Answer:
w = -531 kJ
1. Work was done by the system.
Explanation:
Step 1: Given data
- Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).
- Change in the internal energy of the system (ΔU°): 156 kJ
Step 2: Calculate the work done (w)
We will use the following expression.
ΔU° = q + w
w = ΔU° - q
w = 156 kJ - 687 kJ
w = -531 kJ
By convention, when w < 0, work is done by the system on the surroundings.
Answer:
While Bohr's atomic model hypothesizes that electrons move in particular energy levels around the nucleus, the electron cloud model suggests that electrons move in an unpredictable pattern but are more likely to be in certain regions than others.
Answer:
1.60.
Explanation:
- The no. of millimoles of HCl = MV = (0.15 M)(20.0 mL) = 3.0 mmol.
- The no. of millimoles of KOH = MV = (0.10 M)(20.0 mL) = 2.0 mmol.
<em>Since the no. of millimoles of HCl is larger than that of KOH. The solution is acidic.</em>
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∴ M of remaining HCl [H⁺] remaining = (NV)HCl - (NV)KOH/V total = (3.0 mmol) - (2.0 mmol) / (40.0 mL) = 0.025 M.
∵ pH = - log[H⁺]
<em>∴ pH = - log[H⁺] </em>= - log(0.025) = <em>1.602 ≅ 1.60.</em>
Answer:
The correct option is D
Explanation:
Normally, beta-oxidation of fatty acid occurs in the mitchondrial matrix, however, when the fatty acid chains are too long, the beta-oxidation occurs in the peroxisomes <u>where the oxidation is not attached to ATP synthesis but rather transferred (i.e high energy electrons are transferred) to O₂ to form hydrogen peroxide</u> (H₂O₂). This is the major difference between the beta-oxidation that occurs in the peroxisomes to that which occurs in the mitochondria.
