How many protons and electrons are present in a vanadium atom
2 answers:
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Answer:
599.26 grams of potassium sulfate will be produced.
Explanation:
Moles of chromium (III) sulfate =
According to reaction, 1 mole of chromium (III) sulfate gives 3 moles of potassium sulfate.
Then 1.1480 moles of chromium (III) sulfate will give:
Mass of 3.4440 moles of potassium sulfate:
= 3.4440 mol × 174 g/mol = 599.26 g
599.26 grams of potassium sulfate will be produced.
Answer:
- Mass of monobasic sodium phosphate = 1.857 g
- Mass of dibasic sodium phosphate = 1.352 g
Explanation:
<u>The equilibrium that takes place is:</u>
H₂PO₄⁻ ↔ HPO₄⁻² + H⁺ pka= 7.21 (we know this from literature)
To solve this problem we use the Henderson–Hasselbalch (<em>H-H</em>) equation:
pH = pka +
In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21
If we use put data in the <em>H-H </em>equation, and solve for [HPO₄⁻²], we're left with:
From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M
We replace the value of [HPO₄⁻²] in this equation:
0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M
1.616 * [H₂PO₄⁻] = 0.1 M
[H₂PO₄⁻] = 0.0619 M
With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:
[HPO₄⁻²] + 0.0619 M = 0.1 M
[HPO₄⁻²] = 0.0381 M
With the concentrations, the volume and the molecular weights, we can calculate the masses:
- Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.
- Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.
- mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 g
- mass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 g