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BartSMP [9]
3 years ago
9

If we want to change a gas to its liquid state, should we add or remove energy from the gas?

Chemistry
1 answer:
Nat2105 [25]3 years ago
5 0
You need to lose some energy from your very excited gas atoms. The easy answer is to lower the surrounding temperature. When the temperature drops, energy will be transferred out of your gas atoms into the colder environment. When you reach the temperature of the condensation point, you become a liquid.
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Which process would require the least amount of energy?
Aleksandr [31]
Answer:<span> a) the process for the first ionization energy
</span>
<span>Every time you take a electron you're requiring more and more energy. Expelling the first one will require less energy than expelling the second and the second will require less than the third, and so on.
When you take the first one, the atom becomes positive and with that the negative forces of the electron will be more attracted to the positive charge. The more electrons that are lost, the more positive this ion will become, causing it to be more difficult to separate the electrons from the atom.
</span>
8 0
3 years ago
When methyloxirane is treated with HBr, the bromide ion attacks the less substituted position. However, when phenyloxirane is tr
konstantin123 [22]

Answer:

See explanation and picture below

Explanation:

First, in the case of methyloxirane (Also known as propilene oxide) the mechanism that is taking place there is something similar to a Sn2 mechanism. Although a Sn2 mechanism is a bimolecular substitution taking place in only step, the mechanism followed here is pretty similar after the first step.

In both cases, the H atom of the HBr goes to the oxygen in the molecule. You'll have a OH⁺ in both. However, in the case of methyloxirane the next step is a Sn2 mechanism step, the bromide ion will go to the less substitued carbon, because the methyl group is exerting a steric hindrance. Not a big one but it has a little effect there, that's why the bromide will rather go to the carbon with more hydrogens. and the final product is formed.

In the case of phenyloxirane, once the OH⁺ is formed, the next step is a Sn1 mechanism. In this case, the bond C - OH⁺ is opened on the side of the phenyl to stabilize the OH. This is because that carbon is more stable than the carbon with no phenyl. (A 3° carbon is more stable than a 2° carbon). Therefore, when this bond opens, the bromide will go there in the next step, and the final product is formed. See picture below for mechanism and products.

4 0
3 years ago
What is the mass of 5 mole of ammonia . Calculate the number of NH₃ molecules, nitrogen atom and hydrogen atoms in it..
Aloiza [94]

Molar mass of NH_3

\\ \sf\longmapsto 14u+3(1u)

\\ \sf\longmapsto 14u+3u

\\ \sf\longmapsto 17g/mol

We know.

No of moles=Given mass/Molar mass

\\ \sf\longmapsto Given\;Mass=17(5)

\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g

Now

Lets write the balanced equation

\\ \sf\longmapsto N_2+3H_2=2NH_3

  • There is 2moles of Ammonia
  • 3moles of H_2
  • 1mole of N_2

Now

\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}

For Hydrogen

\\ \sf\longmapsto 3\times 6.023\times 10^{23}

\\ \sf\longmapsto 18.069\times 10^{23}

\\ \sf\longmapsto 1.8\times 10^{22}molecules

For Ammonia

\\ \sf\longmapsto 2\times 6.023\times 10^{23}

\\ \sf\longmapsto 12.046\times 10^{23}

\\ \sf\longmapsto 1.2\times 10^{22}molecules

For Nitrogen

\\ \sf\longmapsto 1\times 6.023\times 10^{23}

\\ \sf\longmapsto 6.023\times 10^{23}molecules

3 0
2 years ago
A cast fossil is formed when
Liono4ka [1.6K]
The answer is D)
<span>when the remains of a living thing are preserved and turned into rock </span>
8 0
3 years ago
Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%, respectively)
Reika [66]

Answer:

Average atomic mass = 79.9034 amu

Explanation:

The formula for the calculation of the average atomic mass is:

Average\ atomic\ mass=(\frac {\%\ of\ the\ first\ isotope}{100}\times {Mass\ of\ the\ first\ isotope})+(\frac {\%\ of\ the\ second\ isotope}{100}\times {Mass\ of\ the\ second\ isotope})

Given that:

<u>For first isotope: </u>

% = 50.69 %

Mass = 78.9183 amu

<u>For second isotope: </u>

% = 49.31 %

Mass = 80.9163 amu

Thus,  

Average\ atomic\ mass=\frac{50.69}{100}\times {78.9183}+\frac{49.31}{100}\times {80.9163}\ amu

Average\ atomic\ mass=40.0036+39.8998\ amu

<u>Average atomic mass = 79.9034 amu</u>

4 0
3 years ago
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