Answer:
1. B, D,
2.A, F
Explanation:
1. According to the law of conservation of mass, In a course of chemical reaction, matter can neither be created nor destroyed but can be changed from one form to another. This means the amount of matter at the begining and ending of a reaction must be thesame.
2. Chemical reaction is not easily reversible. when gas is produced, provided the reaction system is an open system, the gas cannot be recovered and the reactants cannot be recovered from the products. likewise color change are attributed to chemical reaction
Ideal Gas law PV=nRT
P- pressure(atm)
V-volume( liter)
R- gas constant
T- temperature(kelvin)
n - number of moles
The calculation is (Measurement in m³ x 1000³)
Therefore
Measurement in m³ x 1 000 000
Its not a fraction as there are 1000mm in a m, so to convert from m to mm you must multiply

Here the base is a benzoate ion, which is a weak base and reacts with water.

The equation indicates that for every mole of OH- that is produced , there is one mole of C6H5COOH produced.
Therefore [OH-] = [C6H5COOH]
In the question value of PH is given and by using pH we can calculate pOH and then using pOH we can calculate [OH-]
pOH = 14 - pH
pH given = 9.04
pOH = 14-9.04 = 4.96
pOH = -log[OH-] or ![[OH^{-}] = 10^{^{-pOH}}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%2010%5E%7B%5E%7B-pOH%7D%7D%20)
![[OH^{-}] = 10^{^{-4.96}}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%2010%5E%7B%5E%7B-4.96%7D%7D%20)
![[OH^{-}] = 1.1\times 10^{-5}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%201.1%5Ctimes%2010%5E%7B-5%7D%20)
The base dissociation equation kb = 
![kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}](https://tex.z-dn.net/?f=%20kb%20%3D%5Cfrac%7B%5BC6H5COOH%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BC6H5COO%5E%7B-%7D%5D%7D)
H2O(l) is not included in the 'kb' equation because 'solid' and 'liquid' are taken as unity that is 1.
Value of Kb is given = 
And value of [OH-] we have calculated as
and value of C6H5COOH is equal to OH-
Now putting the values in the 'kb' equation we can find the concentration of C6H5COO-
![kb =\frac{[C6H5COOH][OH^{-}]}{[C6H5COO^{-}]}](https://tex.z-dn.net/?f=%20kb%20%3D%5Cfrac%7B%5BC6H5COOH%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BC6H5COO%5E%7B-%7D%5D%7D)
![1.6\times 10^{-10} = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{[C6H5COO^{-}]}](https://tex.z-dn.net/?f=%201.6%5Ctimes%2010%5E%7B-10%7D%20%3D%20%5Cfrac%7B%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B%5BC6H5COO%5E%7B-%7D%5D%7D%20)
![[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}](https://tex.z-dn.net/?f=%20%5BC6H5COO%5E%7B-%7D%5D%20%3D%20%5Cfrac%7B%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%5B1.1%5Ctimes%2010%5E%7B-5%7D%5D%7D%7B1.6%5Ctimes%2010%5E%7B-10%7D%7D%20)
or 
So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L
Moles of NaC6H5COO would be = 
Moles of NaC6H5COO (sodium benzoate) = 0.38 mol