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.
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl
Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.
To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH
Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572. That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH
Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH
Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH
I hope this helps.
If molecules are in a closed container then we expect the pressure to increase as the kinetic energy increases. This is because the atoms of an element collide with the walls of the container and increase the pressure.
If we use the formula 
, where P is the pressure, V is the volume, n is the number of moles, R the ideal gas constant and T is the temperature. According to the formula, P is directly proportional to temperature. An increase in temperature leads to an increase in pressure.
Since we know that temperature is the average kinetic energy of molecules present. It means as we increase the temperature we increase the kinetic energy of the molecules which in turn leads to an increase in the pressure.
Answer:
The answer is decreased temperature and increased salinity
Explanation:
It is what is known as the thermohaline circulation
The thermohaline circulation moves the water slowly. This water moves mainly due to differences in its relative density. Much denser water sinks over water that is less dense. Two factors impact the density of seawater: temperature and salinity.
Cold water is denser than hot water:
-Water cools when it loses heat, it occurs at high latitudes.
-Water is heated when it receives energy from the sun, at low latitudes.
Saltier water is much denser than water that has less salt:
-Sea water becomes salty if the evaporation rate increases.
-Sea water becomes less salty if there is a water inlet over the sea.
