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3 years ago

Which sub-layer thins out into space, where there is no air

Chemistry

2 answers:

Tomtit [17]3 years ago

7 0

Answer:

The exosphere

Explanation:

In Earth atmosphere, the closer to the ground is the troposphere, then the stratosphere, then the mesosphere, in which the highest clouds and radar can reach. Then in the thermosphere, where the auroras and the satellites are form, the air glow starts to disappear until we reach the space in the last layer called exosphere.

KengaRu [80]3 years ago

3 0

The "exosphere" is the most distant and tenuous "layer" of our atmosphere.

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What is the molarity of a solution that contains 0.202 mol KCl in 7.98 L solution?

mr Goodwill [35]

Explanation:

molarity = no. of moles of solute/solution in litres

molarity =0.202/7.98

=0.025 M

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3 years ago

6.285×10^3 mg = _____? _____ kg

TiliK225 [7]

One kilogram is equal to one thousand grams. Further, one gram is equal to 1000 mg. The conversion is as shown below,
(6.285 x 10³ mg) x (1 g / 1000 mg) x (1 kg / 1000 g)
The numerical value of the operation above is 0.006285 kg.

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4 years ago

Read 2 more answers

PLEASE HELP!!!!!!!!!!!!

Nikitich [7]

Answer:

1 : NA = 1 | 2 : N = 2 | 3 :

N = 1 | O = 5 |

O = 3 | |

Explanation:

im sorry but ion know the last one

5 0

3 years ago

Just as the depletion of stratospheric ozone today threatens life on Earth today, its accumulation was one of the crucial proces

inessss [21]

Answer:

(a) rate = -(1/3) Δ[O₂]/Δt = +(1/2) Δ[O₃]/Δt

(b) Δ[O₃]/Δt = 1.07x10⁻⁵ mol/Ls

Explanation:

By definition, the reaction rate for a chemical reaction can be expressed by the decrease in the concentration of reactants or the increase in the concentration of products:

aX → bY (1)

$rate= -\frac{1}{a} \frac{\Delta[X]}{ \Delta t} = +\frac{1}{b} \frac{\Delta[Y]}{ \Delta t}$

where, a and b are the coefficients of de reactant X and product Y, respectively.

(a) Based on the definition above, we can express the rate of reaction (2) as follows:

3O₂(g) → 2O₃(g) (2)

$rate = -\frac{1}{3} \frac{\Delta[O_{2}]}{\Delta t} = +\frac{1}{2} \frac{ \Delta[O_{3}]}{ \Delta t}$ (3)

(b) From the rate of disappearance of O₂ in equation (3), we can find the rate of appearance of O₃:

$rate = +\frac{1}{2} \frac{\Delta[O_{3}]}{ \Delta t} = -\frac{1}{3} \frac{\Delta[O_{2}]}{ \Delta t}$

$+\frac{\Delta[O_{3}]}{ \Delta t} = -\frac{2}{3} \frac{\Delta[-1.61 \cdot 10^{-5}]}{ \Delta t}$

$\frac{\Delta[O_{3}]}{ \Delta t} = 1.07 \cdot 10^{-5} \frac{mol}{Ls}$

So the rate of appearance of O₃ is 1.07x10⁻⁵ mol/Ls.

Have a nice day!

6 0

4 years ago

During which season is the sun highest in the sky?

Scilla [17]

It’s summer ywwwwwwwww

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3 years ago