<h3>
Answer:</h3>
0.89 J/g°C
<h3>
Explanation:</h3>
Concept tested: Quantity of heat
We are given;
- Mass of the aluminium sample is 120 g
- Quantity of heat absorbed by aluminium sample is 9612 g
- Change in temperature, ΔT = 115°C - 25°C
= 90°C
We are required to calculate the specific heat capacity;
- We need to know that the quantity of heat absorbed is calculated by the product of mass, specific heat capacity and change in temperature.
That is;
Q = m × c × ΔT
- Therefore, rearranging the formula we can calculate the specific heat capacity of Aluminium.
Specific heat capacity, c = Q ÷ mΔT
= 9612 J ÷ (120 g × 90°C)
= 0.89 J/g°C
Therefore, the specific heat capacity of Aluminium is 0.89 J/g°C
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
<em>4 moles of Al produce 2 moles of Al₂O₃</em>
<em />
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>
Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
<h3>
78.2% </h3>
Answer:
Although hydrogen has the highest calorific value, it is not used as a domestic fuel. It is because hydrogen is a highly combustible and it reacts explosively when it comes in contact with air. And hence as a result, storing of the hydrogen gas is difficult and is dangerous at the same time.
The answer is 159.69 g/mol
