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4 years ago

Which electron configuration represents the electrons of an atom of neon in an excited state?

Chemistry

2 answers:

Nikitich [7]4 years ago

4 0

The answer is C) 2-7-1, because the second shell isnt full and it already has a third shell.

ruslelena [56]4 years ago

3 0

Answer: Option (C) is the correct answer.

Explanation:

Atomic number of neon is 10 and its electronic configuration is $1s^{2}2s^{2}2p^{6}$ .

As neon has a vacant 3s orbital also. Therefore, in the excited state an electron will jump into the excited energy level. So, in excited state of neon one electron will jump into the 3s orbital.

Hence, its electronic distribution in the valence shell will become 2-7-1.

Thus, we can conclude that 2-7-1 electron configuration represents the electrons of an atom of neon in an excited state.

You might be interested in

When a solution of nitric acid is added to a to solution of calcium hydroxide, the salt formed has what formula?

SpyIntel [72]

Nitric acid and calcium hydroxide react to form calcium nitrate and water.

The molecular equation is as follows:

2HNO3 (aq) + Ca(OH)2 (aq) = Ca(NO3)2 (aq) + 2H2O (l)

So the formula of calcium nitrate, the salt formed in this reaction is Ca(NO3)2 .

Water here seem to be the only real product. It is formed from the OH+ of the hydroxide and the H+ of the acid which combine to form two water molecules.

Calcium and Nitrate ions remain unchanged before and after the reaction and are thus known as spectator ions.

8 0

3 years ago

A mixture of carbon dioxide and hydrogen gases is maintained in a 6.68 L flask at a pressure of 2.14 atm and a temperature of 19

matrenka [14]

Answer:

The mass of hydrogen gas in the mixture: w₂ = 0.433 g

Explanation:

According to the ideal gas equation:

for an ideal gas, $P.V = n_{total}.R.T$

and $n_{total}= n_{1}+n_{2}$

Here, P: total pressure of the gases = 2.14 atm

V: total volume of the gases = 6.68 L

T: temperature = 19 °C = 19+273.15 = 292.15K (∵ 0°C = 273.15K)

R: gas constant = 0.08206 L·atm·K⁻¹·mol⁻¹

$n_{total}$ : total number of moles of gases

To calculate the total number of moles of gases:

$n_{total} = \frac{P.V}{R.T} = \frac{2.14 atm\times 6.68 L}{0.08206 LatmK^{-}mol^{-}\times 292.15K}$ = 0.5963 moles

Let, the number of moles of carbon dioxide be n₁ and number of moles of hydrogen be n₂

Given: mass of carbon dioxide: w₁ = 16.8 g, mass of hydrogen: w₂ = ?g

molar mass of carbon dioxide: m₁ = 44.01 g/mol, molar mass of hydrogen: m₂= 2.016 g/mol

Therefore, $n_{total}$ = n_{1}+n_{2} = (w₁ ÷ m₁) + (w₂ ÷ m₂)

⇒ 0.5963 mol = (16.8 g ÷ 44.01 g/mol) + (w₂ ÷ 2.016 g/mol)

⇒ 0.5963 mol = (0.3817mol) + (w₂ ÷ 2.016 g/mol)

⇒ 0.5963 mol - 0.3817mol = (w₂ ÷ 2.016 g/mol)

⇒ 0.2146 mol = (w₂ ÷ 2.016 g/mol)

⇒ w₂ = 0.433 g

Therefore, the mass of hydrogen gas in the mixture: w₂ = 0.433 g

4 0

4 years ago

A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi

attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

$E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}$

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

$E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\$