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3 years ago

Draw the structure of bromous acid, hbro2. optimize formal charges.

Chemistry

1 answer:

ArbitrLikvidat [17]3 years ago

8 0

STRUCTURE OF BROMOUS ACID: H–O–Br=O

In this structure, all the elements have a formal charge of zero. The formal charge of each element is calculated below: 

H: 1 – 1/2(2) – 0 = 0
O: 6 – 1/2(4) – 4 = 0
Br: 7 – 1/2(6) – 4 = 0
O: 6 – 1/2(4) – 4 = 0

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3 years ago

Suppose a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O). The total pressure of

julia-pushkina [17]

Considering the definition of mole fraction, the mole fraction of O₂ in the mixture is 0.434.

<h3>Definition of mole fraction</h3>

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

In other words, the mole fraction expresses the concentration of solute in a solution as the ratio of moles of substance to total moles of solution:

$mole fraction=\frac{moles of substance}{moles of solution}$

<h3>Mole fraction of O₂ in this mixture</h3>

In this case, you know a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O).

So, the total moles of the solution can be calculated as:

Total moles = moles of oxygen (O₂) + moles of nitrous oxide (N₂O)

Then:

Total moles= 4.60 moles + 6 moles

Total moles= 10.60 moles

Finally, the more fraction of O₂ can be calculated as follow:

$Mole fraction of O_{2} =\frac{moles of O_{2}}{total moles}$

$Mole fraction of O_{2} =\frac{4.60 moles}{10.6o moles}$

Solving:

Mole fraction O₂ = 0.434

Finally, the mole fraction of O₂ in the mixture is 0.434.

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2 years ago

Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)

Brilliant_brown [7]

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)

We can write the above reaction in two reactions, one for oxidation and the other for reduction:

Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻ (2)

Reduction reaction

Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)

We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).

Therefore, the reducing agent in reaction (1) is lithium (Li).

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