<span>The atomic weight of 13C should be pretty close to 13.0. (If you have the exact mass, use it in the problem.) So,
9.00 g / 13.0 g/mol = 0.692 moles
Therefore, the answer should be 0.692 moles are in 9.00 g of 13C.</span>
Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
it would be A ,inorganic Compound
Answer:
<h3>1. B</h3><h3>2. A</h3><h3>3. B</h3><h3>4. B</h3><h3>5. C</h3><h3>I HOPE IT HELPS :) 100% sureness</h3>