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jarptica [38.1K]

3 years ago

15

A balanced chemical equation explicitly provides you with the

1 answer:

scZoUnD [109]3 years ago

4 0

Answer:

A balanced chemical equation explicitly provides you with the

a. conditions under which the reaction occurs.

b. molar masses of the reactants and products.

c. conversion factors needed to solve a stoichiometry problem.

d. mole ratios needed to solve a stoichiometry problem.

The correct answer to the question is

d. mole ratios needed to solve a stoichiometry problem.

Explanation:

A chemical reaction is sad to be balanced if it contains the complete list of reactants and products present in their relative proportions. The amounts of reactants and products can be calculated from a balanced chemical reaction equation by applying a process known as stoichiometry

An example of stoichiometry is the formation of two moles of water from one mole of oxygen and two moles of hydrogen

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What is the percentage yield of O2 if 12.3 g of KClO3 (molar mass 123 g) is decomposed to produce 3.2 g of O2 (molar mass 32 g)

My name is Ann [436]

Answer:

The percentage yield of O2 is 66.7%

Explanation:

Reaction for decomposition of potassium chlorate is:

2KClO₃ → 2KCl + 3O₂

The products are potassium chloride and oxygen.

Let's find out the moles of chlorate.

Mass / Molar mass = Moles

12.3 g / 123 g/mol = 0.1 mol

So ratio is 2:3, 2 moles of chlorate produce 3 mol of oxygen.

Then, 0.1 mol of chlorate may produce (0.1 .3)/ 2 = 0.15 moles

Let's convert the moles of produced oxygen, as to find out the theoretical yield.

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(Produced Yield / Theoretical yield) . 100 =

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3 years ago

A first-order reaction is 45% complete at the end of 43 minutes. What is the length of the half-life of this reaction

shtirl [24]

The half-life of the reaction is 50 minutes

Data;

Time = 43 minutes
Type of reaction = first order
Amount of Completion = 45%

<h3>Reaction Constant</h3>

Let the initial concentration of the reaction be X $_0$

The reactant left = (1 - 0.45) X $_0$ = 0.55 X $_0$ = X

For a first order reaction

$\ln(\frac{x}{x_o}) = -kt\\ k = \frac{1}{t}\ln (\frac{x_o}{x}) \\ k = \frac{1}{43}\ln (\frac{x_o}{0.55_o})\\ k = 0.013903 min^-^1$

<h3>Half Life </h3>

The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.

$x = \frac{x_o}{2} \\t = t_\frac{1}{2} \\t_\frac{1}{2} = \frac{1}{k}\ln(\frac{x_o}{x_o/2})\\$

Substitute the values

$t_\frac{1}{2} = \frac{1}{k}\ln(2)=\frac{0.6931}{0.013903}\\t_\frac{1}{2}= 49.85 min = 50 min$

The half-life of the reaction is 50 minutes

Learn more on half-life of a first order reaction here;

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Answer:

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