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Natali5045456 [20]
3 years ago
11

The tendency of an element to react is closely related to

Chemistry
1 answer:
Bad White [126]3 years ago
8 0
C. The number of Valence electrons,
Every atom tries to follow the Octet rule i.e To have 8 electrons in its Valence shell.
Every atom tries to accomodate 8 electrons in its Valence shell to stabilize themselves, Metals usually have 1-3 eletrons in their Valence shell which they donate to non metals so their Valnce shell has 8 electroons, (The previous will now be the Valence shell and it will be full)
Similarly Non metals have 4-7 electrons, they accept electrons from metals so they can have 8 electrons in their Valence shell.
Noble gases already have 8 electrons in their Valance shell, so they do not react and stable.

Hope it helps :)

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Answer:

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Explanation:

i dont know how water vapor would hit the ocean and go underground? so i think its F

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What is the formal charge of nitrogen in no2cl
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Answer:

The formal charge on nitrogen in NO_{2}Cl is +1.

Explanation:

The structure of NO_{2}Cl is as follows.

(In attachment)

Formal\, charge = Valence\, electron - (nonbonding\, electrons + \frac{Bonding\,electrons}{2})

From the structure, Nitrogen has no non bonding electrons. Nitrogen has four bonds and each bond corresponds to 2 electrons. Hence, nitrogen have eight bonding electrons and five valence electrons.

Formal\,charge\,on\,nitrogen = 5-[0+ \frac{8}{2}]= +1

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The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2
Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

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