Answer:
1) Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make 0.157 mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.
2) 2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.
Explanation:
- Firstly, we should balance the equation of heptane combustion.
- We can balance the equation by applying the conservation of mass to the equation.
- The balanced equation is: <em>2NH₄OH + H₂SO₄ → (NH₄)₂SO₄ + 2H₂O.</em>
- This means that every 2.0 moles of ammonium hydroxide (NH₄OH) will produce 1.0 mole of ammonium sulfate (NH₄)₂SO₄ when it is neutralized by sulfuric acid.
- We need to calculate the no. of moles in 11.0 g of ammonium hydroxide that is neutralized using the relation: <em>n = mass/molar mass.
</em>
n of 11.0 g of ammonium hydroxide (NH₄OH) = mass/molar mass = (11.0 g)/(35.04 g/mol) = 0.314 mol.
<u><em>Using cross multiplication:
</em></u>
2.0 moles of NH₄OH make → 1.0 mole of (NH₄)₂SO₄.
0.314 mol of NH₄OH make → ??? moles of (NH₄)₂SO₄.
∴ The no. of moles of (NH₄)₂SO₄ that will be made from neutralizing (11.0 g) of NH₄OH = (0.314 mol)(1.0 mol)/(2.0 mol) = 0.157 mol.
<em>∴ Ammonium hydroxide is neutralized by sulfuric acid to produce ammonium sulfate and water. It will make </em><em>0.157</em><em> mol ammonium sulfate when you neutralize 11.00 g ammonium hydroxide.</em>
D ) 0.30 atm, you just subtract 1.5 with 1.2 I forgot what the law was called
Answer:
Can you send the picture of this question
The molarity of (HNO₃) that was used if 2.00 L must be used to prepare 4.5 L of a 0.25M HNO₃ solution is 0.563 M
<u><em>calculation</em></u>
This is calculated usind M₁V₁=M₂V₂ formula
where,
M₁( molarity ₁) = ?
V₁( volume ₁) = 2.00 L
M₁ (molarity ₂) = 0.25M
V₂( volume₂) = 4.5 L
make M₁ the subject of the formula by diving both side of the formula by V₁
M₁ is therefore = M₂V₂/V₁
M₁ =[ (0.25 M x 4.5 L) / 2.00 L ] =0.563 M
