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3 years ago

Find the Protons and Electrons for nitrogen -3 ion with 7 neutrons.

Chemistry

2 answers:

chubhunter [2.5K]3 years ago

3 0

Answer:

protons: 7; electrons: 10

Explanation:

nitrogen atomic number=# of protons = 7 (doesn't change either in ion or in neutral)

# of electron in neutral N=7

N-3= three electrons are added, thus 7+3=10

allochka39001 [22]3 years ago

3 0

I need points before I fail..
Protons 7 and electrons 10

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4 years ago

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Answer : The value of $K_c$ for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

$Br_2(aq)+Cl_2(aq)\rightleftharpoons 2BrCl(aq)$

The expression of $K_c$ will be,

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

First we have to calculate the concentration of $Br_2,Cl_2\text{ and }BrCl$ .

$\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M$

$\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M$

Now we have to calculate the value of $K_c$ for the given reaction.

$K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}$

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$K_c=0.36$

Therefore, the value of $K_c$ for the given reaction is, 0.36

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