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Answer:
Mass = 713.4 ×10⁻⁴ g
Explanation:
Given data:
Number of moles of calcium phosphate = 2.3×10⁻⁴ mol
Mass of calcium phosphate = ?
Solution:
Formula:
Number of moles = mass/molar mass
Molar mass of calcium phosphate is 310.18 g/mol
by putting values,
2.3×10⁻⁴ mol = mass / 310.18 g/mol
Mass = 2.3×10⁻⁴ mol × 310.18 g/mol
Mass = 713.4 ×10⁻⁴ g
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Answer: There are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.
Explanation:
Given: 
= 1.0 M, 
= ?
= 0.1 M, 
= 200 mL
Formula used is as follows.
Substitute values into the above formula as follows.
Thus, we can conclude that there are 20 milliliters of 1.0 M HCl is required to be diluted to make 200 mL of a 0.1 M solution.
Answer:
We'll have 8.0 moles Fe3O4 and 4.0 moles CO2
Explanation:
Step 1: data given
Number of moles Fe2O3 = 12.0 moles
Number of moles CO = 12.0 moles
Step 2: The balanced equation
3Fe2O3 +CO → 2Fe3O4 + CO2
Step 3: Calculate the limiting reactant
For 3 moles Fe2O3 we need 1 mol CO to produce 2 moles Fe3O4 and 1 mol CO2
Fe2O3 is the limiting reactant. It will completely be consumed (12.0 moles).
CO is in excess. There will react 12.0 / 3 = 4.0 moles
There will remain 12.0 - 4.0 = 8.0 moles
Step 4: Calculate moles products
For 3 moles Fe2O3 we need 1 mol CO to produce 2 moles Fe3O4 and 1 mol CO2
For 12.0 moles Fe2O3 we'll have 2/3 * 12.0 = 8.0 moles Fe3O4
For 12.0 moles Fe2O3 we'll have 12.0 / 3 = 4.0 moles CO2
We'll have 8.0 moles Fe3O4 and 4.0 moles CO2
