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2 years ago

The atom that contains only one electron in the highest occupied energy sublevel

Chemistry

2 answers:

nydimaria [60]2 years ago

5 0

Explanation:

The highest occupied energy sub-level is defined as number of electrons present in the outermost energy level of an atom.

For example, for alkali metals general electronic configuration is $ns^{1}$ .

This means that in the highest occupied energy sub-level of an alkali metal there will be only one electron.

Whereas for alkaline earth metals general electronic configuration is $ns^{2}$ . This means there are 2 electrons present in the highest occupied energy sub-level of an alkaline earth metal.

Thus, we can conclude that atom that contains only one electron in the highest occupied energy sublevel are alkali metals.

Georgia [21]2 years ago

4 0

All atoms that are considered as alkali metals have one valence electron on the outermost shell.

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What is an acid and what is a base

Mice21 [21]

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Explanation:

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3 years ago

Read 2 more answers

29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.

arsen [322]

The answer for the following mention bellow.

Therefore the final temperature of the gas is 260 k

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 150.0 kPa

Final pressure ( $P_{2}$ ) = 210.0 kPa

Initial volume ( $V_{1}$ ) = 1.75 L

Final volume ( $V_{2}$ ) = 1.30 L

Initial temperature ( $T_{1}$ ) = -23°C = 250 k

To find:

Final temperature ( $T_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

We know;

$\frac{P*V}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ × $\frac{V_{1} }{V_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $T_{1}$ ) represents the initial temperature of the gas

( $T_{2}$ ) represents the final temperature of the gas

So;

$\frac{150 * 1.75}{210 * 1.30}$ = $\frac{260}{T_{2} }$

( $T_{2}$ ) =260 k

Therefore the final temperature of the gas is 260 k

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