Missing question: <span> The vapor pressure of water at 26.4 °C is 25.81 mm Hg.
p(total) = p(H</span>₂O) + p(O₂).
p(total) = 817 torr; <span> total pressure of the mixture.
</span>p(H₂O) = 25.81 torr, <span>partial pressure of water.
</span>p(O₂) = 817 torr - 25.81.
p(O₂) = 791.19 torr; partial pressure of oxygen.
Answer: I think it is ‘Genetically diverse’
Explanation:
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>
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