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3 years ago

What happens when the electron moves from the first energy level to the second energy level?

Chemistry

2 answers:

STatiana [176]3 years ago

5 0

If an electron moves up from the first orbit to the higher energy levels, energy will be absorbed by the electron itself and no emission line produced.
But if it moves from the orbits 6,5,4 and 3 to orbit 2, energy will be released by the electron and different emission lines wll be produced.

ki77a [65]3 years ago

4 0

Your answer is: Energy is absorbed by the atom.

this is correct hope it helps:)

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2 years ago

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3 years ago

an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a tem

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Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

$P\cdot V = n\cdot R\cdot T$ ,

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$ .

where

$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$n$ is the number of moles of particles in this gas,
$R$ is the ideal gas constant, and
$T$ is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of $R$ will be $8.314$ if $P$ , $V$ , and $T$ are in SI units. Convert these values to SI units:

$P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$ ;
$V$ shall be in cubic meters, $\rm m^{3}$ ;
$T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K$ .

Apply the ideal gas law:

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}$ .

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3 years ago