Answer:
5 moles of oxygen are required.
Explanation:
Given data:
Moles of O₂ required = ?
Moles of H₂ present = 10 mol
Solution:
Chemical equation:
O₂ + 2H₂ → 2H₂O
Now we will compare the moles of oxygen and hydrogen.
H₂ : O₂
2 : 1
10 : 1/2×10 = 5 mol
5 moles of oxygen are required.
To calculate the mean, you add up all of the data values, and then divide that sum by the *number* of values.
For instance, if you wanted to find the mean score at a home run derby, and you’re given the following numbers for home runs scored by each player:
5, 4, 6, 5, 3, 1
You could calculate the mean by adding all of the score up
5 + 4 + 6 + 5 + 3 + 1 = 24
And dividing by the number of hitters (in this case, 6)
24 / 6 = 4
So the *mean score* of the home run derby would be 4.
<h3>Answer:</h3>
#1. Ca²⁺
# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
<h3>Explanation:</h3>
The question above concerns solubility of salts or ions in water.
The solution given contains Ag+, Ca2+, and Co2+ ions.
- In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
- In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.
- The net ionic equation for the reaction is;
Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)
- From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.
- In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.
- The net ionic equation for the reaction is;
3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
- According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.
Answer:
C is the excess reactant.
Explanation:
Reaction is C + O2 --> CO2
1mol of C required to react with 1mol O2
Therefore 15 - 10 = 5moles of C will be in excess
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