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3 years ago

5. An element whose highest occupied s sublevel and a nearby d sublevel contain electrons

Chemistry

1 answer:

Mumz [18]3 years ago

5 0

Answer:

Transition Element

Explanation:

Transition elements are defined as those elements which can form at least one stable ion and has partially filled d-orbitals. They are also characterized by forming complex compounds and having different oxidation states for a single metal element.

Transition metals are present between the metals and the non metals in the periodic table occupying groups from 3 to 12. There general electronic configuration is as follow,

(n-1)d ¹⁻¹⁰ns ¹⁻²

The general configuration shows that for a given metal, the d sublevel will be in lower energy level as compared to corresponding s sublevel. For example,

Scandium is present in fourth period hence, its s sublevel is present in 4rth energy level so its d sublevel will be present in 3rd energy level respectively.

Hence, we can conclude that for transition metals the electron are present in highest occupied s sublevel and a nearby d sublevel .

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3 years ago

Which lists the elements in order from least conductive to most conductive?

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Answer: Option (A) is the correct answer.

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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

OleMash [197]

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, $\rm m^{3}$ ;
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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