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Mumz [18]
3 years ago
11

The electrons in the outermost shell are called valence electrons. How many valence electrons does the

Chemistry
1 answer:
kipiarov [429]3 years ago
5 0

Answer:7

Explanation:All halogens have 7 valence electrons. Valence electrons are the total number of electrons present in the outermost shell of an atom (i.e. Seven valence electrons. There are thus 7 valence electrons for Fluorine, 2 in the 2s orbital, and 5 in the 2p orbitals.

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Help I don’t understand it and I need the answers
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Demonstration 1 showed a chemical change because a new gaseous substance was formed, and demonstration 2 showed a physical change because liquid water became water vapor.

Explanation:

Chemical changes alter the chemical makeup of a subject, and a physical change only alters the appearance of a subject, not the chemical makeup.

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A gas takes up a volume of 10 liters, has a pressure of 7.5 atm, and a
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I’m trying to do something similar to that

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A mineral is made from what in the ground
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<span>Minerals are solid substances that occur naturally. They can be made from a single element (like gold or copper) or from a combination of elements.</span>
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What general trend applies to solubility?
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Answer:

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Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
zhenek [66]

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}

Here A and A cancel each other

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

We know that : a + b = 3

1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

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

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

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

new rate r' =

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

7 0
3 years ago
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