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ycow [4]
3 years ago
10

Ionization energy is the energy needed to eject an electron from an atom or ion. Calculate the ionization energy, I E , of the o

ne‑electron ion He + . The electron starts in the lowest energy level, n = 1 .
Chemistry
1 answer:
irakobra [83]3 years ago
3 0

Answer:

Ionization energy of He^+ is 54.4 eV.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}eV

where,

E_n = energy of  orbit

n = number of orbit

Z = atomic number

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{2^2}{1^2}eV=-54.4 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty }=-13.6\times \frac{2^2}{\infty ^2}eV=0

Ionization energy of He^+ =

E=E_{\infty }=E_1=0 - (-54.4 eV)=54.4 eV

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Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

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2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

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\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

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The vapor pressure for the glucose solution at 17.19 Torr.

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Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

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