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fiasKO [112]
3 years ago
5

Which of the following electron configurations describes an element that is a noble gas?

Chemistry
2 answers:
butalik [34]3 years ago
5 0

Answer:

The answer to your question is: letter D.

Explanation:

Noble gases are located in group VIIIA of the periodic table, this means that they have 8 eight electrons in their outermost shell.

Due to this characteristic, they are stable and do not react with other elements.

a. 1s22s22p4    The outermost shell of this electron configuration has 6 electrons, then this element has 6 electrons not 8. This configuration is of an element of the group VIA.

b. [Ne]2s22p2  The outermost shell of this element has 4 electrons, so this is not the configuration of a noble gas.

c. [Ar] 3s1  This element only has one electron in its outermost shell, so this is the electron configuration of an alkaline metal.

d. 1s22s22p6 This element has 8 electrons in its outermost shell, so this is the electron configuration of a noble gas.

alexira [117]3 years ago
5 0

Answer:

D

Explanation:

got it right on the chem quiz

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What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97.6˚C) and 1 atm
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Answer : The value of change in entropy for freezing process is, -18.07 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{freezing}}{T_f}

where,

\Delta S = change in entropy

\Delta H_{fus} = change in enthalpy of fusion = 3.17 kJ/mol

As we know that:

\Delta H_{fus}=-\Delta H_{freezing}=-3.17kJ/mol=-3170J/mol

T_f = freezing point temperature = -97.6^oC=273+(-97.6)=175.4K

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

\Delta S=\frac{\Delta H_{freezing}}{T_m}

\Delta S=\frac{-3170J/mol}{175.4K}

\Delta S=-18.07J/mol.K

Therefore, the value of change in entropy for freezing process is, -18.07 J/mol.K

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