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Novay_Z [31]
3 years ago
9

How many electrons would Boron with a +2 charge have?

Chemistry
1 answer:
irina [24]3 years ago
4 0

5 electrons

Boron atomic number 5 has five electrons in its ground state.

Commonly Boron will lose 3 electrons leaving 2 electrons in its most common ionic form.

Explanation:

The atomic number gives the number of protons. Protons which have a positive charge are balanced by an equal number of electrons in a neutral atom.

Boron number 5 has five protons and therefore as a neutral atom also has five electrons.

Boron has an electron configuration of

1s22s22p1

The most stable electron configuration for Boron is

1s2

+ 3 charges. By losing three electrons Boron can achieve the stable electron structure of Helium

Brainliest? :D

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Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

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B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

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1 + b = 3

b =3 -1  = 2

b = 2

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r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

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[B']=1/2[B]

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r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

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<u>r' = 1/4 r</u>

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