Answer:
108.43 grams KNO₃
Explanation:
To solve this problem we use the formula:
Where
- ΔT is the temperature difference (14.5 K)
- Kf is the cryoscopic constant (1.86 K·m⁻¹)
- b is the molality of the solution (moles KNO₃ per kg of water)
- and<em> i</em> is the van't Hoff factor (2 for KNO₃)
We <u>solve for b</u>:
- 14.5 K = 1.86 K·m⁻¹ * b * 2
Using the given volume of water and its density (aprx. 1 g/mL) we <u>calculate the necessary moles of KNO₃</u>:
- 275 mL water ≅ 275 g water
- moles KNO₃ = molality * kg water = 3.90 * 0.275
- moles KNO₃ = 1.0725 moles KNO₃
Finally we <u>convert KNO₃ moles to grams</u>, using its molecular weight:
- 1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃
The total number of atoms make up the products :
D) 1 carbon, 4 hydrogen, and 4 oxygen
<h3>Further explanation
</h3>
Complete combustion of Hydrocarbons with Oxygen will produce CO₂ and H₂O compounds.
If O₂ is insufficient there will be incomplete combustion produced by CO and H and O
Hydrocarbon combustion reactions (especially alkanes)
For combustion of methane (CH₄) and two molecules of oxygen (O₂).
The number of atoms make up the products
CO₂ : 1 carbon, 2 oxygen
2H₂O : 4 hydrogen , 2 oxygen
Answer:b
Explanation:
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Answer:
0.665 moles of CO₂
Explanation:
The balance chemical equation for the combustion of Ethane is as follow:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
Step 1: <u>Calculate moles of C₂H₆ as;</u>
Moles = Mass / M.Mass
Putting values,
Moles = 10.0 g / 30.07 g/mol
Moles = 0.3325 moles
Step 2: <u>Calculate Moles of CO₂ as;</u>
According to balance chemical equation,
2 moles of C₂H₆ produced = 4 moles of CO₂
So,
0.3325 moles of C₂H₆ will produce = X moles of CO₂
Solving for X,
X = 0.3325 mol × 4 mol ÷ 2 mol
X = 0.665 moles of CO₂
