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lawyer [7]
1 year ago
12

How many moles of h+ ions are present in 2.8 l of 0.25 m hydrobromic acid solution?

Chemistry
1 answer:
Romashka [77]1 year ago
4 0

0.7 mol of H⁺ ions are present in 2.8 l of 0.25 m hydrobromic acid solution.

Hydrobromic acid is a strong acid, we can assume that all acid molecules dissociate completely to yield H+ ions and dissociated anion.

<h3>The equation for the dissociation of HBr :</h3>

                             Hbr <em>(s) →  </em>H⁺ <em>(aq)  </em>+  Br⁻<em> (aq)</em>

<em></em>

moles H⁺ = (2800ml) ( \frac{1L}{1000ML})  (\frac{0.25 mol Hbr}{L}) ( \frac{1 mol H}{1 mol Hbr})

               = 0.7 mol

Therefore, 0.7 mol of H⁺ ions is present.

Learn more about H⁺ ions here:

brainly.com/question/12697532

#SPJ4

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The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make
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Answer:

the energy of the third excited rotational state \mathbf{E_3 = 16.041 \ meV}

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = \dfrac{m_1 \times m_2}{m_1 + m_2}

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The rotational level Energy can be expressed by the equation:

E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)

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J = 3 ( i.e third excited state)  &

I = \mu r^2_o

E_J= \dfrac{h^2}{8  \pi  \mu r^ 2 \mur_o } \times J ( J +1)

E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8  \times  \pi ^2  \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2  } \times 3 ( 3 +1)

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1 J = \dfrac{1}{1.6 \times 10^{-19}}eV

E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV

E_3 = 16.041  \times 10 ^{-3} \ eV

\mathbf{E_3 = 16.041 \ meV}

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