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timama [110]
3 years ago
10

If a neutral atom has 12 electrons, how many can be used for creating bonds?

Chemistry
1 answer:
ZanzabumX [31]3 years ago
5 0

Answer:

C. 2.

Explanation:

  • According to octet rule:

The electronic configuration of the neutral atom is: 2,8,2.

So, it has 2 valence electrons that are able to create bonds.

<em>Thus, the right choice is: C. 2.</em>

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3. A helium laser emits light with a wavelength of 6.33 x 10^-7m. What is
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Answer:

4.52 x 10¹⁴ cycles/s

Explanation:

From c = f·λ => f = c/λ = (3.0 x 10⁸ m/s)/(6.63 x 10⁻⁷m) = 4.52 x 10¹⁴ cycles/s.

f = frequency = ?

λ = wavelength = 6.63 x 10⁻⁷ meter

c = speed of light in vacuum = 3.0 x 10⁸ meters/s

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3 years ago
Wich substances is most likely to form in a precipitation reaction?
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Answer:

A precipitation reaction refers to the formation of an insoluble salt when two solutions containing soluble salts are combined. The insoluble salt that falls out of solution is known as the precipitate, hence the reaction's name.

Explanation:

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3 years ago
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

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