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11111nata11111 [884]

1 year ago

8

A coin collector has a bag of 50 pennies. What is the percent abundance of the post-1982 pennies? ASAP ANSWER PLS.

1 answer:

Novosadov [1.4K]1 year ago

5 0

The percent abundance of the post-1982 pennies is 56%.

<h3>What is percent abundance?</h3>

Percent abundance is the percentage amount of all naturally occurring isotopes of an element.

<h3> Percent abundance of the post-1982 pennies</h3>

The percent abundance of the post-1982 pennies is calculated as follows;

percent abundance = number of post-1982 pennies / total number of pennies x 100%

percent abundance = (28) / (50) x 100%

percent abundance = 56%

Thus, the percent abundance of the post-1982 pennies is 56%.

Learn more about percent abundance here: brainly.com/question/6844925

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A 9.79 mol sample of freon gas was placed in a balloon. Adding 3.50 mol of freon gas to the balloon increased its volume to 21.8

aivan3 [116]

Answer:

16.06 L was the initial volume of the balloon.

Explanation:

Initial moles of freon in ballon = $n_1=9.79 mol$

Initial volume of freon gas in ballon = $V_1=?$

Moles of freon gas added in the balloon = n = 3.50 mole

Final moles of freon in ballon = $n_2=n_1+n=9.79 mol+3.50 mol=13.29 mol$

Final volume of freon gas in ballon = $V_2=21.8 L$

Using Avogadro's law:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$ ( at constant pressure and temperature)

$V_1=\frac{V_2\times n_1}{n_2}=\frac{21.8 L\times 9.79 mol}{13.29 mol}=16.06L$

16.06 L was the initial volume of the balloon.

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2 years ago

A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a

deff fn [24]

Answer:

A) $E_{a} = 350KJ/mol$ , $E_{a} = 50KJ/mol$ , $E_{a} = 50KJ/mol$

A = 1.5× $10^{-7}s^{-1}$ , A = 1.9× $10^{-7} s^{-1}$ , A=1.5× $10^{-7} s^{-1}$

B) 4.469

Explanation:

From Arrhenius equation

$K=Ae^{\frac{E_{a} }{RT} }$

where; K = Rate of constant

A = Pre exponetial factor

$E_{a}$ = Activation Energy

R = Universal constant

T = Temperature in Kelvin

Given parameters:

$E_{a} =165KJ/mol$

$T_{1}=505K$

$T_{2}=525K$

$R=8.314JK^{-1}mol^{-1}$

taking logarithm on both sides of the equation we have;

$InK=InA-\frac{E_{a} }{RT}$

since we have the rate of two different temperature the equation can be derived as:

$In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )$

$In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1} }.(\frac{1}{505} -\frac{1}{525} )$

$In(\frac{K_{2} }{K_{1} } )$ = 19846.04×7.544× $10^{-5}$ = 1.497

$\frac{K_{2} }{K_{1} }$ = $e^{1.497}$ = 4.469

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