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3 years ago

Arrange the elements in order of increasing atomic radius. Use the periodic table to help you.

Chemistry

2 answers:

Alexandra [31]3 years ago

5 0

Answer:

F, S, Mg, Ba

Explanation:

Fluorine has a small atomic radius. These go up until Ba, which has a larger radius

Alik [6]3 years ago

4 0

Answer: Order of atomic radius follows:

$F$

Explanation:

Atomic radius of an atom is defined as the total distance from the nucleus to the outermost shell of the atom.

As moving from top to bottom, there is an addition of shell around the nucleus and the outermost shell gets far away from the nucleus and hence, the distance between the nucleus and outermost shell increases. Thus, increasing the atomic radii of the atom.

As moving from left to right in a period, more and more electrons get added up in the same shell and the attraction between the last electron and nucleus increases, which results in the shrinkage of size of an atom. Thus, decreasing the atomic radii of the atom on moving towards right of the periodic table.

For the given elements:

Magnesium is present in Group 2, Period 3 of the periodic table.

Barium is present in Group 2, Period 6 of the periodic table.

Fluorine is present in Group 17, Period 2 of the periodic table.

Sulfur is present in Group 16, Period 3 of the periodic table.

The increasing order of atomic radius follows:

$F$

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Answer:

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3 years ago

reaction is found to have a rate constant of 12.5 s-1 at 25.0 Celsius. When you heat the reaction up by ten degrees Celsius, the

lisabon 2012 [21]

Answer : The activation energy for the reaction is, 52.9 kJ/mol

Explanation :

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$T_1$ = initial temperature = $25.0^oC=273+25.0=298.0K$

$T_2$ = final temperature = $25.0^oC+10=35.0^oC=273+35.0=308.0K$

$K_1$ = rate constant at $25.0^oC$ = $12.5s^{-1}$

$K_2$ = rate constant at $35.0^oC$ = $2\times K_1=2\times 12.5s^{-1}=25.0s^{-1}$

$Ea$ = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

Now put all the given values in this formula, we get:

$\log (\frac{25.0s^{-1}}{12.5s^{-1}})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{298.0K}-\frac{1}{308.0K}]$

$Ea=52903.05J/mole=52.9kJ/mol$

Therefore, the activation energy for the reaction is, 52.9 kJ/mol

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Answer:

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we need to put

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4 years ago

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