Chemistry, 20.01.2020 19:31, Lianabel0517

Which of the following is chemically inert (unreactive)? select one:

(a) carbon (atomic number 6)
(b) sodium (atomic number 11)
(c) neon (atomic number 10)
(d) oxygen (atomic number 8)

Answers: 3

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Answers

answered: dsanders20

The chemically inert element is Neon.

Explanation:

Reactivity of an element is defined as the tendency of an element to loose or gain electrons.

Elements are divided into various categories:

Metals are the elements which loose electrons. They are present on the left side of the periodic table. They are present from Group 1 to Group 14 of the periodic table.Non-metals are the elements which gain electrons. They are present on right side of the periodic table. They are present from group 15 to group 17 of the periodic table.Noble gases are present in the farthermost right side of the periodic table. They are present in group 18. They are inert in nature or are unreactive. Transition metals are present in the center of the periodic table from group 3 to group 12.Rare earth metals are present in group 3 from the left. All the elements present in lanthanides series are considered as rare Earth metals.

For the given options:

Option A: Carbon

Carbon is present in Group 14 and Period 2 of the periodic table. It is a metal.

Option B: Sodium

Sodium is present in Group 1 and Period 3 of the periodic table. It is a metal.

Option C: Neon

Neon is present in Group 18 and Period 2 of the periodic table. It is a noble gas.

Option D: Oxygen

Oxygen is present in Group 16 and Period 2 of the periodic table. It is a non-metal.

Hence, the chemically inert element is Neon.

answered: Guest

the answer is true.

this statement is true.

answered: Guest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

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Which of the following is chemically inert (unreactive)? select one:

(a) carbon (atom...