proton?

四月一日的小皮

你怎么知道要balance e 而不是 proton?
silent91 发表于 20-6-2010 07:51 PM

通常full equation是没有e－的。
还有，更改下：

2MnO4- + 16H+ + 5H202 -> 2Mn2+ + 8H2O + 10H+ + 5O2
－－－－－－－－－－－－－－－－－－－－－－－－－
2MnO4- + 6H+ + 5H202 -> 2Mn2+ + 8H2O + 5O2

Lov瑜瑜4ever

一个水平平放，内径均匀的细长玻璃管中，有一小段水柱封闭住了一段空气柱。在大气压强=760mm Hg，温度=12C，空气柱长100mm。如果大气压强不变，当温度=35C，空气柱有多长？
在12C和35C时，水蒸气的saturated vapour pressure分别是10.5mm Hg和42.0mm Hg.

分析：空气柱实际是空气和水蒸气的混合物
s.v.p和物质的种类有关系，随温度的升高而增大，但却和物质的体积没有关系，所以水蒸气的s.v.p不遵守ideal gas eqn,所以能应用公式(p1v1)/T1=(p2v2)/T2来计算的只是其中的空气。由于saturated的水蒸气的存在，根据dalton law of partial pressure,与大气压强相平衡的空气柱压强，实际是空气的压强和水蒸气的s.v.p的sum
在T1=273+12=285K时,p1=760-10.5=749.5mm Hg
设玻璃管的横截面积=A mm^2,这时v1=100A mm^3
在T2=273+35=308K时,p2=760-42=718mm Hg
体积为v2
从(p1v1)/T1=(p2v2)/T2可知
v2=113A mm^3
所以35C时空气柱长113mm

那么再来看看这题
The volume of a closed room is 60m^3 and the temp. is fixed at 30C. The partial vapour pressure of water in the room is 2.33x10^2 Nm^-2 initially.If a bucket of water was put in the room, calculate the mass of water that evaporated when equilibrium was reached. The s.v.p for water at 30C=4.24x10^3 Nm^-2.

Let the mass of water that has evaporated=mg
initial vapour pressure=2.33x10^2 Nm^-2
Final vapour pressure=4.24x10^3 Nm^-2
so, The vapour pressure due to this water=4.24x10^3-2.33x10^2=4007Nm^-2
From the ideal gas equation,
pV=nRT
4007x60=(m/18)x8.31x(273+30)
so, m=1719g

s.v.p不是不遵守ideal gas equation的吗？
做么上面的问题不能用，而下面的问题却能用的？

Lov瑜瑜4ever

还有那个lattice point就是particle来的？

silent91

electrolysis of PbBr2(aq)

我想问anode 和cathode 的half equation 还是跟f5的时候一样吗？
有老师讲eq不同了～
例如：在anode : 2H20 ------> 02 + 4e + 4H+   


AgBr 在electrochemical cell 里是否会产生electrode potential?
因为又一个老师讲说会occur precipitation 所以 standard e cell 会等于零～
在同一个时间，运用在Nerst eq 的时候，学校老师+ 参考书都是写AgBr有standard e cell value 的～
谁对谁错？

qwer0909

想问一下，calculate the concentration , in mol dm-3, of the monobasic HX in KA2, given tat 1 mole of HX is 36.5g, ka2 contain 3.5gdm-3 of HX
谢谢

silent91

想问一下，calculate the concentration , in mol dm-3, of the monobasic HX in KA2, given tat 1 mole of ...
qwer0909 发表于 25-7-2010 09:25 PM

ans is 0.096mol^-3 ?