Answer:
the correct answer is an Orange baseball.
Answer:
The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.
Explanation:
- Firstly, we need to calculate the no. of moles of 4.00 g of liquid butane C₄H₁₀:
n = mass/molar mass = (4.0 g)/(58.12 g/mol) = 0.0688 mol.
∴ 0.0688 mol of butane requires a gain in enthalpy of 1.67 kJ to be evaporized.
<u><em>Know using cross multiplication:</em></u>
0.0688 mol of butane to be vaporized requires → 1.67 kJ.
1.0 mol of butane to be vaporized requires → ??? kJ.
∴ 1.0 mol of butane to be vaporized requires = (1.0 mol)(1.67 kJ)/(0.0688 mol) = 24.265 kJ.
<em>∴ The molar enthalpy of vaporization in kJ/mol for butane = 24.265 kJ/mol.</em>
Answer:
Bromine Pentachloride (BrCl5)
Explanation:
Answer: Catalyst provides an alternative path to the reaction with a lower activation energy. Catalyst binds with the active sides of the substrate. It weakens the bonds of the substrate. When the bonds are weaker, less amount of energy is required for the breakdown of the bonds. So lower amount of activation energy is required to start a reaction hence reaction takes place in lesser amount of time.