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goldenfox [79]
3 years ago
15

Which words describe an air mass likely to be over the northern parts of the Pacific Ocean? wet and cold, warm and dry, cold and

dry, wet and warm?
Chemistry
2 answers:
liubo4ka [24]3 years ago
6 0

Over the northern parts of the Pacific Ocean, the Maritime Polar air mass exists. This means that the air mass likely to be over the northern parts of the Pacific Ocean would be wet and cold.

s2008m [1.1K]3 years ago
3 0

Answer:

wet and cold

Explanation:

An air mass blowing inland from the sea is usually wet and cold. This is because of the fact that it is saturated with water and cooled by rapid exchanges with the sea. Hence any time air is coming inland from the sea, we usually experiences cool moist breeze unlike air leaving the land and moving towards the sea which a is normally hot and dry.

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The molecules will speed up.
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Write balanced net ionic equations and the corresponding equilibrium equations for the stepwise dissociation of the triprotic ac
Elza [17]

Answer:

H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1

H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2

HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3

Explanation:

Hello!

In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:

H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1

H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2

HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3

Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:

Ka1 > Ka2 > Ka3

Best regards!

3 0
2 years ago
Consider the equation for the formation of a kidney stone.
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Answer:

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The half-life for the radioactive decay of iridium-192 is 74.2 days. Calculate the amount in grams of Ir-192 that will be left f
scZoUnD [109]

Answer: There are

a) 2.12 g of Ir-192 after 199.2 days

b) 0.52 g of Ir-192 after 350 days

Explanation:

Radioactive decay follows a first-order kinetics since <em>the half life does not depend on the initial amount,</em> then the equation which describes this process is:

N(t)=N_{o}*e^{-\frac{-0.693*t}{t_{1/2} } }

Where N(t) is the amount given a certain t time, and N_{o} is the initial amount. t_{1/2} is the half life.

Then, the radioactive decay equation for this problem is:

N(t)=13.65 g*e^{-\frac{-0.693*t}{74.2 days} }

Note that the half life and the given time t must be on the same units, in this case days. Finally you calculate the amount for a) 199.2 days and b) 350 days:

a) N(t)=13.65 g*e^{-\frac{-0.693*199.2 days}{74.2 days} }=2.12 g

b) N(t)=13.65 g*e^{-\frac{-0.693*350 days}{74.2 days} }=0.52 g

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2

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