Answer:
B. 111 J
Explanation:
The change in internal energy is the sum of the heat absorbed and the work done on the system:
ΔU = Q + W
At constant pressure, work is:
W = P ΔV
Given:
P = 0.5 atm = 50662.5 Pa
ΔV = 4 L − 2L = 2 L = 0.002 m³
Plugging in:
W = (50662.5 Pa) (0.002 m³)
W = 101.325 J
Therefore:
ΔU = 10 J + 101.325 J
ΔU = 111.325 J
Rounded to three significant figures, the change in internal energy is 111 J.
Answer:
<u>reaction are equal.</u>
Explanation:
The law of the conservation of mass states that the total mass before and after a reaction are equal.
In a reaction, the mass of the reactants are always equal to the mass of the products. Nothing ever disappears from the equation, although substances may become gas and "disappear" the air.
C=10⁻⁶ mol/L
pH=14-pOH
pOH=-lg[OH⁻]
pH=14+lg10⁻⁶=14-6=8
B. pH = 8
Answer:
Heat transfer in step 2 = 47.75 J
Explanation:
Internal energy = heat + work done
U = Q + W
In a cyclic process the total internal energy change of the system = 0.
In the process there are two steps. The total heat exchange in the process is the sum of heat exchanges in the two processes.
We have to find the heat exchange in step 2.
In step 1,
W = 1.25 J Q = -37 J
= -37 + 1.25 = -35.75 J
In step 2, the internal energy change will be negative of that in step 1.
U = 35.75 J
W = -12 J
U = Q + W
35.75 = Q -12
Q = 47.75 J
Heat transfer in step 2 = 47.75 J
For the first one the answer is B. and the second one is D.
