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Rufina [12.5K]
2 years ago
5

.) Neon and HF have approximately the same molecular masses. a.)Explain why the boiling points of Neon and HF differ b.)Compare

the change in the boiling points of Ne, Ar, Kr, and Xe with the change of the boiling points of HF, HCl, HBr, and HI, and explain the difference between the changes with the increasing atomic or molecular mass.
Chemistry
1 answer:
lisov135 [29]2 years ago
4 0

Answer:

See explanation

Explanation:

a) The magnitude of intermolecular forces in compounds affects the boiling points of the compound. Neon has London dispersion forces as the only intermolecular forces operating in the substance while HF has dipole dipole interaction and strong hydrogen bonds operating in the molecule hence HF exhibits a much higher boiling point than Ne though they have similar molecular masses.

b) The boiling points of the halogen halides are much higher than that of the noble gases because the halogen halides have much higher molecular masses and stronger intermolecular forces between molecules compared to the noble gases.

Also, the change in boiling point of the hydrogen halides is much more marked(decreases rapidly)  due to decrease in the magnitude of hydrogen bonding from HF to HI. The boiling point of the noble gases increases rapidly down the group as the molecular mass of the gases increases.

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2 years ago
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What is the average mass of a single chlorine atom in grams
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The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: E = R_y/n^2 In this equation R_y stands
Katarina [22]

Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

E=\frac{R_y}{n^2}

R_y=2.18\times 10^{-18} J =  Rydberg energy

n =  principal quantum number of the orbital

Energy of 11th orbit = E_{11}

E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J

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E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

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=E'=0.38\times 10^{-20} J

\lambda =\frac{hc}{E'} (Planck's' equation)

\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}

\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

3 0
2 years ago
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