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Vlada [557]
2 years ago
8

Consider molecular, complete ionic, and net ionic equations.

Chemistry
1 answer:
8090 [49]2 years ago
3 0

Explanation:

(a) The molecular equations shows the equation in which all the species of the reactants and the products are in molecules and the net charge is zero.

The complete ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and are represented as ions.

The net ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and do not show the spectator ions which are same in the reactants and the products.

(b) If there is no spectator ions in the reaction, then the complete and the net ionic equations would be identical.

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2 years ago
A 250. ml sample of 0.0328m hcl is partially neutralized by the addition of 100. ml of 0.0245m naoh. find the concentration of h
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Answer: 0.0164 molar concentration of hydrochloric acid in the resulting solution.

Explanation:

1) Molarity of 0.250 L HCl solution : 0.0328 M

Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0328=\frac{\text{Number of moles}}{0.250 L}

Moles of HCl in 0.250 L solution = 0.0082 moles

2) Molarity of 0.100 L NaOH solution : 0.0245 M

Molarity=\frac{\text{Number of moles}}{\text{Volume of solution in liters}}=0.0245=\frac{\text{Number of moles}}{0.100 L}

Moles of NaOH in 0.100 L solution = 0.00245 moles

3) Concentration of hydrochloric acid in the resulting solution.

0.00245 moles of NaOH will neutralize 0.00245 moles of HCl out of 0.0082 moles of HCl.

Now the new volume of the solution = 0.100 L +0.250 L = 0.350 L

Moles of HCl left un-neutralized = 0.0082 moles - 0.00245 moles =  0.00575 moles

Molarity=\frac{\text{number of moles}}{\text{volume of solution in L}}

Molarity of HCl left un-neutralized :\frac{0.00575 moles}{0.350L}=0.0164 M

0.0164 molar concentration of hydrochloric acid in the resulting solution.

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