<h3>
Answer:</h3>
2.125 g
<h3>
Explanation:</h3>
We have;
- Mass of NaBr sample is 11.97 g
- % composition by mass of Na in the sample is 22.34%
We are required to determine the mass of 9.51 g of a NaBr sample.
- Based on the law of of constant composition, a given sample of a compound will always contain the sample percentage composition of a given element.
In this case,
- A sample of 11.97 g of NaBr contains 22.34% of Na by mass
A sample of 9.51 g of NaBr will also contain 22.345 of Na by mass
% composition of an element by mass = (Mass of element ÷ mass of the compound) × 100
Mass of the element = (% composition of an element × mass of the compound) ÷ 100
Therefore;
Mass of sodium = (22.34% × 9.51 g) ÷ 100
= 2.125 g
Thus, the mass of sodium in 9.51 g of NaBr is 2.125 g
Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use <span>Henderson–Hasselbalch equation:
pH = pKa + log(c(</span>NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
pH = 4,02.
<span>[H</span>₃O⁺] = 10∧(-4,02).
<span>[H</span>₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.
Answer:
its ability to lose electron
