Answer:
Concentration of hydroxide-ion at equivalence point = 
Explanation:

1 mol of
reacts with 1 mol of KOH to produce 1 mol of 
At equivalence point, all
gets converted to
.
Moles of
produced at equivalence point is equal to moles of KOH added to reach equivalence point.
So, moles of
produced = 
Total volume of solution at equivalence point = (25.00+43.76) mL = 68.76 mL
Concentration of
at equivalence point = 
produced at equivalence point is due to hydrolysis of
. We have to construct an ICE table to calculate concentration of
at equivalence point.

I:0.0897 0 0
C: -x +x +x
E: 0.0897-x x x
![\frac{[HC_{3}H_{5}O_{2}][OH^{-}]}{[C_{3}H_{5}O_{2}^{-}]}=K_{b}(C_{3}H_{5}O_{2}^{-})=\frac{10^{-14}}{K_{a}(HC_{3}H_{5}O_{2})}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHC_%7B3%7DH_%7B5%7DO_%7B2%7D%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BC_%7B3%7DH_%7B5%7DO_%7B2%7D%5E%7B-%7D%5D%7D%3DK_%7Bb%7D%28C_%7B3%7DH_%7B5%7DO_%7B2%7D%5E%7B-%7D%29%3D%5Cfrac%7B10%5E%7B-14%7D%7D%7BK_%7Ba%7D%28HC_%7B3%7DH_%7B5%7DO_%7B2%7D%29%7D)
species inside third bracket represent equilibrium concentrations
So, 
or, 
So,
M = 
So, concentration of hydroxide-ion at equivalence point = x M = 
Answer:
Volume will decreased by decreasing the amount of gas.
Explanation:
The amount of gas and volume are directly related to each other. By decreasing the amount of gas volume will decrease by keeping the temperature and pressure constant.
Mathematical expression:
V ∝ n
V = Kn
we put the K by removing the sign of proportionality.
K = proportionality constant
n = amount of gas
V = volume of gas
The volume- amount relation is called Avogadro law.
By increasing the amount of gas volume will also goes to increase.
It is a noble gas, it is flammable, it combines to form water
Answer:
Explanation:
Since temperature is to be maintained ; constant temperature and a such, the process is ISOTHERMAL
- to calculate mass of then gas = P1V/RT1
- m1 = 10 x 10^5 x 0.3 / 287 x 330
- similarly for m2 = P2V/RT2
- hence required mass = m1 - m2
b) Considering a well-insulated cylinder ;
Cp = 29J/molK
Temperature T2 = T1 [ P1/P2]^(1-γ)/γ
where γ = 1.4
plugging the values into the equation ; T2 = 171k
hence mass of the gas m1 = P1V/RT1
= 10 x 10^5 x 0.3/287 x 171
= 6.11kg
mass of m2 = P2V/RT2
= 10^5 x 0.3/287 x 171
= 0.611kg
hence required mass = m1 - m2
= 5.50kg