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发表于 7-6-2008 12:46 AM
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有问题叫你suggest a simple chemical test to differentiate between ２ compounds 时，如果须要提高temp的话最好是写heat，表写reflux。 |
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发表于 7-6-2008 01:04 AM
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回复 224# morphv02c01 的帖子
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那通常我们都用HEAT咯???很少用REFLUX....? |
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发表于 7-6-2008 01:29 AM
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原帖由 幸福蛋糕 于 7-6-2008 01:04 AM 发表 
那通常我们都用HEAT咯???很少用REFLUX....?
老实说 HEAT/REFLUX可用在synthesis时,因两个都给相同yield 的product,在于快或慢... |
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发表于 7-6-2008 01:32 AM
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原帖由 morphv02c01 于 7-6-2008 12:46 AM 发表
有问题叫你suggest a simple chemical test to differentiate between ２ compounds 时，如果须要提高temp的话最好是写heat，表写reflux。
simple chemical test是指可在lab进行到的...reflux不能轻易在lab进行到...所以用heat会较适合...
[ 本帖最后由 秋分 于 7-6-2008 02:29 AM 编辑 ] |
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发表于 7-6-2008 02:15 AM
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原帖由 秋分 于 7-6-2008 01:32 AM 发表 
simple chemical test是指可在lab进行到的...reflux不能轻已在lab进行到...所以用heat会较适合...
Thank you for making furthur explanation on my statement. |
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发表于 7-6-2008 02:28 AM
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原帖由 morphv02c01 于 7-6-2008 02:15 AM 发表
Thank you for making furthur explanation on my statement.
不客气...对了,麻烦用中文,要不然会被扣分... |
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发表于 7-6-2008 09:20 PM
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发表于 7-6-2008 09:26 PM
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你们有在做PAST YEAR PAPER吗???有的话..我们可以一起讨论... |
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发表于 7-6-2008 10:03 PM
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5) 6.0 cm^3 of a gaseous hydrocarbon was sparked with 34.5cm^3 of oxygen. The volume of the gaseous products after cooling to room temperature was 22.5 cm^3. The volume was reduced to 4.5 cm^3 on passing the gases through aqueous potassium hydroxide. What is the molecular formula of hydrocarbon?
有谁可以教我怎样算出 volume of oxygen reacted? |
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发表于 7-6-2008 10:32 PM
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原帖由 darksider 于 7-6-2008 10:03 PM 发表
5) 6.0 cm^3 of a gaseous hydrocarbon was sparked with 34.5cm^3 of oxygen. The volume of the gaseous products after cooling to room temperature was 22.5 cm^3. The volume was reduced to 4.5 cm^3 on pass ...
react 过后的22.5 cm3 是CO2 和 O2
KOH用来吸CO2，所以吸了过后剩下的4.5cm3就是O2
所以氧气总共用了 34.5-4.5=30.0 cm3 |
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发表于 7-6-2008 11:02 PM
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原帖由 mengbai 于 2008/6/7 10:32 PM 发表 
react 过后的22.5 cm3 是CO2 和 O2
KOH用来吸CO2，所以吸了过后剩下的4.5cm3就是O2
所以氧气总共用了 34.5-4.5=30.0 cm3
它的ｐｒｏｄｕｃｔ不是 co2 + h2o 吗？ |
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发表于 8-6-2008 02:36 AM
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原帖由 darksider 于 7-6-2008 11:02 PM 发表
它的ｐｒｏｄｕｃｔ不是 co2 + h2o 吗？
H2o(L) it is in liquid state |
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发表于 8-6-2008 02:37 AM
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原帖由 幸福蛋糕 于 7-6-2008 09:26 PM 发表
你们有在做PAST YEAR PAPER吗???有的话..我们可以一起讨论...
哈哈抱歉我去年已考了... |
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发表于 8-6-2008 09:35 AM
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原帖由 秋分 于 2008/6/8 02:36 AM 发表
H2o(L) it is in liquid state
ＫＯＨ吸了ＣＯ２（２２．５ｃｍ＾３），所剩的４．５ｃｍ＾３是unreacted的o2?
[ 本帖最后由 darksider 于 8-6-2008 09:44 AM 编辑 ] |
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发表于 8-6-2008 11:07 AM
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原帖由 darksider 于 7-6-2008 11:02 PM 发表
它的ｐｒｏｄｕｃｔ不是 co2 + h2o 吗？
对
product是 CO2和H2O，可是这个题目还有剩O2
所以react过后，有CO2，H2O，和O2
H2O没有体积
所以22.5是CO2和O2 |
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发表于 8-6-2008 12:34 PM
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原帖由 mengbai 于 2008/6/8 11:07 AM 发表
对
product是 CO2和H2O，可是这个题目还有剩O2
所以react过后，有CO2，H2O，和O2
H2O没有体积
所以22.5是CO2和O2
谢，我刚做了，不知对吗？
My attempted solution:
Volume of CO2 produced : 22.5 cm^3 - 4.5 cm^3 = 18 cm^3
Volume of unreacted oxygen : 22.5 cm^3 - 18 cm^3 = 4.5 cm^3
Volume of oxygen reacted : 34.5 cm^3 - 4.5 cm^3 = 30.0 cm^3
CxHy + (x + y/4) O2 ---> x CO2 + y/2 H2O
mole of CxHy = (6.0 cm^3 / 1000 ) / 24 cm^3
= 0.00025 mol
mole of O2 : x + y/4 = (30 cm^3 / 1000) / 24 dm^3
x + y/4 = 0.00125 mol ------------1
Mole of CO2 : x = (18 cm^3 /1000) / 24 dm^3
x = 0.00075 mol
From 1 , 0.00075 mol + y/4 = 0.00125 mol
y/4 = 0.0005 mol
y = 0.002 mol
Therefore , mol of C : mol of O2 : mol of CO2 : mol of H2O
0.00025 mole : 0.00125 mol : 0.00075 mol : 0.0001 mol
1 mole : 5 mol : 3 mol : 4 mol
Since X = 3 mol , y = 8 mol , the molecular formula of this compound is C3H8. |
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发表于 10-6-2008 06:13 PM
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发表于 10-6-2008 07:07 PM
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原帖由 幸福蛋糕 于 2008/6/10 06:13 PM 发表
IDEAL GAS 那课必须懂什么??
You need to understand kinetic molecular theory of gases, boyle's law,charles' law , dalton's law of partial pressure,ideal gas equation,deviation of real gases from ideal gas behaviour,van der waals' equation - [ P + ( (n^2)a / V^2 ) ] x ( V-nb) = nRT and argument of van der waals' that 2 assumptions in kinetic molecular theory are erroneous.
Ideal gas equation : PV = nRT
How to find R?
Assuming the gas is at standard temperature and pressure, its volume will be 22.4L , its pressure will be 1.0 atm and its temperature will be 273.15k (0 degree celcius) and its mole is one.
Substitute them into the ideal gas equation.
1.0 atm x 22.4 L = 1 mole x R x 273.15 k
R = 1.0 atm x 22.4 L
----------------------
1 mole x 273.15 k
R = 0.00821 atm.L mole^-1 k^-1 (3.s.f)
When using ideal gas equation to find out the one of the variable, always use L for volume , atm for pressure , K for temperature , n for mole and atm.L mole^-1 k^-1 for gas constant.
1L = 1 dm^3 = 1x10^3 cm^3 = 1x10^-3 m^3
1atm = 1.01 x 10^5 Pa = 760 mm Hg = 760 torr = 14.7 psi
For deviation of real gases from ideal gas behaviour, refer to this page.
http://library.thinkquest.org/C006669/data/Chem/gases/vanderwaals.html |
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发表于 10-6-2008 09:04 PM
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 恩,,虽然有点乱...但还是谢谢...
我问你,你学到HYBRIDISATION了吗?? |
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发表于 11-6-2008 06:17 PM
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