F and O
because they are more reactive than Cl
Answer:
8.0356 * 10^-5 moles of NaHCO3
Explanation:
Sulphuric acid = H2SO4
Sodium bicarbonate = NaHCO3
The reaction between both compounds is given by;
2NaHCO3(aq) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)
In the reactin above;
2 mol of NaHCO3 neutralizes 1 mol of H2SO4
At stp, 1 mol occupies 22.4 L;
1 mol = 22.4 L = 22400 mL
x mol = 0.9 mL
x = 0.9 / 22400 = 4.0178 * 10^-5 moles of H2SO4
Since 2 mol = 1 mol from the equation;
x mol = 4.0178 * 10^-5
x mol = 2 * 4.0178 * 10^-5
x = 8.0356 * 10^-5 moles of NaHCO3
Answer : The pH of buffer is 9.06.
Explanation : Given,
Concentration of HBrO = 0.34 M
Concentration of KBrO = 0.89 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[KBrO]}{[HBrO]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BKBrO%5D%7D%7B%5BHBrO%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of buffer is 9.06.
The answer is the third choice because is oxidation half reactions, only the oxidation state of the reducing agent changes; in this third choice, Fe2+ is oxidized and becomes Fe3+. Just keep in mind that half reactions usually include the change in electrons:
Fe2+ -> Fe3+ +e-<span />
