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podryga [215]
3 years ago
12

WILL MARK BRAINLIEST IF GOOD EXPLANATION...

Chemistry
1 answer:
daser333 [38]3 years ago
7 0

Answer:

a) 64.27%

b) 58%

c) ethanol is the limiting reactant

d) ethanol is the limiting reactant

Explanation:

We have to note that the expected yield is the theoretical yield while the actual mass or amount of product formed is the actual yield.

a) theoretical yield=68.3g

Actual yield= 43.9 g

Percentage yield= 43.9/68.3 ×100

Percentage yield= 64.27%

b) theoretical yield= 0.0722 moles

Actual yield = 0.0419

Percentage yield= 0.0419/0.0722 × 100

Percentage yield= 58%

c) note that the limiting reactant yields the least number of moles of product

Ethanol will be the limiting reactant since it is not present in excess.

d) from the reaction equation;

1 mole of acetic acid produces 1 mole of ethyl acetate

0.58 moles of ethanol produces 0.58 moles of ethyl acetate

1 mole of acetic acid yields 1 mole of ethyl acetate

Hence 0.82 moles of acetic acid yields 0.82 moles of ethyl acetate

Hence ethanol is the limiting reactant.

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Explanation:

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A piece of metal that has a density of 5.2 g/cm3 and a mass of 100 g was placed in a full jar of water. How many mL of water was
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100g / (5.2g/cm3)

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Wat are the five assumption of kinetic therory of gases​
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Answer:

The five assumption of Kinetic molecular theory are given below.

Explanation:

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6 0
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A certain radioactive isotope decays at a rate of 0.2​% annually. Determine the ​half-life of this​ isotope, to the nearest year
pychu [463]

Answer:

The half-life of the radioactive isotope is 346 years.

Explanation:

The decay rate of the isotope is modelled after the following first-order linear ordinary differential equation:

\frac{dm}{dt} = -\frac{m}{\tau}

Where:

m - Current isotope mass, measured in kilograms.

t - Time, measured in years.

\tau - Time constant, measured in years.

The solution of this differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope. It is known that radioactive isotope decays at a yearly rate of 0.2 % annually, then, the following relationship is obtained:

\%e = \frac{m(t)-m(t+1)}{m(t)}\times 100\,\% = 0.2\,\%

1 - \frac{m(t+1)}{m(t)} = 0.002

1 - \frac{m_{o}\cdot e^{-\frac{t+1}{\tau} }}{m_{o}\cdot e^{-\frac{t}{\tau} }}=0.002

1 - e^{-\frac{1}{\tau} } = 0.002

e^{-\frac{1}{\tau} } = 0.998

-\frac{1}{\tau} = \ln 0.998

The time constant associated to the decay is:

\tau = -\frac{1}{\ln 0.998}

\tau \approx 499.500\,years

Finally, the half-life of the isotope as a function of time constant is given by the expression described below:

t_{1/2} = \tau \cdot \ln 2

If \tau \approx 499.500\,years, the half-life of the isotope is:

t_{1/2} = (499.500\,years)\cdot \ln 2

t_{1/2}\approx 346.227\,years

The half-life of the radioactive isotope is 346 years.

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