Question

How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3410 L bulb at 788.0 K if the pressure is 1.220 atm What is the complete ground state electron configuration for the neon atom

Answer 1

Answer:

n= 0.08186

{He}2s^2 2p^6

Explanation:

PV=nRT

n=PV/RT

n= (1.220 atm)(4.3410 L) / (0.0821 atm*L/mol*K)(788.0 K)

n=0.08186

As for the electron configuration:

Ne:

{He} 2s^2 2p^6

or long hang:

1s^2 2s^2 2p^6

Answer 2

• The moles of gaseous boron trifluoride is 0.08186 moles.
• The ground state electronic configuration of neon is:

$[Ne]=1s^22s^22p^{10}$

Given:

• A gaseous boron trifluoride in a 4.3410 L bulb at 788.0 K, if the pressure is 1.220 atm.
• Neon atom.

To find:

• The moles of gaseous boron trifluoride in a container.
• The electronic configuration of neon in the ground state.

1.

The pressure of the gaseous boron trifluoride = P = 1.220 atm

The volume of the gas in bulb = V = 4.3410 L

The moles of the  gaseous boron trifluoride = n

The temperature of gaseous boron trifluoride = T = 788.0 K

Using an ideal gas equation:

$PV = nRT\\\\1.220 atm\times 4.3410 L=n\times 0.0821 atm L/mol K\times 788.0 K\\\\n=\frac{1.220 atm\times 4.3410 L}{0.0821 atm L/mol K\times 788.0 K}\\\\n=0.08186 mol$

The moles of gaseous boron trifluoride is 0.08186 moles.

2.

The atomic number of neon atom = 10

The electronic configuration in the ground state is the most stable arrangement of the electrons in the lowest energy levels.

The ground state electronic configuration of neon is:

$[Ne]=1s^22s^22p^{10}$

Learn more about the ideal gas equation and electronic configuration here:

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