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aleksklad [387]
2 years ago
14

If the half life of a radioactive isotope is 1 day, then how much of the original isotope remains at the end of two days?

Chemistry
2 answers:
ikadub [295]2 years ago
8 0

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25%

Explanation:

In a half life of a radioactive isotope is 1 day,it means it loses its half mass each day

We a formula for N half life

\sf  { \frac{1}{2} }^{n}

where n is the number of days

Here the isotope is kept for 2 days

so it's left over mass will be

\sf { (\frac{1}{2}) }^{2}  \\  \\  \\  \frac{1}{2}  \times  \frac{1}{2}  \\  \\  \sf  \frac{1}{4}

It's left over mass 1/4th of the original mass

Now, we need to find it's percentage by multiplying with 100

\sf  \frac{1}{4}  \times 100 \%\\  \\  \sf   \cancel\frac{100}{4} \% \\  \\  \sf 25\%

<u>So</u><u> </u><u>2</u><u>5</u><u>%</u><u> </u><u>mass</u><u> </u><u>will</u><u> </u><u>be</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>2</u><u> </u><u>day</u><u> </u><u>of</u><u> </u><u>half</u><u> </u><u>life</u><u> </u><u>radioactive</u><u> </u><u>isotope</u><u>.</u><u> </u>

JulijaS [17]2 years ago
8 0
25% because the half life of 100 is 50 and the half life of 50 is 25
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As you move from left to right across a period, what happens to the atomic radil?
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3 years ago
A sample of xenon gas at a pressure of 1.14 atm and a temperature of 20.4 °C, occupies a volume of 16.9 liters. If the gas is al
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Answer:

P2= 0.696atm

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8 0
3 years ago
A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure
Lesechka [4]

Answer:

the final volume of the gas is V_2 = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure P_{ext}:

P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}

P_{ext} = 1.03 \ atm

The workdone W = P_{ext}V

The change in volume ΔV= \dfrac{W}{P_{ext}}

ΔV = \dfrac{130.1 \ J  \times \dfrac{1 \ L  \ atm}{ 101.325 \ J}  }{1.03 \ atm }

ΔV = \dfrac{1.28398717 }{1.03  }

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial  volume = 61.5 mL

The change in volume V is \Delta V = V_2 -V_1

-  V_2= -  \Delta V  -V_1

multiply through by (-), we have:

V_2=   \Delta V+V_1

V_2 =  1250 mL + 61.5 mL

V_2 = 1311.5 mL

∴ the final volume of the gas is V_2 = 1311.5 mL

5 0
3 years ago
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