So what are u asking............????????
Answer:
8.55 × 10³ cal
Explanation:
Step 1: Given and required data
- Specific heat of water (c): 1 cal/g.°C
- Initial temperature: 22.7 °C
- Final temperature: 38.8 °C
Step 2: Calculate the temperature change (ΔT)
ΔT = Final temperature - Initial temperature = 38.8 °C - 22.7 °C = 16.1 °C
Step 3: Calculate the heat required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 1 cal/g.°C × 531 g × 16.1 °C = 8.55 × 10³ cal
Answer:
677.7 mmHg
Explanation:
The first empirical study on the behaviour of a mixture of gases was carried out by John Dalton. He established the effects of mixing gases at different pressures in the same vessel.
Dalton's law states that,the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases present in the mixture of gases. When a gas is collected over water, the gas also contains some water vapour. The partial pressure of the gas will now be given as; total pressure of gas mixture - saturated vapour pressure of water (SVP) at that temperature.
Given that;
Total pressure of gas mixture = 692.2 mmHg
SVP of water at 17°C = 14.5 mmHg
Therefore, partial pressure of oxygen = 692.2-14.5
Partial pressure of oxygen = 677.7 mmHg
Answer:
c.all atoms have the same arrangement for electrons
