Answer:
485.76 g of CO₂ can be made by this combustion
Explanation:
Combustion reaction:
2 C₄H₁₀(g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g)
If we only have the amount of butane, we assume the oxygen is the excess reagent.
Ratio is 2:8. Let's make a rule of three:
2 moles of butane can produce 8 moles of dioxide
Therefore, 2.76 moles of butane must produce (2.76 . 8)/ 2 = 11.04 moles of CO₂
We convert the moles to mass → 11.04 mol . 44g / 1 mol = 485.76 g
Answer:
Explanation:
In an aqueous solution of potassium sulfate (K₂SO₄), the solute is K₂SO₄ and the solvent is water. The percentage by mass describes the grams of solute there are dissolved per 100 grams of solution. It can be calculated as:
mass percentage = (mass of solute/total mass of solution) x 100%
For example, in an aqueous solution which is 2% by mass of K₂SO₄, there are 2 grams of K₂SO₄ per 100 g of solution.
Alkali metals.
Elements found in group 1 of the periodic table.
<u>Answer:</u> The freezing point of solution is 5.35°C
<u>Explanation:</u>
The equation used to calculate depression in freezing point follows:
To calculate the depression in freezing point, we use the equation:
Or,
where,
Freezing point of pure solution = 5.5°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal freezing point elevation constant = 4.90°C/m
= Given mass of solute (naphthalene) = 2.60 g
= Molar mass of solute (naphthalene) = 128.2 g/mol
= Mass of solvent (benzene) = 675 g
Putting values in above equation, we get:
Hence, the freezing point of solution is 5.35°C
