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.
<span>The correct answer is 'freezing point depression'. Colligative properties depend on the concentration of molecules of a solute. Examples of other colligative properties are boiling point elevation or vapour pressure lowering. The salt causes ice on the side walk to melt because it lowers the freezing point. </span>
Answer:
V₂ = 15.00 atm
Explanation:
Given data:
Initial pressure = 5.00 atm
Initial volume = 3.00 L
Final pressure = 760 mmHg ( 760/760 = 1 atm)
Final volume = ?
Solution:
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = 5.00 atm × 3.00 L / 1 atm
V₂ = 15.00 atm
The correct answer would be 32/16s
Answer: option B.
carbon + oxygen → carbon dioxide
Explanation:
