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ivanzaharov [21]
3 years ago
7

Complete these structures by adding electron dots as needed.

Chemistry
1 answer:
LiRa [457]3 years ago
7 0
This is hard to show but here is how you would determine these. NOTE each dot is an electron. 
<span>Question 1) </span>
<span>F-H </span>
<span>1) determine the valance electrons for each. F has 7 and H has 1 </span>
<span>2) one electron from both F and H form the bond "-" which means that you still have 6 electrons to place around F and none to place around H. Place the 6 in sets of 2 around the F </span>
<span>.. </span>
<span>F-H </span>
<span>¨ </span>
<span>Question 2) </span>

<span>2) H-O-H </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>O has 6 valence electrons minus 2 used in the bonds to the H's = 4 electrons to place </span>

<span>H-O-H: place two dots above and below the oxygen </span>

<span>Question 3) </span>
<span>3) O=N----H : NOTE: a double bond requires O and N to share two of their electrons each </span>
<span>O has 6 valence electrons minus 2 used in the bonds to N = 4 electrons to place </span>
<span>N has 5 valence electrons minus 3 used in the bonds to O and H = 2 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to N = 0 electrons to place </span>
<span>place the 2 dots on top and bottom of oxygen. </span>
<span>place 2 above the N </span>
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Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

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P(N₂O₄) = 200 - 98.7

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Answer:

1. The electronic configuration of X is: 1s2 2s2 sp6 3s2

2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

3. The formula of the compound form by X and Y is given as: XY

Explanation:

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The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6

The formula of the compound form by X and Y is given below :

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Their valency will cancel out thus forming XY

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