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1 year ago

Which statement best explains why the elements in Group 18 do not have electronegativity values?

Chemistry

1 answer:

Gnesinka [82]1 year ago

5 0

Answer:

1) The elements have filled valence levels.

Explanation:

Since they have filled valence levels, they're stable and don't need to electrons to fill their valence shells since they're already full.

2) False, They do have electrons

3) False, He does have only one electron shell, but going down the periods, every next element have one more electron shell than a preceding one has.

4)False, they're actually the smallest atoms of their respective period

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Which of the following is an example of a chemical change?

labwork [276]

A chemical change changes the bonds by forming or breaking them, and cannot be brought back to it's original form.

b. milk souring

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2 years ago

4. Define oxidation and reduction in terms of the gain or loss of electrons.

Andreyy89

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1 year ago

Why is it important to analyze data?

gayaneshka [121]

Answer:

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2 years ago

What is changed when matter undergoes a physical change?

Furkat [3]

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2 years ago

A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th

kondaur [170]

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

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are $P(NO_{2} )$ = 98.7 MPa and P(N₂O₄) = 101.3 MPa

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2 years ago