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3 years ago

Dan described two layers of Earth's atmosphere.

Chemistry

2 answers:

tester [92]3 years ago

6 0

Answer:

The layer A is the thermosphere and the layer B is the exosphere.

Explanation:

One of the five layers of the atmosphere of the Earth is called thermosphere. It is the fourth layer of the Earth's atmosphere and is situated below the exosphere and above the mesosphere. The thermosphere is the biggest layer of the Earth's atmosphere, and the charged particles found in the thermosphere makes it easier for long distance communication through radio.

The exosphere is the topmost layer of the Earth's atmosphere, here the atmosphere merges with the interplanetary space and thins out. It is situated directly above the thermosphere. It is the layer, which distinguishes Earth's atmosphere from outer space.

saul85 [17]3 years ago

4 0

There are following layers in earth atmosphere

a) Troposhere: the innermost layer or lowest part of atmosphere

b)stratosphere: next to troposhere

c)mesosphere:above stratosphere, here lowest temperature is observed

d)Thermosphere and Ionosphere: This layer allows long distance radio communication

e)Exosphere: this layer separates outer space from earth's surface.

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What volume in litres will 38gr of F2 occupy at 0.999 bar and 273 K

Oliga [24]

Answer:

$V=22.68L$

Explanation:

Hello,

In this case, we use the ideal gas equation to compute the volume as shown below:

$PV=nRT\\\\V=\frac{nRT}{P}$

Nonetheless we are given mass, for that reason we must compute the moles of gaseous fluorine (molar mass: 38 g/mol) as shown below:

$n=38 g *\frac{x}{y} \frac{1mol}{38 g} =1mol$

Thus, we compute the volume with the proper ideal gas constant, R:

$V=\frac{1mol*0.083\frac{bar*L}{mol*K}*273K}{0.999bar} \\\\V=22.68L$

Best regards.

6 0

3 years ago

You have two containers at 25°C and 1 atm. One has 22.4 L of hydrogen gas, and the other has 22.4 L of oxygen gas.

olga_2 [115]

Both containers have the same number of molecules= 5.62 x 10²³

<h3>Further explanation</h3>

Given

22.4 L of hydrogen gas

22.4 L of oxygen gas

25°C and 1 atm.

Required

true statement

Solution

Conditions at T 25 ° C and P 1 atm are stated by RTP (Room Temperature and Pressure). Vm in this condition = 24 liters / mol

and from Avogadro's Law :

<em>At the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles </em>

So two gas have the same molecules

1 mol = 6.02 x 10²³ molecules

24 L = 1 mol, so for 22.4 L :

$\tt \dfrac{22.4~L}{24~L}\times 6.02\times 10^{23}~molecules=5.62\times 10^{23}~molecules$

6 0

3 years ago

Molecule Energy Released

charle [14.2K]

Carbohydrate; contains a low ratio of oxygen atoms.

4 0

3 years ago

Read 2 more answers

the half life of the radioactive element strontium-90 is 29.1 years. If 16 grams of strontium-90 are initially present, how many

8_murik_8 [283]

Answer:

4 g after 58.2 years

0.0156 After 291 years

Explanation:

Given data:

Half-life of strontium-90 = 29.1 years

Initially present: 16g

mass present after 58.2 years =?

Mass present after 291 years =?

Solution:

Formula:

how much mass remains =1/ 2n (original mass) ……… (1)

Where “n” is the number of half lives

to find n

For 58.2 years

n = 58.2 years /29.1 years

n= 2

or 291 years

n = 291 years /29.1 years

n= 10

Put values in equation (1)

Mass after 58.2 years

mass remains =1/ 22 (16g)

mass remains =1/ 4 (16g)

mass remains = 4g

Mass after 58.2 years

mass remains =1/ 210 (16g)

mass remains =1/ 1024 (16g)

mass remains = 0.0156g

8 0

4 years ago

Read 2 more answers

A 4 L sample of gas at 298 K and 2 atm contains 0.250 mol of gas. If we add another 0.250 mol of gas at the same pressure and te

andrey2020 [161]

Answer:

$V_2=8L$

Explanation:

Hello there!

In this case, considering the Avogadro's gas law, which treats the volume and moles in a directly proportional way via:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

Which can be solved for the final volume, V2, as shown below:

$V_2=\frac{V_1n_2}{n_1}$

Thus, by plugging in the given data, we obtain:

$V_2=\frac{4L*(0.250mol+0.250mol)}{0.250mol}\\\\V_2=8L$

Regards!

4 0

3 years ago