Na is cation so it loses electron to be positive and become stable losing one valence shells one electron so it's oxidation number is +1 ie A is correct
Answer: The given statement is TRUE.
Explanation:
An equilibrium reaction is one in which rate of forward reaction is equal to the rate of backward reaction.
Equilibrium constant is defined as the ratio of the product of the concentration of products to the product of the concentration of reactants each raised to their stochiometric coefficient.
For example for the given equilibrium reaction;

![K_{eq}=\frac{[H_2]^2[O_2]}{[H_2O]^2}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5BO_2%5D%7D%7B%5BH_2O%5D%5E2%7D)
Thus the given statement that in calculating the equilibrium constant for a reaction, the coefficients of the chemical equation are used as exponents for the factors in the equilibrium expression is True.
Answer:
B and C
Explanation:
When we have to do a buffer solution we always have to choose the reaction that has the <u>pKa closer to the desired pH value</u>. When we find the pKa values we will obtain:
![pKa_1=-Log[6.9x10^-^3]=2.16](https://tex.z-dn.net/?f=pKa_1%3D-Log%5B6.9x10%5E-%5E3%5D%3D2.16)
![pKa_2=-Log[6.2x10^-^8]=7.20](https://tex.z-dn.net/?f=pKa_2%3D-Log%5B6.2x10%5E-%5E8%5D%3D7.20)
![pKa_3=-Log[4.8x10^-^13]=12.31](https://tex.z-dn.net/?f=pKa_3%3D-Log%5B4.8x10%5E-%5E13%5D%3D12.31)
The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which
is the <u>acid</u> and
is the <u>base</u>. Therefore the answer for the first question is B and the answer for the second question is C.