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3 years ago

Match the statements with the atmospheric layer they describe.

Chemistry

2 answers:

vekshin13 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]3 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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What is an experimental control

attashe74 [19]

The experimental control is the standard used as a comparison for the experimental groups.

For example, you may be trying to find out how different types of disinfectants affect bacterial growth. The control group would receive no disinfectant whereas the experimental groups would be the ones on which the disinfectants were tested.

Hope this makes sense!

3 0

3 years ago

How does the A Hreaction relate to the A He of molecules involved in a reaction?

igor_vitrenko [27]

Answer:

B. ΔHreaction = ΔH°f reactants- ΔH°f products

Explanation:

Using Hess's law, it is possible to sum ΔH of several related reactions to find ΔH of a particular reaction.

Having in mind Hess's law, ΔH°f is defined as the change in enthalpy during the formation of 1 mole of substance from its constituent elements (That is, pure elements, mono or diatomics, that have a ΔH° = 0).

For example, in ΔH°f of H₂O, the equation is:

H₂(g) + 1/2O₂(g) → H₂O(g)

The constituent elements with ΔH°f = 0 are H₂(g) and O₂(g).

Now, using Hess's law, you can sum the ΔH°f of substance in a reaction as, for example:

NaOH + HCl → H₂O + NaCl. ΔHr

The ΔH°f for each substance in the reaction is:

NaOH: Na + 1/2H₂ + 1/2O₂ → NaOH (1)

HCl: 1/2H₂ + 1/2Cl₂ → HCl (2)

H₂O: H₂ + 1/2O₂ → H₂O (3)

NaCl: Na + 1/2Cl₂ → NaCl (4)

The algebraic sum of (3) + (4) is -(ΔH°f reactants):

H₂ + 1/2O₂ + Na + 1/2Cl₂ → NaCl + H₂O ΔH°f reactants

This reaction - {(1)+(2)} ΔH°f products

NaOH + HCl → H₂O + NaCl.

ΔHr = ΔH°f reactants- ΔH°f products

In the example, we obtain this relationship that can be expanded for all reactions. Thus, right answer is:

<h3>B. ΔHreaction = ΔH°f reactants- ΔH°f products</h3>

8 0

3 years ago

This graph shows two curves pertaining to a hydrogen s orbital.

fgiga [73]

Answer 1) : According to the complete question attached in the answer,

The radial wave function which is denoted by $R_{nl}(r)$ shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.

Also, the radial probability distribution curve denoted as $R^{2}_{nl}(r)$ shown in blue color is observed to touch zero, and shows the place of radial node.

Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by 'n'.

It is observed that for any atomic orbital, the total number of nodes will be n-1 .

Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.

Hence, there will be only radial nodes, which is

(n−1 = total number of radial nodes in s orbitals)

According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)

This represents the 5s orbital.

Answer 2) The radial nodes are observed in I'm seeing radial nodes at

$1.9a_{0}$ , $6.4a_{0}$ , $13.9a_{0}$ and $27.0a_{0}$ .

where $a_{0}$ represents the hydorgen bohr atomic radius = 0.0529177 nm

Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.

3 0

3 years ago

Read 2 more answers

Pick the selection that increases in energy.

RUDIKE [14]

Answer:

i think mix im um 1s 6f 4d is correct answer

8 0

3 years ago

Calculate number of CO2 molecules in 0.250 mol of co2

Anna11 [10]

Carbon's molar mass is 12gm

Oxygen's molar mass is 16gm

So CO2 molar mass is 44gm

Hence 1mol of CO2 contains 44gm
There 1/4mol of CO2 contains 44/4=11gm.

Hope it helps

8 0

3 years ago