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3 years ago

An atom bonds with another atom. What is the best classification for this reaction?

Chemistry

2 answers:

Galina-37 [17]3 years ago

7 0

<h3><u>Answer;</u></h3>

Chemical bond

<h3><u>Explanation;</u></h3>

<em><u>Chemical bond is an attraction that is formed between atoms or molecules which enables formation of chemical compounds. </u></em>
<em><u>Chemical bonds result from attraction of atoms to each other through the sharing of electrons, exchanging of electrons or the transfer of electrons or electrostatic forces.</u></em>
Chemical bonds may be classified as metallic bond, ionic bond or covalent bond.
Ionic bonds occur when electrons are transferred from one atom to another, and the result is attraction between the resulting oppositely charged ions. Covalent bonds occurs as a result of sharing of electrons between non-metal atoms. Metallic bond occur in metals as a result of positively charged nucleus and a sea of electrons in the atoms.

OlgaM077 [116]3 years ago

4 0

Explanation:

When an atom bonds with another atom then it results in the formation of a chemical bond.

This formation of bond could be due to transfer or sharing of electrons.

When a chemical bond is formed due to transfer of electrons then this type of bond is known as ionic bond.

When a chemical bond is formed due to sharing of electrons then this type of bond is known as covalent bond.

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m

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Answer:

$m_{H_2O}=4.86gH_2O$

Explanation:

Hello,

In this case, the described chemical reaction is:

$C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O$

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

$n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6$

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

$n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6$

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

$m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O$

Best regards.

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3 years ago