I would say Alaska because that's where most of the subarctic climate is found.
A and B are experiencing winter. The picture which isn't available here in this question is attached below.
Option C.
<h3><u>Explanation:</u></h3>
The earth is tilt by an angle of 23.2° to the vertical plane. This makes the seasonal variation of earth, because in some time of the year, the northern hemisphere faces the sun directly, experiencing summer and then southern hemisphere is away from summer experiencing winter and vice versa. The summer occurs when the place directly faces the sun. And the winter happens when the place obliquely faces the sun or doesn't face the sun at all.
Here in this diagram, we can see that the points A and B are the north pole and the part in northern hemisphere respectively which aren't facing the sun directly, whereas C and D are facing the sun. Thus the southern hemisphere is experiencing summer and the northern hemisphere the winter.
Answer:

Explanation:
1. Calculate the rate constant
The integrated rate law for first order decay is

where
A₀ and A_t are the amounts at t = 0 and t
k is the rate constant

2. Calculate the half-life

Answer:
(a) The system does work on the surroundings.
(b) The surroundings do work on the system.
(c) The system does work on the surroundings.
(d) No work is done.
Explanation:
The work (W) done in a chemical reaction can be calculated using the following expression:
W = -R.T.Δn(g)
where,
R is the ideal gas constant
T is the absolute temperature
Δn(g) is the difference between the gaseous moles of products and the gaseous moles of reactants
R and T are always positive.
- If Δn(g) > 0, W < 0, which means that the system does work on the surroundings.
- If Δn(g) < 0, W > 0, which means that the surroundings do work on the system.
- If Δn(g) = 0, W = 0, which means that no work is done.
<em>(a) Hg(l) ⇒ Hg(g)</em>
Δn(g) = 1 - 0 = 1. W < 0. The system does work on the surroundings.
<em>(b) 3 O₂(g) ⇒ 2 O₃(g)
</em>
Δn(g) = 2 - 3 = -1. W > 0. The surroundings do work on the system.
<em>(c) CuSO₄.5H₂O(s) ⇒ CuSO₄(s) + 5H₅O(g)
</em>
Δn(g) = 5 - 0 = 5. W < 0. The system does work on the surroundings.
<em>(d) H₂(g) + F₂(g) ⇒ 2 HF(g)</em>
Δn(g) = 2 - 2 = 0. W = 0. No work is done.