Answer:
The new equilibrium total pressure will be increased to one-half to initial total pressure.
Explanation:
From the information given :
The equation of the reaction can be represented as;

From above equation:
2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide
So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.
So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.
Let the total pressure at the initial equilibrium be 
and the total pressure at the final equilibrium be 
According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Thus;
P ∝ 1/V
P = K/V
PV = K
where K = constant
So;
PV = constant
Hence;

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

also;
Thus ;



Dividing both sides by 


From ;




Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.
Answer : The mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.
Explanation :
Law of conservation of mass : It states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
The balanced chemical reaction will be,

As we are given:

According to the law of conservation of mass,
Total mass of
= Mass of
+ Mass of C
Total mass of
= 2.67 + 1 = 3.67 g
Now we have to calculate the mass of
and C.

and,

Therefore, the mass of carbon and oxygen produced is 8.83 g and 23.6 g respectively.
Because it is a mix of H2O (water) and NaCl (salt)
Answer:
Ka = 
Explanation:
Initial concentration of weak acid =
pH = 6.87
![pH = -log[H^+]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%2B%5D)
![[H^+]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D)
![[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-6.87%7D%3D1.35%20%5Ctimes%2010%5E%7B-7%7D%5C%20M)
HA dissociated as:

(0.00045 - x) x x
[HA] at equilibrium = (0.00045 - x) M
x = 
![Ka = \frac{[H^+][A^{-}]}{[HA]}](https://tex.z-dn.net/?f=Ka%20%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)

0.000000135 <<< 0.00045
