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ValentinkaMS [17]
3 years ago
8

Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O

H− ions in the solution after the reaction has gone to completion. Assume that there is no volume change when adding the grams of NaOH
Chemistry
1 answer:
Kipish [7]3 years ago
3 0

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = \frac{15.0 g}{40 g/mol}=0.375 mol

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

Molarity=\frac{Moles}{Volume(L)}

n=0.250 M\times 0.150 L=0.0375 mol

NaOH+HNO_3\rightarrow NaNO_3+H_2O

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

\frac{1}{1}\times 0.0375 mol of NaOH

Moles of NaOH left unreacted  in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

1\times 0.3375 moles=0.3375 mol of hydroxide ion

The molarity of hydroxide ion in solution ;

=\frac{0.3375 mol}{0.150 L}=2.25 M

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

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the white precipitate would be calcium carbonate. CaCo₃

note/ this is a common eqn u need to remember.

X - CO₂ (carbón dioxide)
Y - CaCo₃ (calcium carbonate)

sodium carbonate is a basic salt
7 0
3 years ago
An airplane starts at rest and Excelerator down the runway for 20 seconds. At the end of the runway, its velocity is 80 m/s Nort
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According to the formula you have given us to work with . . .

1). The airplane's acceleration is

(80 m/s north - zero) / (20 sec) = 4 m/sec^2 north

2).  For the cyclist:

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Multiply each side by 20s :  V-final = 0.5 m/s^2 south x (20sec) = 

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8 0
3 years ago
What is the volume in ML
BabaBlast [244]

Answer:1 mL

Explanation:

4 0
3 years ago
If 2g of zinc granules was reacted with excess dilute HCl to evolve hydrogen gas which came to completion after 5 minutes. Calcu
nordsb [41]

Answer:

25 g/hr

Explanation:

Remember that the rate of reaction refers to the rate at which reactants are used up or or the rate at which products appear.

Hence;

Rate of reaction = mass of reactant used up/time taken

Mass of reactant used up= 2g

Time taken = 5 minutes or 0.08 hours

Rate of reaction = 2g/0.08 hours = 25 g/hr

4 0
3 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
Ksenya-84 [330]

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.

3 0
3 years ago
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