The total change in internal energy would simply be
calculated using the formula:
ΔU = -P (V2 – V1) + ΔH
where ΔU is the change in internal energy; P is constant
pressure = 30 atm = 3,039,750 Pa; V2 is final volume = 2 L = 0.002 m^3; V1 is
initial volume = 7.20 L = 0.0072 m^3; while ΔH is the heat = -74,400 J (heat
released so negative)
Therefore:
ΔU =-3,039,750 Pa * (0.002 m^3 - 0.0072 m^3) + (- 74,400
J)
<span>ΔU = - 58,593.3 J = - 58.6 kJ</span>
The correct answer would be B for sure
Answer:i think its 6 what do you think
Explanation:
This compound should be classified as a suspension fluid. This happens when a fluid is denser than another fluid.
Answer: KI> Al(NO3)3> HF> CH3OH
Explanation:
Potassium iodide is a halide of group 1 which is very soluble in water and whose ions are very mobile in solution. It will readily dissociate and conduct electricity making the bulb brightest when KI is the electrolyte. Salts of group 13 are not as ionic as those of group 1 but dissociate appreciably and conduct electricity. HF is a halogen acid but self ionizes and shows some degree of conductivity. Methanol will have the lowest conductivity making the bulb dimmest when methanol is the electrolyte.
