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worty [1.4K]
3 years ago
12

117 milligrams (mg) of purified product was isolated from a chemical reaction. This experimental yield of product represents a 8

9.0% yield for the reaction. Calculate the theoretical yield, in milligrams (mg), for this reaction. Enter your answer as digits only, no units, using the proper number of significant figures.
Chemistry
1 answer:
Brut [27]3 years ago
5 0

Answer:

131 mg

Explanation:

Percent yield = 89%

Actual yield = 117 mg

Percent yield is given by

\text{Percent yield}=\dfrac{\text{Actual yield}}{\text{Theoretical yield}}\times 100\\\Rightarrow 89=\dfrac{117}{\text{Theoretical yield}}\times 100\\\Rightarrow \text{Theoretical yield}=\dfrac{117}{89}\times 100\\\Rightarrow \text{Theoretical yield}=131.46\approx 131\ \text{mg}

The theoretical yield, for this reaction is 131 mg.

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Read 2 more answers
A buffer solution contains 0.240 M ammonium chloride and 0.499 M ammonia. If 0.0565 moles of perchloric acid are added to 250 mL
snow_lady [41]

Answer:

The pH of the resulting solution is 9.02.

Explanation:

The initial pH of the buffer solution can be found using the Henderson-Hasselbalch equation:

pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})  

pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.240}{0.499}) = 4.43        

pH = 14 - pOH = 14 - 4.43 = 9.57

Now, the perchloric acid added will react with ammonia:

n_{NH_{3}} = 0.499moles/L*0.250 L - 0.0565 moles = 0.0683 moles

Also, the moles of ammonium chloride will increase in the same quantity according to the following reaction:

NH₃ + H₃O⁺ ⇄ NH₄⁺ + H₂O    

n_{NH_{4}Cl} = 0.240 moles/L*0.250L + 0.0565 moles = 0.1165 moles

Finally, we can calculate the pH of the resulting solution:

pOH = pKb + log(\frac{[NH_{4}Cl]}{[NH_{3}]})  

pOH = -log(1.78\cdot 10^{-5}) + log(\frac{0.1165 moles/0.250 L}{0.0683 moles/0.250 L}) = 4.98  

pH = 14 - 4.98 = 9.02

Therefore, the pH of the resulting solution is 9.02.

I hope it helps you!

5 0
3 years ago
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