The empirical formula of the compound obtained from the question given is NaBrO₃
<h3>Data obtained from the question </h3>
- Sodium (Na) = 15.24%
- Bromine (Br) = 52.95%
- Oxygen (O) = 31.81%
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as illustrated below:
Divide by their molar mass
Na = 15.24 / 22.99 = 0.663
Br = 52.95 / 79.90 = 0.663
O = 31.81 / 16 = 1.988
Divide by the smallest
Na = 0.663 / 0.663 = 1
Br = 0.663 / 0.663 = 1
O = 1.988 / 0.663 = 3
Thus, the empirical formula of the compound is NaBrO₃
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Here is your answer
B. NaCl
_________________
In option A. Na isn't present.
In option C. there are two atoms of Na
So, option B is correct
HOPE IT IS USEFUL
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No. This is a double displacement reaction.
This is because the only spectator ion is potassium ion, K+. Sulfuric acid is a strong acid, but only for the first H+ which is produced. The remaining HSO4^- ion is a weak acid, and is only minimally ionized. (*) Potassium hydroxide is a strong base and is completely ionized. As for the products, K2SO4 is completely ionized, but water is not. H+ and OH- combine to make molecular water
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2HOH(l) ….. “molecular” equation
H+ + HSO4^- + 2K+ + 2OH- → 2K+ + SO4^2- + 2HOH(l) …. ionic equation
H+ + HSO4^- + 2OH → SO4^2- + 2HOH(l) ……. net ionic equation