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3 years ago

The formula is used to calculate the yield of a reaction.

Chemistry

2 answers:

creativ13 [48]3 years ago

6 0

Answer;:

Theoretical yield depends on limiting reagent.

Theoretical yield: no of moles of limiting reagent x molar mass of product

Percentage yield= observed yield x100/ theoretical yield

frosja888 [35]3 years ago

5 0

The theoretical yield for a reaction is calculated based on the limiting reagent. This allows researchers to determine how much product can actually be formed based on the reagents present at the beginning of the reaction. The actual yield will never be 100 percent due to limitations.

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What does Le Châtelier's principle state?

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Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

Aleksandr-060686 [28]

Answer : The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$ and the pH of a buffer is, 2.95

Explanation : Given,

$K_a=7.1\times 10^{-4}$

Concentration of $HNO_2$ (weak acid)= 0.26 M

Concentration of $KNO_2$ (conjugate base or salt)= 0.89 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (7.1\times 10^{-4})$

$pK_a=4-\log (7.1)$

$pK_a=3.15$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}$

Now put all the given values in this expression, we get:

$pH=3.15+\log (\frac{0.89}{0.26})$

$pH=2.95$

The pH of a buffer is, 2.95

Now we have to calculate the $H_3O^+$ ion concentration.

$pH=-\log [H_3O^+]$

$2.95=-\log [H_3O^+]$

$[H_3O^+]=1.12\times 10^{-3}M$

The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$

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3 years ago

Moles of Zn(NO3)2 in 131.50g of this substance.

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number of moles= mass / molecular mass
mass=131.50g
molecular mass of Zn(NO3)2 =189.39 g/mol
number of moles = 131.50 g/189.39 g/mol =0.69433 mol

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The number of molecules decrease

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Answer:

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