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FromTheMoon [43]
4 years ago
10

22.5 mL of an HNO3 solution were titrated with 31.27 mL of a .167 M NaOH solution to reach the equivalence point. What is the mo

larity of the HNO3 solution
Chemistry
1 answer:
natulia [17]4 years ago
3 0

Answer:

M₂ = 0.23 M

Explanation:

Given data:

Volume of HNO₃ = 22.5 mL

Volume of NaOH = 31.27 mL

Molarity of NaOH = 0.167 M

Molarity of HNO₃ = ?

Solution:

M₁ = Molarity of NaOH

V₁ = Volume of NaOH

M₂ = Molarity of HNO₃

V₂ = Volume of HNO₃

M₁V₁ = M₂V₂

0.167 M × 31.27 mL = M₂ × 22.5 mL

5.2 M. mL = M₂ × 22.5 mL

M₂ =  5.2 M. mL /22.5 mL

M₂ = 0.23 M

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Answer:

The answer to your question is given after the questions so I just explain how to get it.

Explanation:

a)

Get the molecular weight of Phosphoric acid

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b)

Molarity = \frac{moles}{volume}

Molarity = \frac{0.00046}{0.35}

Molarity = 0.0013 or 1.31 x 10⁻³

c)

Formula            C₁V₁ = C₂V₂

                              V₁ = C₂V₂ / C₁

Substitution

                              V₁ = (0.0013)(1) / 0.01

Simplification and result

                              V₁ = 0.0013 / 0.1

                              V₁ = 0.13 l = 130 ml            

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An electron jumps down from the fourth energy level to the first energy level and releases a light wave of 10.8 nm in wavelength
nata0808 [166]

Answer:

-20.2  * 10^-18 J

Explanation:

From;

E = hc/λ

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From;

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E4 =  (-1.8 * 10^-17) + (-2. 176* 10^-18)

E4 = -20.2  * 10^-18 J

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