Answer: 6.64 moles of carbon.
Explanation:
Given data:
Number of moles of C = ?
Number of moles of CCl₂F₂ = 6.64 mol
Solution:
In one mole of CCl₂F₂ there is one mole of carbon two moles of chlorine and two moles of fluorine are present.
In 6.6 moles of CCl₂F₂ :
Moles of carbon = 6.64 × 1 = 6.64 moles of carbon.
Moles of chlorine = 6.64× 2 = 13.28 moles of chlorine
Moles of fluorine = 6.64× 2 = 13.28 moles of fluorine
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"windweathered" because the use of wind weathering would bring the sand all around and it would be wind weathered
Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
Answer:
electric balance, goggles, beaker
Explanation:
I did it and got it right
