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2 years ago

Which forms when a humid air mass rises into a cooler temperature area?

Chemistry

2 answers:

Zielflug [23.3K]2 years ago

3 0

Option A: Clouds

In the morning, air is cool and as sun begins to rise it starts increasing the temperature of air. By time, the air becomes warmer and warmer. Depending upon the surrounding conditions, air in different areas heat up at different rates.

Due to this heating, thermal formation takes place, this is due to uneven heating of surface of earth. The thermal formation at surface causes difference in temperature of surface of the earth and air around it. The warm air has tendency to rise thus, the air in the thermal rise and expand. Due to expansion it cools down, this process continues till the temperature of thermal air reaches equals to the temperature of surrounding air. This results in the formation of cloud.

Thus, when a humid air mass rises into a cooler temperature area, clouds formation takes place

Karo-lina-s [1.5K]2 years ago

3 0

The correct answer is A

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Name two control substance?

guapka [62]

The two control bases would be water and salt.

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g A laboratory analysis of an unknown compound found the following composition: C 75.68% ; H 8.80% ; O 15.52%. What is the empir

sammy [17]

Answer:

THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O

Explanation:

The empirical formula for the unknown compound can be obtained by following the processes below:

1 . Write out the percentage composition of the individual elements in the compound

C = 75.68 %

H = 8.80 %

O = 15.52 %

2. Divide the percentage composition by the atomic masses of the elements

C = 75 .68 / 12 = 6.3066

H = 8.80 / 1 = 8.8000

O = 15.52 / 16 = 0.9700

3. Divide the individual results by the lowest values

C = 6.3066 / 0.9700 = 6.5016

H = 8.8000 / 0.9700 = 9.0722

O = 0.9700 / 0.9700 = 1

4. Round up the values to the whole number

C = 7

H = 9

O = 1

5 Write out the empirical formula for the compound

C7H90

In conclusion, the empirical formula for the unknown compound is therefore C7H9O

7 0

3 years ago

Name a possible product of this reaction in the presence of ether and AlCl3: methylbenzene + 1-chlorodecane.a. 1-methyl-2-decylb

Vikki [24]

Answer:

None of these

Explanation:

Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.

Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :

Step -1 : Generation of stable carbocation.

Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.

Step-2: Attack of the ring to the carbocation

The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.

The product formed is shown in mechanism does not mention in any of the options.

So, None of these is the answer