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erik [133]
2 years ago
7

Why is the mesosphere colder than the troposphere

Chemistry
1 answer:
mezya [45]2 years ago
3 0
The mesosphere is the coldest of the atmospheric layers. In fact it is colder than the Polar regions of the Earth cold enough to freeze water vapor into ice clouds. As you increase in altitude, temperature decreases The thermosphere is the layer to which this pattern does not apply.
You might be interested in
identify the reagents you would use to convert each of the following compounds into pentanoic acid: (a) 1-pentene (b) 1-bromobut
Morgarella [4.7K]

a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

In the second step, the 1-pentylborane is oxidized to pentanoic acid using hydrogen peroxide (H₂O₂) or other suitable oxidizing agents. The oxidation is catalyzed by an acid, such as hydrochloric acid (HCl), and proceeds via a proton transfer from the 1-pentylborane to the hydrogen peroxide. The end result is the conversion of 1-pentane to pentanoic acid.

The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

b)The conversion of 1-Bromo butane to pentanoic acid using sodium cyanide (NaCN) proceeds via a nucleophilic substitution reaction. The reaction mechanism involves the following steps:

1. Attack of the nucleophile, NaCN, on the carbon atom of 1-Bromo butane to form a tetrahedral intermediate.

2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

To know more about reagents, click below:

brainly.com/question/26283409

#SPJ4

3 0
1 year ago
This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial press
Solnce55 [7]

Answer:

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A ( phosphine in this case ) after a time, t

           [A]₀ is the initial concentration of A

           k is the rate constant, and

           t is the time

We also know that for a first order reaction

           k = 0.693/ t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t/= 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 /  35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀  = - kt

thus:

(pPH₃ )t  = 2.29 atm x e^(- 0.0198 s⁻¹ x 70.5 s ) = 0.57 atm

3 0
3 years ago
What is the smallest part of Substance called?
enot [183]
It is definitely a molecule 
3 0
3 years ago
Read 2 more answers
A plot of 1/[BrO-] vs. time is linear and the slope is equal to 0.056 M-1s-1. If the initial concentration of BrO- is 0.65 M, ho
finlep [7]

Answer:

time taken for one-half of the BrO⁻ ion to react is t= 27.45 secs

Explanation:

equation of reaction

3BrO⁻(aq) → BrO₃⁻(aq) + 2Br⁻(aq) (second order reaction)

given

the slope of the graph is 0.056M⁻¹s⁻¹ = k(constant)

initial concentration [A]₀ = 0.65M

for second order reaction,we can calculate the time taken for one-half of the BrO- ion to react using:

\frac{1}{[A]} =\frac{1}{[A]}₀ ⁺ k × t

where initial concentration [A]₀ = 0.65M

[A] = [A]₀÷2 = 0.325M

\frac{1}{0.325M} = \frac{1}{0.65M} + 0.056M⁻¹s⁻¹ × t

3.077= 1.54 + 0.056t

3.077-1.54=0.056t

1.537=0.056t

t= 27.45 secs

4 0
2 years ago
I can't figure out how to remotely come up with an answer close to any of the multiple choice options.
noname [10]
Hey:) Have you tried googling your question?
8 0
3 years ago
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