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3 years ago

What is the pH of a 0.235 M Ba(OH)2 solution?

Chemistry

1 answer:

Fynjy0 [20]3 years ago

6 0

Answer:

13.672

Explanation:

2 moles of OH in Ba(OH)₂

0.235 M × 2 mol OH = 0.47 M OH

pOH = -log[OH-] ; pOH = -log[0.47 M OH] ; 0.328 = pOH (consider significant figures)

14 - pOH = pH because pH + pOH = 14

14 - 0.328 = 13.672 [there are three decimal places after the OH concentration, so there should be in the answer as well - 13.<u>672</u>]

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Calculate the wavelength, in nanometers, of the light emitted by a hydrogen atom when its electron falls from the n = 7 to the n

Gre4nikov [31]

Answer:

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Explanation:

The energy of nth energy levels of the H atom is given as:

$E_n = -2.18 \times 10^{-18} \times \frac{1}{n^2} J$

Energy of the seventh energy level = $E_7$

$E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J$

$E_7=-2.18 \times 10^{-18} \times \frac{1}{7^2} J=-4.4490\times 10^{-20} J$

Energy of the seventh energy level = $E_4$

$E_4=-2.18 \times 10^{-18} \times \frac{1}{4^2} J$

$E_4=-2.18 \times 10^{-18} \times \frac{1}{16} J=-1.3625\times 10^{-19} J$

Energy of the light emitted will be equal to the energy difference of the both levels.

$E=E_7-E_4=-4.4490\times 10^{-20} J-(-1.3625\times 10^{-19} J)$

$E=9.176\times 10^{-20} J$

Wavelength corresponding to energy E can be calculated by using Planck's equation:

$E=\frac{hc}{\lambda }$

$\lambda =\frac{hc}{E}=\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{9.176\times 10^{-20} J}=2.166\times 10^{-6} m=2166 nm$

The wavelength of the light emitted by a hydrogen atom for the given transition is 2166 nm.

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3 years ago

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m_a_m_a [10]

Answer:

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Answer:

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Explanation:

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