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LekaFEV [45]
3 years ago
5

Identify the neutral element represented by this excited-state electron configuration, then write the ground-state electron conf

iguration for that element.
Excited State: 1s2 2s2 2p2 3s1
Element Symbol: ?
Ground State: ?
Chemistry
1 answer:
sineoko [7]3 years ago
5 0

Answer:

1s² 2s² 2p³

Nitrogen

Explanation:

Excited state configuration: 1s² 2s² 2p² 3s¹

Unknown:

Ground state configuration = ?

Element symbol = ?

Solution:

Let us start by understanding what a ground state configuration entails:

A ground state configuration shows the lowest allowed energy levels of an atom. The excited state denotes when electrons have moved to higher energy levels away from their ground state.

The superscript in the configuration depicts the number of electrons in each of the sublevels.

We can use this number to identify the atom we are dealing with:

 Total number of electrons = 2 + 2 + 2 + 1 = 7 electrons

The element with 7 electrons on the periodic table is Nitrogen.

Now, the ground state configuration:

In writing the electronic configuration of an atom, certain rules must be complied with.

  • Aufbau's principle states that the sublevels with lower energies are filled up before those with higher energies. The order of filling is:

        1s 2s 2p 3s 3p 4s etc

To fill the s-orbital = 2 electrons

                p-orbital = 6 electrons

The outermost shell electrons are usually the ones excited.

  Now, the 3s¹ shell is an excited one taking 1 electron from the second energy level;

 The ground state configuration is 1s² 2s² 2p³

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Kipish [7]
Molecular chemical equation: 
K₂C₂O₄(aq)+Pb(OH)₂(aq) → 2KOH(aq) + PbC₂O₄<span>(s).
Ionic equation:
2K</span>⁺(aq) + C₂O₄²⁻(aq) + Pb²⁺(aq) + 2OH⁻(aq) → 2K⁺(aq) +2OH⁻(aq)+PbC₂O₄(s)
Net ionic eqation:
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7 0
2 years ago
At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249
oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= \frac{1249 g}{396.35 g/mol}

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

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3 years ago
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Answer:

This is known as the coefficient factor

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Dennis_Churaev [7]

Answer:

Explanation:

complete combustion reaction of  ethane is given by the reaction

2C2H6+7O2..............4CO2+6H2O

no of moles in 34 grams of O2=34/32=1.063

7mole of O2 produced 6 moles of H2O

therefore 1.063 moles of O2 produced=1.063*6/7=0.9 moles

now 0.9 moles of H2O contain how much grams=0.9*18=16.2 grams

3 0
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The creation of carbon monoxide is an effect. What is one cause?
myrzilka [38]
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sorry hope that helps though 
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