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Taya2010 [7]
2 years ago
4

A certain oxygen atom has the electron configuration 1s22s22px22py2. how many unpaired electrons are present? 2 incorrect: your

answer is incorrect. is this an excited state of oxygen? yes no in going from this state to the ground state would energy be released or absorbed? neither, the oxygen atom is in the ground state. energy would be absorbed. energy would be released.
Chemistry
2 answers:
nexus9112 [7]2 years ago
8 0

Answer:

They are all paired together. It is in an excited state and releases energy.

Explanation:

Before I answer this question, I think you need a basic understanding of orbital configuration and how everything there works.

Oxygen's electron configuration is 1s2 2s2 2p4, where the 1s2 and 2s2 blocks are completely filled, and the 2p4 block is only partially filled (4 electrons distributed into 3 boxes). For some reason this page won't let me add a picture, but please google "orbital configuration of oxygen" and it will come up in the images section.

However in this case, if the 3 boxes of the 2p corresponded to the letters x, y, and z, you would find that the x and y boxes are completely filled (2px2 2py2).

This shows that there are no unpaired electrons in the 3 boxes, because all 4 electron are pushed into the 2px2 and 2py2 box.

This is different from the regular, non-excited state of oxygen, where 2px is filled, 2py has one electron, and 2pz also has one electron (written as 2px2 2py1 2pz1).

From this, we can see that this is in an excited state because the 2py2 box has two electrons, different from the 2px2 2py1 2pz1, where the electrons are each in their own box. Because it is different, it is in an excited state.

You probably know that in an excited state, energy is released because the electron emits a photon when it goes back to its ground state.

Alexxandr [17]2 years ago
4 0
<span>Answer: From the electron configuration, this might be the ground state. There are no unpaired electrons. There is a certain stability from spin-paired electrons that may over-ride Hund's Rule, and it may take energy to go to unpaired electrons, which might be induced in an excited state. UNPAIRED ELECTRON: a single electron in an orbital. For oxygen, this would have the electron configuration 1s2,2s2, 2px2, 2py1,2pz1.</span>
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Answer:

−2399.33 kJ

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∴ the enthalpy change for 1.0 mole of NH₄NO₃ in this reaction = −2399.33 kJ

7 0
2 years ago
To find the Ce4+ content in a solid sample, 4.3718 g of the solid sample were dissolved and treated with excess iodate to precip
gulaghasi [49]

Answer:

3.43 %

Explanation:

We need  to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.

0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2

0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce

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5 0
2 years ago
Read 2 more answers
A solution contains one or more of the following ions: Ag + , Ca 2 + , and Co 2 + . Ag+, Ca2+, and Co2+. Lithium bromide is adde
Eva8 [605]

Answer:

Since with LiBr no precipitation takes place. So, Ag+ is absent

When we add Li2SO4 to it, precipitation takes place.

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When Li3PO4 is added, again precipitation takes place.Reaction is:

Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate

A. Ca2+ and Co2+ are present in solution

B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)

C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)

8 0
2 years ago
Some hydrogen gas is enclosed within a chamber being held at 200∘C∘C with a volume of 0.0250 m3m3. The chamber is fitted with a
Semenov [28]

Answer:

The final volume is 39.5 L = 0.0395 m³

Explanation:

Step 1: Data given

Initial temperature = 200 °C = 473 K

Volume = 0.0250 m³ = 25 L

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The pressure reduce to 0.950 *10^6 Pa

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Step 2: Calculate the volume

P1*V1 = P2*V2

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⇒with V1 = the initial volume = 25 L

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V2 = 39.5 L = 0.0395 m³

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3 0
2 years ago
How many chloride ions are in 0.486 moles of chloride ions?​
svetlana [45]

Answer:

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you might think that it will affect the mass but the mass of an electron is almost negligible so we will ignore that

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Amunt of ions in 0.486 moles = 2.9 * 10^23 ions

Hence, option 1 is correct

6 0
2 years ago
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