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My name is Ann [436]
3 years ago
7

The hydrolysis of ethyl acetate (CH3COOC2H5) by sodium hydroxide yields ethanol (C2H5OH) and sodium acetate (NaO2CCH3) by the re

action shown, which follows second order kinetics. In an experiment 500 mL of 0.1 M ethylacetate in water is mixed with 500 mL of 0.1 M NaOH in water. After 20 minutes the concentration of sodium acetate is measured and is 0.0435M. Determine the rate constant for the reaction.
CH3COOC2H5 + NaOH C2H5OH + NaO2CCH3

A) 0.002 L mol-1s-1 B) 0.11 s-1 C) 0.11 L mol-1 s-1 D) 0.3 L mol-1 s-1 E) 0.5 L mol-1 s-1
Chemistry
1 answer:
devlian [24]3 years ago
6 0

Answer:

0.02

Explanation:

work using rate law

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The hydrolysis in aqueous HCl of compound A can lead to the formation of a carboxylic acid and an alcohol.

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The picture shows the structures of compound A, benzoncaine and the possible products of the proposed reaction.

The acidic hydrolysis is the inverse of the esterification reaction. Therefore, the ester group of compund A will react to form the equivalent carboxylic acid and alcohol.

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The thallium (present as Tl2SO4) in a 9.486-g pesticide sample was precipitated as thallium(I) iodide. Calculate the mass percen
tino4ka555 [31]

Answer: The mass percentage of Tl_2SO_4 is 5.86%

Explanation:

To calculate the mass percentage of Tl_2SO_4 in the sample it is necessary to know the mass of the solute (Tl_2SO_4 in this case), and the mass of the solution (pesticide sample, whose mass is explicit in the letter of the problem).

To calculate the mass of the solute, we must take the mass of the TlI precipitate.  We can establish a relation between the mass of TlI and Tl_2SO_4 using the stoichiometry of the compounds:

moles\ of\ TlI = \frac{0.1824 g}{331.27\frac{g}{mol} } = 5.51*10^{-4}\ mol.

Since for every mole of Tl in TlI there are two moles of Tl in Tl_2SO_4, we have:

moles\ of\ Tl_2SO_4 = 2 * moles\ of\ TlI = 1,102*10^{-3}\ mol

Using the molar mass of Tl_2SO_4 we have:

mass\ of\ Tl_2SO_4 = 1,102*10^{-3}\ mol * 504.83\ \frac{g}{mol}= 0.56\ g

Finally, we can use the mass percentage formula:

mass\ percentage = (\frac{solute\ mass}{solution\ mass} )*100 = (\frac{mass\ of\ Tl_2SO_4}{pesticide\ sample\ mass})*100 = (\frac{0.56g}{9.486g})*100 = 5.86\%

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