Answer:
C.AgNO3 + NaCl → AgCl + NaNO3
Explanation:
1.Natural selection is the differential survival and reproduction of individuals due to differences in phenotype. It is a key mechanism of evolution, the change in the heritable traits characteristic of a population over generations.
2.Because predators were able to spot the light moths more easily, the dark moths were more likely to survive and reproduce. ... The peppered moth case is an example of natural selection. In this case, changes in the environment caused changes in the characteristics that were most beneficial for survival
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
Answer: = D
Explanation:
The atomic mass increases from Ne to O2 to Cl2 hence the boiling point also increases, therefore
Ne < O2 < Cl2
Answer:
6.25 g
Explanation:
From the question given above, the following data were obtained:
Half-life (t½) = 68.8 years
Time (t) = 344 years
Original amount (N₀) = 200 g
Amount remaining (N) =?
Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 68.8 years
Time (t) = 344 years
Number of half-lives (n) =
n = t / t½
n = 344 / 68.8
n = 5
Thus, 5 half-lives has elapsed.
Finally, we shall determine the amount of the Uranium-232 that remains. This can be obtained as follow:
Original amount (N₀) = 200 g
Number of half-lives (n) = 5
Amount remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2⁵ × 200
N = 1/32 × 200
N = 200 / 32
N = 6.25 g
Thus, the amount of Uranium-232 that remains is 6.25 g