Answer:
AlN₃O₉
Explanation:
Assume that you have 100 g of the compound.
Then you have 12.7 g Al, 19.7 g N, and 67.6 g O.
1. Calculate the <em>moles</em> of each atom
Moles of Al = 12.7 × 1/26.98 = 0.4707 mol Al
Moles of N = 19.7 × 1/14.01 = 1.406 mol N
Moles of O = 67.6 × 1/16.00 = 4.225 mol O
2. Calculate the <em>molar ratios</em>.
Al: 0.4707/0.4707 = 1
N: 1.406/0.4707 = 2.987
O: 4.225/0.4707 = 8.976
3. Determine the <em>empirical formula</em>
Round off all numbers to the closest integer.
Al: 1
N: 3
O: 9
The empirical formula is AlN₃O₉.
If the volume of 425 grams was 48.0 cm³, simply divide
g/cm³ = 425 g/48 cm³ = 8.85 g/cm³
If using water in water displacement, 1 mL = 1 cm³
8.85 g/cm³ = 8.85 g/mL
This density is most closely aligned with that of B. Copper
Hope I helped!
Can not be D or B and A is not enough so C is right.
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>
