<span>polyethylene or polythene are the main ingredients </span>
The equilibrium vapour pressure is typically the pressure exerted by a liquid .... it is A FUNCTION of temperature...
Explanation:
By way of example, chemists and physicists habitually use
P
saturated vapour pressure
...where
P
SVP
is the vapour pressure exerted by liquid water. At
100
∘
C
,
P
SVP
=
1
⋅
a
t
m
. Why?
Well, because this is the normal boiling point of water: i.e. the conditions of pressure (i.e. here
1
⋅
a
t
m
) and temperature, here
100
∘
C
, at which the VAPOUR PRESSURE of the liquid is ONE ATMOSPHERE...and bubbles of vapour form directly in the liquid. As an undergraduate you should commit this definition, or your text definition, to memory...
At lower temperatures, water exerts a much lower vapour pressure...but these should often be used in calculations...especially when a gas is collected by water displacement. Tables of
saturated vapour pressure
are available.
Answer:
What is the new volume if the temperature is constant? V=2.50L. P = lookPa. P2=40k Pa. V2 = x. PV = P2 ... If a sample of gas occupies 6.8 L at 327°C
Answer:
a) The equilibrium will shift in the right direction.
b) The new equilibrium concentrations after reestablishment of the equilibrium :
![[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M](https://tex.z-dn.net/?f=%5BSbCl_5%5D%3D%280.370-x%29%20M%3D%280.370-0.0233%29%20M%3D0.3467%20M)
![[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BSbCl_3%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)
![[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M](https://tex.z-dn.net/?f=%5BCl_2%5D%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2Bx%29%20M%3D%286.98%5Ctimes%2010%5E%7B-2%7D%2B0.0233%29%20M%3D0.0931%20M)
Explanation:
a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
On increase in amount of reactant
If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of 
is increasing .So, the equilibrium will shift in the right direction.
b)
Concentration of = 0.195 M
Concentration of 
= 
Concentration of 
=
On adding more 
to 0.370 M at equilibrium :
Initially
0.370 M 
At equilibrium:
(0.370-x)M 
The equilibrium constant of the reaction = 
The equilibrium expression is given as:
![K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BSbCl_3%5D%5BCl_2%5D%7D%7B%5BSbCl_5%5D%7D)
On solving for x:
x = 0.0233 M
The new equilibrium concentrations after reestablishment of the equilibrium :
Number 1: (A.)
Number 2: (A.)
Number 3: (B.)
I'm probably wrong but that is what i think
