It depends on the process.
Like for example if the process is isothermal(temperature is constant), you can use,
PV = constant or P1V1 = P2V2 where P1V1 are initial conditions and P2V2 are final.
For adiabatic process,
PV^gamma = constant or P1V1 ^gamma = P2V2 ^gamma.
where gamma = Cp
------
Cv
Cp = specific heat at constant pressure and Cv = specific at constant volume.
Value of Gamma will be given in question.
Hope this helps!
C. Represents an oxidation-reduction reaction
Answer: 0.0508mL
Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.
therefore concentration of the base is 1.0156/20 = 0.0508 mL
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V
Water decomposes when electrolyzed to produce hydrogen and oxygen gas. If 2.5 grams of water were decomposed 1.04 grams of oxygen will be formed.
BCA table:
2
O ⇒
+ 
B 0.13 0 + 0
C -0.13 0.065 + 0.065
A 0 0.065
Explanation:
Balanced equation for water decomposition into hydrogen and oxygen gases
2
O ⇒
+ 
B 0.13 0 + 0
C -0.13 0.065 + 0.065
A 0 0.065
Number of moles of water = 
mass = 2.5 grams
atomic mass= 18 grams
number of moles can be known by putting the values in the formula,
n = 
= 0.13 moles
2 moles of water gives one mole of oxygen on decomposition
so, 0.13 moles of water will give x moles of oxygen on decompsition
= 
x = 0.065 moles of oxygen will be formed.
moles to gram will be calculated as
mass =number of moles x atomic mass
= 0.065 x 16
= 1.04 grams of oxygen.