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3 years ago

If you were to remove an electron from a sodium atom that has eleven protons, what would be the electrical charge of the ion?

Chemistry

1 answer:

damaskus [11]3 years ago

4 0

Answer:

positive one (+1)

Explanation:

sodium having atomic mass is twenty three (23) and atomic number is eleven (11).

Sodium atom having only one electron in it's outer most shell and it is easy for atom to lose this electron from outer most shell to make itself stable.

So after losing this electron positive charge on the upper right side of the atom will occur with the number of electron lose that is Na+1 .

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3 years ago

You just got home from a run on a hot Atlanta afternoon. You grab a 1.00 liter bottle of water and drink three quarters of it in

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Answer:

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3 years ago

.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f

Tasya [4]

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.

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

$K_c=\frac{(0.6)^2}{(1)\times (1)}$

$K_c=0.36$

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

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