I believe the answer would be C. slows down, hope this helps:)
Explanation:
According to Le Chatelier's principle, any disturbance caused in an equilibrium reaction will shift the equilibrium in a direction that will oppose the change.
As the given reaction is as follows.

(a) When increase the temperature of the reactants or system then equilibrium will shift in forward direction where there is less temperature. It is possible for an endothermic reaction.
Thus, formation of
will increase.
- (b) When we decrease the volume (at constant temperature) of given reaction mixture then it implies that there will be increase in pressure of the system. So, equilibrium will shift in a direction where there will be decrease in composition of gaseous phase. That is, in the backward direction reaction will shift.
Hence, formation of
will decrease with decrease in volume.
- When we increase the mount of
then equilibrium will shift in the direction of decrease in concentration that is, in the forward direction.
Thus, we can conclude that formation of
will increase then.
Answer:
21.16 MPa
Explanation:
Partial pressure of oxygen = 5.62 MPa
Total gas pressure = 26.78 MPa
But
Total pressure of the gas= sum of partial pressures of all the constituent gases in the system.
This implies that;
Total pressure of the system = partial pressure of nitrogen + partial pressure of oxygen
Hence partial pressure of nitrogen=
Total pressure of the system - partial pressure of oxygen
Therefore;
Partial pressure of nitrogen= 26.78 - 5.62
Partial pressure of nitrogen = 21.16 MPa
<span>Movement of water from an area of lower solute concentration to an area of higher solute concentration</span>
Explanation:
mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g
molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol
moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2
For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).
moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2
Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.
Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.
which is 53 mL.