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Harrizon [31]
3 years ago
10

Acetone is one of the most important solvents in organic chemistry. It is used to dissolve everything from fats and waxes to air

plane glue and nail polish. At high temperatures, it decomposes in a first-order process to methane and ketene (CH2═C═O). At 600°C, the rate constant is 8.7 × 10^−3 s^−1.
a. What is the half-life of the reaction?
b. How much time is required for 32% of a sample of acetone to decompose?
c. How much time is required for 81% of a sample of acetone to decompose?
Chemistry
1 answer:
Yanka [14]3 years ago
4 0

Answer: a. 79.6 s

b. 44.3 s

c. 191 s

Explanation:

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  

t = age of sample

a = let initial amount of the reactant  

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{8.7\times 10^{-3}s^{-1}}=79.6s

b) for completion of 32% of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-32}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{68}

t=44.3s

c) for completion of 81 % of reaction  

t=\frac{2.303}{k}\log\frac{100}{100-81}

t=\frac{2.303}{8.7\times 10^{-3}}\log\frac{100}{19}

t=191s

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How many grams of KCl 03 are needed to produce 6.75 Liters of O2 gas measured at 1.3 atm pressure and 298 K?
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11.48-gram of KCl0_3 are needed to produce 6.75 Liters of O_2  gas measured at 1.3 atm pressure and 298 K

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1.3 atm

V= 6.75 Liters

n=?

R= 0.082057338 \;L \;atm \;K^{-1}mol^{-1}

T=298 K

Putting value in the given equation:

\frac{PV}{RT}=n

n= \frac{1.3 \;atm\; X \;6.75 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 298}

Moles = 0.3588 moles

Now,

Moles = \frac{mass}{molar \;mass}

0.3588 moles = \frac{mass}{32}

Mass= 11.48 gram

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Promethium-147 is sometimes used in luminescent paint. It has a half-life of 956.3 days. If 250 grams (g) of promethium-147 is u
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Answer:

% = 76.75%

Explanation:

To solve this problem, we just need to use the expressions of half life and it's relation with the concentration or mass of a compound. That expression is the following:

A = A₀ e^(-kt)   (1)

Where:

A and A₀: concentrations or mass of the compounds, (final and initial)

k: constant decay of the compound

t: given time

Now to get the value of k, we should use the following expression:

k = ln2 / t₁/₂   (2)

You should note that this expression is valid when the reaction is of order 1 or first order. In this kind of exercises, we can assume it's a first order because we are not using the isotope for a reaction.

Now, let's calculate k:

k = ln2 / 956.3

k = 7.25x10⁻⁴ d⁻¹

With this value, we just replace it in (1) to get the final mass of the isotope. The given time is 1 year or 365 days so:

A = 250 e^(-7.25x10⁻⁴ * 365)

A = 250 e^(-0.7675)

A = 191.87 g

However, the question is the percentage left after 1 year so:

% = (191.87 / 250) * 100

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