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3 years ago

13

Determine the number of equivalents if a 3.89N solution contains 0.76 L of solution

1 answer:

ValentinkaMS [17]3 years ago

5 0

Answer:

2%

Explanation:

oriented C-2, and (3) the minimizing of the number of ... (2) L. A. Mitscher, J. K. Paul, and L. Goldman,Experientia, 19, 195. (1963). ... SOzCeHiBr)3 in 147 ml. of anhydrous methanol containing 0.37 ... bicarbonate and saturated sodium chloride solution, and dried ... determined in 2% chloroform solution; infrared spectra on.

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Why is water in the gas phase when it comes out of a volcano?

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When a car engine burns gasoline, the results of the reaction are similar to when cells burn glucose. both reactions release car

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3 years ago

What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B

natita [175]

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

$A = A_{0}e^{-kt}$

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

$ln[A] = ln[A]_{0}-kt$

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

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3 years ago

A reaction was performed in which 3.6 g 3.6 g of benzoic acid was reacted with excess methanol to make 1.3 g 1.3 g of methyl ben

Anit [1.1K]

<u>Answer:</u> The percent yield of the reaction is 32.34 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For benzoic acid:</u>

Given mass of benzoic acid = 3.6 g

Molar mass of benzoic acid = 122.12 g/mol

Putting values in equation 1, we get:

$\text{Moles of benzoic acid}=\frac{3.6g}{122.12g/mol}=0.0295mol$

The chemical equation for the reaction of benzoic acid and methanol is:

$\text{Benzoic acid + methanol}\rightarrow \text{methyl benzoate}$

By Stoichiometry of the reaction

1 mole of benzoic acid produces 1 mole of methyl benzoate

So, 0.0295 moles of benzoic acid will produce = $\frac{1}{1}\times 0.0295=0.108$ moles of methyl benzoate

Now, calculating the mass of methyl benzoate from equation 1, we get:

Molar mass of methyl benzoate = 136.15 g/mol

Moles of methyl benzoate = 0.0295 moles

Putting values in equation 1, we get:

$0.0295mol=\frac{\text{Mass of methyl benzoate}}{136.15g/mol}\\\\\text{Mass of methyl benzoate}=(0.0295mol\times 136.15g/mol)=4.02g$

To calculate the percentage yield of methyl benzoate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of methyl benzoate = 1.3 g

Theoretical yield of methyl benzoate = 4.02 g

Putting values in above equation, we get:

$\%\text{ yield of methyl benzoate}=\frac{1.3g}{4.02g}\times 100\\\\\% \text{yield of methyl benzoate}=32.34\%$

Hence, the percent yield of the reaction is 32.34 %

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3 years ago