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Mumz [18]
3 years ago
8

Identify the molecules or ions below as Lewis acids, Lewis bases, or neither. If there is more than one possible site in the mol

ecule/ion, focus on the central or the charged atom. H3C-CH2-CH3 __________
Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
8 0

Answer: Neither

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

CH_3CH_2CH_3 is a hydrocarbon containing carbon and hydrogen, which have almost same electronegativity values. Thus is a non polar molecule.

It does not donate electrons as it does not contain any lone pair of electrons. It can not accept electrons as it's octet is already complete. Thus is neither a lewis acid nor a lewis base.

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kvv77 [185]

Answer:

The correct answer is - D. Freezing point depression.

Explanation:

When rock salt is spread over snow-covered icy roads, it generates a liquid layer over it by melting from the surface thereby lowering or depression in the freezing point below the ice.

Therefore, due to this liquid layer comes into the contact with the ice present on the road and causes other ice to melts. This keeps on decreasing the volume of the ice on the road therefore, rock salts spread on the roads during a snowstorm.

7 0
3 years ago
Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi
bija089 [108]

<u>Answer:</u> The value of K_c for the net reaction is \frac{K_b}{K_w}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b

<u>Equation 2:</u>  H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}

The net equation follows:

B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c

As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.

The value of equilibrium constant for net reaction is:

K_c=K_1\times K_2

We are given:  

K_1=K_b

K_2=\frac{1}{K_w}

Putting values in above equation, we get:

K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}

Hence, the value of K_c for the net reaction is \frac{K_b}{K_w}

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3 years ago
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