The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by
1 answer:
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Answer:
3m/s²
Explanation:
Given parameters:
Mass of object = 3.2kg
Force to the right = 16.3N
Force to the left = 6.7N
Unknown:
Acceleration of the object = ?
Solution:
To solve this problem, we use newtons second law of motion;
Net force = mass x acceleration
Net force on object = Force to the right - Force to the left
Net force = 16.3N - 6.7N = 9.6N
So;
9.6 = 3.2 x a
a = = 3m/s²
Answer:
The heat at constant pressure is -3,275.7413 kJ
Explanation:
The combustion equation is 2C₆H₆ (l) + 15O₂ (g) → 12CO₂ (g) + 6H₂O (l)
= (12 - 15)/2 = -3/2
We have;
Where R and T are constant, and ΔU is given we can write the relationship as follows;
Where;
H = The heat at constant pressure
U = The heat at constant volume = -3,272 kJ
= The change in the number of gas molecules per mole
R = The universal gas constant = 8.314 J/(mol·K)
T = The temperature = 300 K
Therefore, we get;
H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ
The heat at constant pressure, H = -3,275.7413 kJ.