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anygoal [31]
3 years ago
7

Element Q is a theoretical nonmetal with atomic number 59. Consider the isotope: Q-95. How many electrons are in an atom of Q-95

if the atom has a charge of -1?
Chemistry
1 answer:
Vinil7 [7]3 years ago
4 0

60

Explanation:

The number of electrons in an atom of Q-95 will be 60 electrons.

The atomic number of an atom is the number of protons in the atom of the isotope

This number is unique to every atom and their isotope

An isotope of an atom is the existence of an atom having the same atomic number but different mass numbers.

Since atomic number the same as number of protons, they are the positively charged particles in an atom.

To have a negative charge, the number of electrons must be more than that of protons.

Charge = number of protons - number electrons

  charge = -1

  -1 = 59  - number of electrons

  Number of electrons = 60 electrons

learn more:

Number of subatomic particles brainly.com/question/2757829

#learnwithBrainly

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Consider the following multistep reaction:
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The question is incomplete, here is the complete question:

Consider the following multistep reaction:

C+D⇌CD (fast)

CD+D→CD₂ (slow)

CD₂+D→CD₃ (fast)

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Based on this mechanism, determine the rate law for the overall reaction.

<u>Answer:</u> The rate law for the reaction is \text{Rate}=k'[C][D]^2

<u>Explanation:</u>

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

C+3D\rightarrow CD_3

The intermediate reaction of the mechanism follows:

<u>Step 1:</u>  C+D\rightleftharpoons CD;\text{ (fast)}

<u>Step 2:</u>  CD+D\rightarrow CD_2;\text{(slow)}

<u>Step 3:</u>  CD_2+D\rightarrow CD_3;\text{(fast)}

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

\text{Rate}=k[CD][D]           ......(1)

As, [CD] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for CD from step 1, we get:

K=\frac{[CD]}{[C][D]}

[CD]=K[C][D]

Putting the value of [CD] in equation 1, we get:

\text{Rate}=k.K[C][D]^2\\\\\text{Rate}=k'[C][D]^2  

Hence, the rate law for the reaction is \text{Rate}=k'[C][D]^2

5 0
3 years ago
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