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Bas_tet [7]
3 years ago
8

Suppose 200.0 mL of a 2.50 M solution of sodium hydroxide is combined with 400.0 mL of a 1.50 M solution of iron(III) nitrate. W

hat is the theoretical yield of precipitate? If 14.8 g of the precipitate actually formed, what is the percent yield?

Chemistry
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer: 83%

Explanation:

The detailed solution is shown in the image attached. First we must work out the balanced reaction equation because accurate solution of the problem must be based on the stoichiometry of the reaction. From the given concentration and volume of reactants, we calculate the amount of substance reacted hence identify the limiting reactant. Lastly we use simple proportion to obtain the theoretical yield of the precipitate. This is now used to calculate the actual yield as shown in the solution attached.

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The answer for the question is c.
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What would happen to the rate of a reaction with rate law rate = k [NO]^2[H2] if
Ede4ka [16]

The rate of a reaction would be one-fourth.

<h3>Further explanation</h3>

Given

Rate law-r₁ = k [NO]²[H2]

Required

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\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}

or  

\large{\boxed{\boxed{\bold{v~=~+\frac{\Delta B}{\Delta t}}}}

The concentration of NO were halved, so the rate :

\tt r_2=k[\dfrac{1}{2}No]^2[H_2]\\\\r_2=\dfrac{1}{4}k.[No]^2[H_2]\\\\r_2=\dfrac{1}{4}r_1

3 0
2 years ago
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gayaneshka [121]

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Explanation :

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q_{rxn} = heat released by the reaction = ?

q_{cal} = heat absorbed by the calorimeter

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Now put all the given values in the above formula, we get:

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yawa3891 [41]

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