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Effectus [21]
4 years ago
9

if the theoretical yield of a acetic acide in a certain experiment is 21.4 grams and the actual yield is 19.1 grams, what is the

percent yield?
Chemistry
1 answer:
Shtirlitz [24]4 years ago
7 0

Answer:

89.2 %

Explanation:

Percent yield = (Yield produced / Theoretical yield) . 100

(19.1 g / 21.4 g) . 100 = 89.2 %

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answer:

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Calculate the final temperature of 35 mL of ethanol initially at 20 ∘C upon absorption of 720 J of heat. (density of ethanol =0.
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Now, we may use the equation:
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4 years ago
0.456 grams of a monoprotic acid is titrated with 45.88 mL of 0.0500 M NaOH. What is the molecular mass (molar mass) of the acid
aivan3 [116]

Answer:

Molar\ mass= 198.78\ g/mol

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For NaOH :

Molarity = 0.0500 M

Volume = 45.88 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 45.88×10⁻³ L

Moles_{NaOH} =0.0500 \times {45.88\times 10^{-3}}\ moles=0.002294\ moles

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The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.002294\ moles= \frac{0.456\ g}{Molar\ mass}

Molar\ mass= 198.78\ g/mol

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