Answer:
1) The correct answer is option b.
2) The correct answer is option a.
Explanation:
1)
At 300 K, the value of the 
The 
and 
is related by :
where,
= equilibrium constant at constant pressure
= equilibrium concentration constant
R = gas constant = 0.0821 L⋅atm/(K⋅mol)
T = temperature = 300 K
= change in the number of moles of gas = 1 - 2 = -1
Now put all the given values in the above relation, we get:
The of the reaction = 24.63
Given = [X] = [Y] = [Z] = 1.0 M
Value of reaction quotient = Q
![Q=\frac{[Z]}{[X][Y]}=\frac{1.0 M}{1.0M\times 1.0 M}=1](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BZ%5D%7D%7B%5BX%5D%5BY%5D%7D%3D%5Cfrac%7B1.0%20M%7D%7B1.0M%5Ctimes%201.0%20M%7D%3D1)
, the equilibrium will move in forward direction that is in the right direction.
2)
At 300 K, the value of the
Given = 
Value of reaction quotient in terms of partial pressure = 
the equilibrium will move in backword direction that is in the left direction.
<span>
Correct Answer:
Option 3 i.e. 30 g of KI dissolved in 100 g of water.
Reason:
Depression in freezing point is a
colligative property and it is directly proportional to molality of solution.
Molality of solution is mathematically expressed as,
Molality = </span>
<span>
In case of
option 1 and 2, molality of solution is
0.602 m. For
option 3, molality of solution is
1.807 m, while in case of
option 4, molality of solution is
1.205 m.
<u><em>Thus, second solution (option 2) has highest concentration (in terms of molality). Hence, it will have lowest freezing point</em></u></span>
Answer:
D. They react readily with oxygen in the air
Answer:
Explanatio
NCO2= 0, 248 /44= 0 ,005636
VCO2= 0,005636* 22,4= 0 ,126254545
The answer to your question is: true.
Hope this helps!
