Answer:
I would go with A
Explanation:
Because the earths equator is warmed by most direct rays of the sun, air a the equator is hotter than air further north or the south. The hotter air rises up at the equator and as colder air moves in to take its place, the wind begins to blow and push the ocean into waves and currents
Answer:
fast
Explanation:
Because kinetic energy is equal to half the product of mass and square of velocity, increase in velocity would increase the kinetic energy as well.
Answer:
5.42
Explanation:
Step 1: Consider the dissociation of NH₄Br
NH₄Br(aq) ⇒ NH₄⁺(aq) + Br⁻(aq)
Br⁻ is the conjugate base of HBr, a strong acid, so it doesn´t react with water. NH₄⁺ is the conjugate acid of NH₃, so it does react with water.
Step 2: Consider the acid reaction of NH₄⁺
NH₄⁺(aq) + H₂O(l) ⇄ NH₃(aq) + H₃O⁺(aq)
Step 3: calculate the acid dissociation constant for NH₄⁺
We will use the following expression.
![K_a \times K_b = K_w\\K_a = \frac{K_w}{K_b} = \frac{1.00 \times 10^{-14} }{1.76 \times 10^{-5}} = 5.68 \times 10^{-10}](https://tex.z-dn.net/?f=K_a%20%5Ctimes%20K_b%20%3D%20K_w%5C%5CK_a%20%3D%20%5Cfrac%7BK_w%7D%7BK_b%7D%20%3D%20%5Cfrac%7B1.00%20%5Ctimes%2010%5E%7B-14%7D%20%7D%7B1.76%20%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%205.68%20%5Ctimes%2010%5E%7B-10%7D)
Step 4: Calculate the concentration of H₃O⁺
We will use the following expression.
![[H_3O^{+} ]= \sqrt{K_a \times C_a } = \sqrt{5.68 \times 10^{-10} \times 0.0255 } = 3.81 \times 10^{-6}M](https://tex.z-dn.net/?f=%5BH_3O%5E%7B%2B%7D%20%5D%3D%20%5Csqrt%7BK_a%20%5Ctimes%20C_a%20%7D%20%3D%20%5Csqrt%7B5.68%20%5Ctimes%2010%5E%7B-10%7D%20%5Ctimes%200.0255%20%7D%20%3D%20%203.81%20%5Ctimes%2010%5E%7B-6%7DM)
Step 5: Calculate the pH
We will use the following expression.
![pH = -log [H_3O^{+} ] = -log (3.81 \times 10^{-6}) = 5.42](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH_3O%5E%7B%2B%7D%20%5D%20%3D%20-log%20%283.81%20%5Ctimes%2010%5E%7B-6%7D%29%20%3D%205.42)