Answer:
The correct matching of the air mass and the letters in the word bank are given as follows;
1. Warm and humid ↔ D
2. Extremely cold and dry ↔ B
3. Cold and dry ↔ A
4. Cold and humid ↔ C
5. Warm and dry ↔ E
Where;
A Represents continental polar
B Represents Artic
C Represents Maritime Polar
D Represents Maritime Tropical
E Represents Continental Tropical
Explanation:
A. A continental polar is one that can be described as a cold and dry climate as the region is located at a further away from the oceanic water bodies that add humidity to the climate
B. The regions of the Artic and the Antarctic have very limited amount of precipitation every year because the air is very cold as well as dry
C. A polar climate is a cold climate region, while a maritime climate is humid.
Therefore, the maritime polar climate combines both cold and humid conditions
D. A warm and humid region has high rainfall and humidity, as such the maritime climate which are humid and the tropical climate, which are warm, combine to give a warm and humid climate
E. The continental tropical climate can be described as warm and dry, compared to the continental water bodies, due to the location being distant from and therefore, the absence of high moisture containing wind that comes from the oceans.
This is an incomplete question, here is a complete question.
The Henry's law constant for oxygen dissolved in water is 4.34 × 10⁹ g/L.Pa at 25⁰C.If the partial pressure of oxygen in air is 0.2 atm, under atmospheric conditions, calculate the molar concentration of oxygen in air-saturated and oxygen saturated water.
Answer : The molar concentration of oxygen is, 
Explanation :
As we know that,
where,
= molar solubility of 
= ?
= partial pressure of = 0.2 atm = 1.97×10⁻⁶ Pa
= Henry's law constant = 4.34 × 10⁹ g/L.Pa
Now put all the given values in the above formula, we get:
Now we have to molar concentration of oxygen.
Molar concentration of oxygen = 
Therefore, the molar concentration of oxygen is,
<span>Xe = VIII = 8 valence electrons
F = VII = 4 (7 ve) = 28 valence electrons</span>
total ve = 8 + 28 = 36 ve
<span>36 - 4(2) = 28 ve
(there are 2 electrons in each bond x 4 bonds)</span>
<span>28 - 4(6) = 4
(We assign the remaining electrons to F atoms)</span>
<span>4 - 2(2) = 0
(Therefore 4 electrons left => we have 2 lone pairs)</span>
The steric number = No. of
σ bonds + #lone pairs
= 4 σ bonds + 2 lone pairs
= 6 => d²sp³ (6 hybrid orbitals)
<span>4 bonds + 2 lone pairs
=> square planar</span>
