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Vinvika [58]
3 years ago
14

Identify the element that has 6 valence electrons and 3 energy levels

Chemistry
1 answer:
evablogger [386]3 years ago
5 0

Answer:

Sulfur

Explanation:

<u>Valence electrons</u> are the group numbers. The group numbers are the numbers that go from left to right on the top. But since you can't have more than 8 valence electrons, the group numbers of 13-18, the valence electrons will actually be from 3-8. So 6 valence electrons would be group 16. That narrows it down a lot. The only elements in this group are oxygen, sulfur, selenium, tellurium, polonium, and livermorium.

Energy level are the period numbers. The period numbers are the numbers that go from top to bottom. So the 3rd energy level is the 3rd period. So when you look at the periodic table, under the 16 group and 3rd period, lies Sulfur.

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Metallic bonding  

Explanation:

Metals have low ionization energies. Therefore, their valence electrons are easily delocalized (attracted to the neighbouring metal atoms). These delocalized electrons are then not associated with a specific metal atom.  Since the electrons are “free”, the metal atoms have become cations, and the electrons are free to move throughout the whole crystalline structure.

We say that a metal consists of an array of cations immersed in a sea of electrons .

The electrons act as a “glue” holding the cations together.

Metallic bonds are the attractive forces between the metal cations and the sea of electrons.

In an NaK alloy, for example, the Na and K atoms contribute their valence electrons to the "sea".  The atoms aren’t bonded to each other, but they are held in place by the metallic bonding.

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Consider the equilibrium
vladimir1956 [14]

Answer:

Kp^{1000K}=0.141\\Kp^{298.15K}=2.01x10^{-18}

\Delta _rG=1.01x10^5J/mol

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

C_2H_6(g)\rightleftharpoons H_2(g)+C_2H_4(g)

Thus, Kp for this reaction is computed based on the given molar fractions and the total pressure at equilibrium, as shown below:

p_{C_2H_6}^{EQ}=2bar*0.592=1.184bar\\p_{C_2H_4}^{EQ}=2bar*0.204=0.408bar\\p_{H_2}^{EQ}=2bar*0.204=0.408bar

Kp=\frac{p_{C_2H_4}^{EQ}p_{H_2}^{EQ}}{p_{C_2H_6}^{EQ}}=\frac{(0.408)(0.408)}{1.184}=0.141

Now, by using the Van't Hoff equation one computes the equilibrium constant at 298.15K assuming the enthalpy of reaction remains constant:

Ln(Kp^{298.15K})=Ln(Kp^{1000K})-\frac{\Delta _rH}{R}*(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=Ln(0.141)-\frac{137000J/mol}{8.314J/mol*K} *(\frac{1}{298.15K}-\frac{1}{1000K} )\\\\Ln(Kp^{298.15K})=-40.749\\\\Kp^{298.15K}=exp(-40.749)=2.01x10^{-18}

Finally, the Gibbs free energy for the reaction at 298.15K is:

\Delta _rG=-RTln(Kp^{298.15K})=8.314J/mol*K*298.15K*ln(2.01x10^{-18})\\\Delta _rG=1.01x10^5J/mol

Best regards.

3 0
3 years ago
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