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erastova [34]
4 years ago
11

Match the statements with the atmospheric layer they describe.

Chemistry
2 answers:
vekshin14 years ago
6 0

The correct matches are:

1. Exosphere - Temperatures reach as high as 2000 C yet it feels very cold

This is the top layer of the atmosphere. The atoms are so dispersed that despite it having very high temperature it doesn't feel like it at all.

2. Thermosphere - Particles that have enough energy can escape into space

The thermosphere is the fourth highest layer of the atmosphere. The atoms in this layer are relatively distant from one another, so the particles that have enough energy manage to escape easily into the exosphere and then the space.

3. Mesosphere - It is the coldest region of the atmosphere

The mesosphere is the third highest layer. In this layer the temperatures constantly drop, and they go down to -85 degrees, making it the coldest layer by far.

4. Stratosphere - Ninety percent ozone is in this layer

The startosphere has a separte zone in it which is dominated by only one gas, the ozone. It is called the ozone layer, the one that protects the Earth from too intense UV radiation, and in fact over 90% of this gas is locate here.

5. Troposphere - It is warm due to the heat from Earth's surface

The troposphere is the densest and lowest of the layers. It is the one that also has Greenhouse gases which manage to trap the heat that is radiated from the surface of the Earth, thus keeping this layer relatively warm.

ella [17]4 years ago
4 0

1. Exosphere - Particles that have enough energy can  escape into space.

2. Thermosphere - Temperatures reach as high as 2000°C,  yet it feels very cold.

3. Mesosphere - It is the coldest region of the atmosphere

4. Stratosphere - Ninety percent of ozone is in this region

5. Troposphere - t is warm due to the heat from  Earth’s surface.

I've taken the Plato test and these are the correct answers. :-)

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djverab [1.8K]

<u>Answer:</u> The Gibbs free energy of the given reaction is -40 kJ

<u>Explanation:</u>

The given chemical equation follows:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

The equation for the standard Gibbs free change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(NH_3(g))})]-[(1\times \Delta G^o_f_{(N_2(g))})+(3\times \Delta G^o_f_{(H_2(g))})]

We are given:

\Delta G^o_f_{(NH_3(l))}=-16.45kJ/mol\\\Delta G^o_f_{(H_2(g))}=0kJ/mol\\\Delta G^o_f_{(N_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol

The equation used to Gibbs free energy of the reaction follows:

\Delta G=\Delta G^o+RT\ln Q_{eq}

where,

\Delta G = free energy of the reaction

\Delta G^o = standard Gibbs free energy = -32.9 kJ/mol = -32900 J/mol  (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[273+25]K=298K

Q_{eq} = Ratio of concentration of products and reactants at any time = \frac{(p_{NH_3})^2}{(p_{H_2})^3\times p_{N_2}}

p_{NH_3}=2.11atm

p_{N_2}=7.92atm

p_{H_2}=2.02atm

Putting values in above equation, we get:

\Delta G=-32900J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(2.11)^2}{(2.02)^3\times 7.92}))\\\\\Delta G=-39553.04J/mol=--39.55kJ=-40kJ

Hence, the Gibbs free energy of the given reaction is -40 kJ

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Answer:

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Explanation:

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mas of water = ?

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From the results, we conclude that the limiting reactant is Oxygen because the experimental yield was higher than the theoretical yield.

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