Answer:
d. 60.8 L
Explanation:
Step 1: Given data
- Heat absorbed (Q): 53.1 J
- External pressure (P): 0.677 atm
- Final volume (V2): 63.2 L
- Change in the internal energy (ΔU): -108.3 J
Step 2: Calculate the work (W) done by the system
We will use the following expression.
ΔU = Q + W
W = ΔU - Q
W = -108.3 J - 53.1 J = -161.4 J
Step 3: Convert W to atm.L
We will use the conversion factor 1 atm.L = 101.325 J.
-161.4 J × 1 atm.L/101.325 J = -1.593 atm.L
Step 4: Calculate the initial volume
First, we will use the following expression.
W = - P × ΔV
ΔV = - W / P
ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L
The initial volume is:
V2 = V1 + ΔV
V1 = V2 - ΔV
V1 = 63.2 L - 2.35 L = 60.8 L
0.004382166 Make sure to round to the right amount of Sig Figs
Answer:
V₂ = 2509.62 cm³
Explanation:
Given data:
Initial volume = 1500 cm³
Initial temperature = -65°C (-65 + 273 = 208 K)
Final temperature = 75°C ( 75 +273 = 348 K)
Final volume = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 1500 cm³ × 348 K / 208 k
V₂ = 522000 cm³.K / 208 k
V₂ = 2509.62 cm³
Answer: 20 mg Te-99 remains after 12 hours.
Explanation: N(t) = N(0)*(1/2)^(t/t1/2)
N(t) = (80 mg)*(0.5)^(12/6)
N(t) = 20 mg remains after 12 hours
