Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
Answer:
- 602 mg of CO₂ and 94.8 mg of H₂O
Explanation:
The<em> yield</em> is measured by the amount of each product produced by the reaction.
The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.
The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

To calculate the yield follow these steps:
<u>1. Mole ratio</u>

<u />
<u>2. Convert 175mg of fluorene to number of moles</u>
- Number of moles = mass in grams / molar mass
<u>3. Set a proportion for each product of the reaction</u>
a) <u>For CO₂</u>
i) number of moles


ii) mass in grams
The molar mass of CO₂ is 44.01g/mol
- mass = number of moles × molar mass
- mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg
b) <u>For H₂O</u>
i) number of moles

ii) mass in grams
The molar mass of H₂O is 18.015g/mol
- mass = number of moles × molar mass
- mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
Answer:
double replacement MgN2O6 + KF
Explanation: