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darksider

大家好，目前有些chemistry的问题，solve到但不明白。
以后肯定也会有难题，希望大家可以指点我。

A molecule of certain protein contains 3 nickel atoms.
In 1 gram of that protein, there is 0.00421 g of nickel.
What is the molecular mass of the protein?

我需要你的解释，为何我们把 (3 x 58.7) / 0.00421 来找answer呢？

TonyDaisie

1g protein 含 0.00421g nickel                  
在moleculat weight 里：？g protein 含 3 x 58.7 的 nickel

用ratio来解决：  1   :   0.00421
                            ?   :   3 x 58.7

1/? = 0.00421/(3 x 58.7)  

darksider

原帖由 TonyDaisie 于 2008/5/25 12:11 PM 发表
1g protein 含 0.00421g nickel                  
在moleculat weight 里：？g protein 含 3 x 58.7 的 nickel

用ratio来解决：  1   :   0.00421
                            ?   :   3 x 58.7

1/? = 0. ...

刚刚也做到了，谢谢

Based on C-12 scale, atom Q has a mass of 125.17 and atom Cl has a mass of 35.71.
What is the mass of C-12 and Cl based on Q scale (Ar 125)?

TonyDaisie

Relative atomic mass = mass of one atom of the element / 1/12 mass of one atom C-12

1u = 1.661 X 10-27 kg  

(在问题里没有提到u 的 convention，我把u当做1/125 mass atom Q）
mass of atom Q = 125.17 X 1.661 X  10-27 kg  
问题一：C-12
mass of C-12 = 12 / ((1/125)125.17) = 11.98
问题二：Cl
mass of Cl = 35.71 / ((1/125)125.17) = 35.66

如果你把u当做一个mass atom Q 的话：
C-12 =  0.096
Cl = 0.285

darksider

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?

darksider

原帖由 TonyDaisie 于 2008/6/2 08:05 PM 发表
Relative atomic mass = mass of one atom of the element / 1/12 mass of one atom C-12

1u = 1.661 X 10-27 kg  

(在问题里没有提到u 的 convention，我把u当做1/125 mass atom Q）
mass of atom Q = 125.1 ...

明白了，谢谢。希望你能帮我解答上面一题，谢！

darksider

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.

Can check if my solution is correct?

TonyDaisie

你不是做到30了吗？？？

30是对的。。。

darksider

原帖由 TonyDaisie 于 2008/6/10 03:15 PM 发表
你不是做到30了吗？？？

30是对的。。。

恩，应该是对了，谢谢。

以后有问题再来请教　

darksider

( A , 5 dm^3 , P atm) -------|--------- ( B , 10dm^3 , 3atm)
The diagram above shows two bulbs containing gas A and gas B respectively at room temperature. When the | is removed and the system is allowed to attain equilibrium, what is the initial pressure of gas A when the final pressure is 5 atm?

kensai

PaVa=naRT
PbVb=nbRT

PaVa+PbVb=(na+nb)RT=Ptotal x Vtotal
照着这样做应该不会错

你的Pa 没有写value出来，所以我不能给你答案

appletlv

原帖由 darksider 于 15-6-2008 12:36 发表
( A , 5 dm^3 , P atm) -------|--------- ( B , 10dm^3 , 3atm)
The diagram above shows two bulbs containing gas A and gas B respectively at room temperature. When the | is removed and the system is all ...

gas B= 10 x 3 = p x 15
                 30= 15p
                   p =2
final p of gas A  = 5- 2=3
gas A = 5 x p = 15 x 3
     45 = 5p
  initial  p for gas A =9