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nadya68 [22]
2 years ago
11

Why doesn’t neon readily form an ionic bond?

Chemistry
2 answers:
Goryan [66]2 years ago
6 0

By definition of noble gases, neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

<h3>Noble gases</h3>

Noble gases are not very reactive, that is, they practically do not form chemical compounds. This means that they do not react with other substances, nor do they even react between atoms of the same gas, as is the case with diatomic gases such as oxygen (O₂).

The chemical stability of the noble gases and therefore the absence of spontaneous evolution towards any other chemical form, implies that they are already in a state of maximum stability.

All chemical transformations involve valence electrons, they are involved in the process of covalent bond formation and the formation of ions. Therefore, the practically null reactivity of the noble gases is due to the fact that they have a complete valence shell, which gives them a low tendency to capture or release electrons.

Since the noble gases do not react with the other elements, they are also called inert gases.

<h3>Neon</h3>

Neon does not easily form an ionic bond because it belongs to the group of noble or inert gases, so its reactivity is practically nil.

Learn more about noble gases:

brainly.com/question/8361108

brainly.com/question/11960526

brainly.com/question/19024000

Oliga [24]2 years ago
6 0

Atoms of neon do not lose or gain electrons of their outermost orbit easily. So they do not form ionic bonds.Neon do not form ionic bonds is that atoms of neon already have a stable octet in their outer shell.

<h3>Hope this helps!!</h3>
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If 1.08 g of sodium sulfate reacts with an excess of phosphoric acid, how much sulfuric acid is produced?
vfiekz [6]

Answer:  0.745 g of H_2SO_4 will be produced from  1.08 g of sodium sulfate

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Na_2SO_4=\frac{1.08g}{142.04g/mol}=0.0076moles  

3Na_2SO_4+2H_3PO_4\rightarrow 2Na_3PO_4+3H_2SO_4

Na_2SO_4 is the limiting reagent as it limits the formation of product and H_3PO_4 is the excess reagent.

According to stoichiometry :

3 moles of Na_2SO_4 produce = 3 moles of H_2SO_4

Thus 0.0076 moles of Na_2SO_4 will require=\frac{3}{3}\times 0.0076=0.0076moles  of H_2SO_4

Mass of H_2SO_4=moles\times {\text {Molar mass}}=0.0076moles\times 98.1g/mol=0.745g

Thus 0.745 g of H_2SO_4 will be produced from  1.08 g of sodium sulfate

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2 years ago
Given 8.25 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100
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Answer:

              10.87 g of Ethyl Butyrate

Solution:

The Balance Chemical Equation is as follow,

   H₃C-CH₂-CH₂-COOH + H₃C-CH₂-OH  →  H₃C-CH₂-CH₂-COO-CH₂-CH₃ + H₂O

According to equation,

    88.11 g (1 mol) Butanoic Acid produces  =  116.16 g (1 mol) Ethyl Butyrate

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           8.25 g Butanoic Acid will produce  =  X g of Ethyl Butyrate

Solving for X,

                      X =  (8.25 g × 116.16 g) ÷ 88.11 g

                      X =  10.87 g of Ethyl Butyrate

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