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hammer [34]
3 years ago
9

How many moles of h+ are associated with the acid h2so3 during neutralization?

Chemistry
1 answer:
Katena32 [7]3 years ago
8 0

H2SO3 or sulfurous acid is actually a strong acid. We know for a fact that strong acids completely dissociate into its component ions in a solution, that is:

 

<span>H2SO3 -->  2H+  +  SO3-</span>

 

<span>So from the equation above, there are 2 moles of H+</span>

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Aluminum has a density of 2.70 g/m. Calculate the mass (in grams) of a piece of aluminum having a volume of 382 mL
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.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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rusak2 [61]

Answer:

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Explanation:

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9.267 g of N

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First of all we divide each by the molar mass of the element

63.57 g / 12 gmol = 5.29 mol of C

6 g of H / 1 g/mol = 6 mol H

9.267 g of N / 14 g/mol =  0.662 mol of N

21.17 g of O / 16 g/mol = 1.32 mol of O

We divide each by the lowest value, in this case 0.662

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6 / 0.662 = 9

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