<h3>Answer:</h3>
2.55 × 10²² Na Atoms
<h3>Solution:</h3>
Data Given:
M.Mass of Na = 23 g.mol⁻¹
Mass of Na = 973 mg = 0.973 g
# of Na Atoms = ??
Step 1: Calculate Moles of Na as:
Moles = Mass ÷ M.Mass
Moles = 0.973 g ÷ 23 g.mol⁻¹
Moles = 0.0423 mol
Step 2: Calculate No, of Na Atoms as:
As 1 mole of sodium atoms counts 6.022 × 10²³ and equals exactly to the mass of 23 g. So, we can write,
Moles = No. of Na Atoms ÷ 6.022 × 10²³ Na Atoms.mol⁻¹
Solving for No. of Na Atoms,
No. of Na Atoms = Moles × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 0.0423 mol × 6.022 × 10²³ Na Atoms.mol⁻¹
No. of Na Atoms = 2.55 × 10²² Na Atoms
<h3>Conclusion: </h3>
2.55 × 10²² sodium atoms are required to reach a total mass of 973 mg in a substance of pure sodium.
There are 2 types of electricity, static electricity and current electricity
<span>AgCl(s) → Ag+(aq) + Cl-(aq) That would be my best guess</span>
Answer:
OH⁻(aq) + NH₄⁺(aq) → NH₃(g) + H₂O(l)
Explanation:
The net ionic equation is a way yo write a chemical equation that list only the species that are involved in the reaction.
For the reaction:
NaOH(aq) + NH₄Br(aq) → NaBr(aq) + NH₃(g) + H₂O(l)
Ionic equation is:
Na⁺(aq) + OH⁻(aq) + NH₄⁺(aq) + Br⁻(aq) → Na⁺(aq) + Br⁻(aq) + NH₃(g) + H₂O(l)
<em>-Ions comes from the species that are in aqueous phase-</em>
And net ionic equation is:
<em>OH⁻(aq) + NH₄⁺(aq) → NH₃(g) + H₂O(l)</em>
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