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2 years ago

Explain why oxygen is a fairly reactive element while neon is not

Chemistry

1 answer:

Blababa [14]2 years ago

8 0

Answer:

Due to presence of unpaired electrons in outermost shell of oxygen, oxygen is a fairly reactive element while neon is not.

Explanation:

Lets look at the electronic configuration of both the elements.

Oxygen, a chalcogen, has EC as = $1s^{2}2s^{2}2p^{4}$

while, for Neon, a noble gas, EC is = $1s^{2}2s^{2}2p^{6}$

In neon, the outermost orbital p is completely filled while in oxygen it is partially filled, with 2 unpaired electrons.

<em>Thus the completely filled orbitals give extra stability to neon and thus less reactivity. In other case, oxygen has 2 unpaired electrons, unstable and has exchange energy, making it more reactive.</em>

Thus, oxygen is a fairly reactive element while neon is not..

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In the reaction below, how many grams of h2o(g) are produced when 2.1 grams o2(g) are consumed? c4h6(g) + o2(g) → co2(g) + h2o(g

amm1812

The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 ......> 8CO2 + 6H2O

number of moles = mass / molar mass
number of moles of oxygen = 2.1 / 32 = 0.065625 moles

Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water

number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams

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3 years ago

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What is the source of fuel that is most widely used today?

34kurt

The <span>source of fuel that is most widely used today is Natural gas. The answer is letter C. The rest of the choices do not answer the question above,</span>

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2 years ago

Ammonia, NH 3 , may react with oxygen to form nitrogen gas and water. 4 NH 3 ( aq ) + 3 O 2 ( g ) ⟶ 2 N 2 ( g ) + 6 H 2 O ( l )

Alex Ar [27]

Answer:

The limiting reactant is NH₃

0.0186moles of N₂ are the one produced by the limiting reactant

0.020 moles of N₂ are the one produced by the reactant in excess

Explanation:

This is the reaction

4NH₃ + 3O₂ → 2N₂ + 6H₂O

We should calculate the moles of each reactant

Mass / Molar mass = Moles

3.55 g / 17g/m = 0.208 moles NH₃

5.33 g / 32g/m = 0.166 moles O₂

4 moles of ammonia react with 3 moles of oxygen

0.208 moles of ammonia react with (0.208 .3)/4 = 0.156 moles O₂

We have 0.166 moles of O₂ and we need 0.156 moles, so O₂ is the reactant in excess.

3 moles of O₂ react with 4 moles of NH₃

0.166 moles of O₂ react with (0.166 . 4)/ 3 = 0.221 moles

We have 0.208 moles NH₃ and we need 0.221, so NH₃ is the limiting reactant.

To know the moles of N₂, let's apply the Ideal Gas Law

P.V =n.R.T

1atm . 0.450L = n . 0.082 . 295K

0.450 / (0.082 .295) = 0.0186 moles

If we have 100 % yield reaction:

4 moles NH₃ make 2 moles N₂

0.208 moles NH₃ make (0.208 .2)/4 = 0.104 moles

So the % yield reaction is.

0.104 moles ___ 100%

0.0186 moles ___ 17.9%

0.0186moles of N₂ are the one produced by the limiting reactant.

3 moles of O₂ produce 2 moles N₂

0.166 moles O₂ produce (0.166 .2)/3 = 0.111 moles

Now, we apply the yield.

100% ____ 0.111 moles

17.9% = 0.020 moles

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2 years ago

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How many significant figures in 368.1371 ?

ICE Princess25 [194]

There are seven significant figures

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2 years ago

How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe2O3 at standard state condaiti

Diano4ka-milaya [45]

Answer:

412.1kJ

Explanation:

For the reaction , from the question -

4Fe (s) + 3O₂ (g) → 2Fe₂O₃ (s)

Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)

In case the compound is in its standard state , enthalphy of formation is zero

Hence ,

for the above reaction ,

ΔHrxn =( 2 * Δ H° (Fe₂O₃ )) - [ ( 4 *Δ H° Fe ) + (3 * Δ H° O₂ )]

The value for Δ H°(Fe₂O₃ ) = - 824.2kJ/mol

Δ H° Fe = 0

Δ H° O₂ = 0

Putting in the above equation ,

ΔH rxn = ( 2 * Δ H° (Fe₂O₃ )) - 0

ΔHrxn = 2× - 824.2 kJ / mol = - 1648.4 kJ/mol

- 1648.4 kJ/mol , this much heat is released by the buring of 4 mol of Fe.

Hence ,

for 1 mol of Fe ,

- 1648.4 kJ/mol / 4 = 412.1kJ

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3 years ago