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3 years ago

7

What happens yo temperature in your layer as u you go higher in altitude?

1 answer:

Neporo4naja [7]3 years ago

5 0

Answer:

Temperatures decrease with increasing height as the ozone layer is left behind and the air thins out with increasing altitude. The lowest portion of the low-pressure mesosphere is heated by the warm air of the upper stratosphere. This heat radiates upward, getting less intense as altitude increases.

Explanation:

Brainliest please?

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What is a possible reason that ecosystem 2 has fewer species than ecosystem 1?

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4 years ago

560cm³ of gas = 3.11g what is the vapour density

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Answer:

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Explanation:

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3 years ago

Need the answer please

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In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌

Oksana_A [137]

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of $CO$ = 0.35 mole

Volume of container = 1 L

Initial concentration of $CO=\frac{moles}{volume}=\frac{0.35moles}{1L}=0.35M$

Initial moles of $H_2O$ = 0.40 mole

Volume of container = 1 L

Initial concentration of $H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M$

equilibrium concentration of $CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M$ [/tex]

The given balanced equilibrium reaction is,

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial conc. 0.35 M 0.40M 0 0

At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}$

$K_c=\frac{x\times x}{(0.40-x)(0.35-x)}$

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

$K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}$

$K_c=0.70$

Thus the value of the equilibrium constant is 0.70.

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