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发表于 5-10-2007 10:29 PM
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发表于 13-10-2007 05:31 PM
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原帖由 avada 于 4-10-2007 01:53 AM 发表 
答案是６。１８g 方法如 huhuxboy 和 rickykhoo 所说
下一题
The rate constant for the following reaction are 8.22x10^-4
at 333K and 3.7x10^-3 at 393K.What is the activation energy of   this rea ...
是chapter chemical reaction of Arrhenius Equation 的question 吗? |
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发表于 13-10-2007 07:10 PM
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原帖由 avada 于 4-10-2007 01:53 AM 发表 
下一题
The rate constant for the following reaction are 8.22x10^-4
at 333K and 3.7x10^-3 at 393K.What is the activation energy of this rea ...
K=Ae^-Ea/RT A=constant, Ea=activation energy
K1=8.22 x 10^-4 T=333K
K2=3.7 x 10^-3 T=393K
lnK = lnA - Ea/RT
ln(8.22 x 10^-4)=lnA - Ea/(8.31 x 333)---------->(1)
ln(3.7 x 10^-3)=lnA - Ea/(8.31 x 393) ---------->(2)
(1) - (2)
-1.5=-5.5173(Ea)
Ea = 27.18kJ |
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发表于 13-10-2007 07:34 PM
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我想提议一个意见。。。。。。。出题的人可以把问题和答案一起放上来吗？？这样就不用等出题的人下次才给答案了。。。 |
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发表于 14-10-2007 08:38 PM
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Calculate the ph of the resulting solution when the following volumes of 0.10moldm-3 NAOH have been added to 25.0 cm3 of 0.10 moldm-3 CH3COOH
1. 24.95 cm3
2. 25.00 cm3 Kb(CH3COO-)=5.56*10-10 mol dm-3
第二题我会做，POST上来给你们TRY下。。
第一题， 除了有CH３COO－（NEUTRALISE） 之外，还有多出来的 CH３COOH ，到底要怎样？ |
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发表于 15-10-2007 06:05 PM
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1.因为CH3COOH是weak acid,和NaOH neutralise后会有 CH3COONa
CH3COONa---->(double arrow) CH3COO- + Na+
所有有CH3COO-,所以这个时候是一个buffer solution啦
mole of acid left=0.05X0.1/1000=5X10^=5X10^-3X10^-3
mole of CH3COO-=2.495X10^-3
Ka=1.8X10^-5
pH= -lg1.8X10^-5 +lg2.495/5X10^-3
=7.44 |
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发表于 15-10-2007 11:34 PM
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不明白可以解释一下吗 ? 为什么说 ACID 的 MOLE 变成 0.05 X...... |
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发表于 15-10-2007 11:40 PM
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原帖由 seiryuu 于 15-10-2007 23:34 发表
不明白可以解释一下吗 ? 为什么说 ACID 的 MOLE 变成 0.05 X......
因为本来有2.5（X10^-3懒惰打了）
然后有2.495和NaOH react
所以就剩2.5-2.495=0.005咯 |
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发表于 17-10-2007 06:40 PM
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发表于 13-11-2007 02:23 AM
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write out the electronic configuration of chromium III ion by using the Hund's rule. |
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发表于 13-11-2007 12:27 PM
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回复 #610 zxteh 的帖子
1s2 2s2 2p6 3s2 3p6 3d3
有点疑问。。问题问的是，只用hund's rule而已吗？
还是用pauli exclusion, hund's rule and aufbau's rule?? |
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发表于 14-11-2007 06:23 PM
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原帖由 JohnChronox 于 13-11-2007 12:27 PM 发表 
1s2 2s2 2p6 3s2 3p6 3d3
有点疑问。。问题问的是，只用hund's rule而已吗？
还是用pauli exclusion, hund's rule and aufbau's rule??
Hund's rule 是最common的 |
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发表于 15-11-2007 11:42 AM
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楼上的，什么是最common?
在chemistry, writing electronic configuration,we need to follow all the 3 rules, so called - pauli exclusion, hund's rule and aufbau's rule
这样才能写出准确的答案。。
如果只用hund's rule，不用其他两个，答案可就差远了。。 |
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发表于 16-11-2007 06:22 PM
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past year question
When 200cm^3 of an aqueous solution of X with the concentration of 10.0gdm^-3 is shaken with 100cm^3 of 1-pentanol , 1.60g of X is extracted . If the aqueous solution is shaken once more with 100cm^3 of 1-pentanol after the first extration ,what is the mass of X that will be extracted ?
[X exist in the same molecular species in both 1-pentanol and water]
A 0.32g
B 0.80g
C 1.34g
D 2.63g |
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发表于 17-11-2007 11:39 AM
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A. let the amount of 1-pentanol used be Y,
hence (10 x 200) / (Y x 100) = 1.6 --1
in 2nd portion, mass of X left = 2.0g - 1.60g = 0.4g
hence 0.4 / (Y x 100/1000) = Z --2
compare equation 1 and 2, we get Z = 1.6 / 5
= 0.32g |
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发表于 17-11-2007 11:43 AM
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原帖由 再见是为了明天 于 16-11-2007 06:22 PM 发表
past year question
When 200cm^3 of an aqueous solution of X with the concentration of 10.0gdm^-3 is shaken with 100cm^3 of 1-pentanol , 1.60g of X is extracted . If the aqueous solution is sha ...
其实还有个更简单的方法。。
the total mass of X is 2.0g, at 1st portion, 1.60g had been extracted, the mass of X remaining is 0.40g
from the answer ABCD, only A is smaller than 0.40g, hence the answer is A |
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发表于 17-11-2007 12:57 PM
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good job...but i prefer to use partition coefficient law
Find out mass of X in aqueous solution 1st :
10 x (200/1000) = 2 g
use formula to find partition coefficient :
[X in pentanol]
--------------- = K
[X in aqueous]
(1.6/100)
---------------- = 8
(2.0-1.6 /200)
After 1st extraction (2.0 - 1.6) = 0.4 g of X remains in 200cm^3 water
Let 'y' g of X be extracted by 2nd 100cm^3 pentanol
By using the formula again :
( y / 100 )
-------------- = 8
( 0.4 - y /200 )
y = 0.32 g |
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发表于 1-2-2008 02:44 PM
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发表于 10-2-2008 11:21 AM
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Of the following solutions, which has the greatest buffering capacity?
A. 0.821mol/dm3 HF and 0.217mol/dm3 NaF
B. 0.821mol/dm3 HF and 0.909mol/dm3 NaF
C. 0.100mol/dm3 HF and 0.217mol/dm3 NaF
D. 0.121mol/dm3 HF and 0.667mol/dm3 NaF |
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发表于 10-2-2008 11:24 AM
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Calculate the pH needed to begin the precipitation of Mg(OH)2 in a solution of 0.01mol/dm3 MgCl2.
A.4.23
B.5.53
C.8.90
D.9.77
E.10.60 |
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