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Inessa [10]
2 years ago
11

a radioisotope has a half life of three hours how much of a 120g sample remains after 9 hours. what will the g amount be after 3

hours, after 6 hours, and after 9 hours?
Chemistry
1 answer:
aksik [14]2 years ago
3 0

\huge\boxed{♔︎Answer♔︎}

60g after 3 hours, 30g after 6 hours and 15g after 9 hours

Explanation:

Weight of the radioactive sample = 120g

half life time period = 3 hours

(a) The weight of sample after 3 hours

\textsf{ No. of half lives} =  \sf  \cancel\frac{3}{3}  = 1

The fraction of sample left

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

Mass of the sample left

\sf \frac{1}{2}  \times 120 =   \cancel\frac{120}{2}  = 60g

<u>6</u><u>0</u><u>g</u><u> </u><u>of</u><u> </u><u>sample</u><u> </u><u>is</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>3</u><u> </u><u>hours</u>

(b) The weight of sample after 6 hours

\textsf{ No. of half lives} =  \sf   \cancel\frac{6}{3}  = 2

The fraction of the sample left

\sf   { \frac{1}{2} }^{2}  =  \frac{1}{2}  \times  \frac{1}{2}  =  \frac{1}{4}

Mass of the sample left

\sf \frac{1}{4}  \times 120 =   \cancel\frac{120}{4}  = 30g

<u>3</u><u>0</u><u>g</u><u> </u><u>of</u><u> </u><u>sample</u><u> </u><u>is</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>6</u><u> </u><u>hours</u>

(c) The weight of sample after 9 hours

\textsf {No. of half lives} =  \sf   \cancel\frac{9}{3}  = 3

The fraction of sample left

\sf { \frac{1}{2} }^{3}  =  \frac{1}{2}  \times  \frac{1}{2}  \times  \frac{1}{2} =  \frac{1}{8}

Mass of sample left

\sf  \frac{1}{8}  \times 120 =  \cancel \frac{120}{8}  = 15g

<u>1</u><u>5</u><u>g</u><u> </u><u>of</u><u> </u><u>sample</u><u> </u><u>is</u><u> </u><u>left</u><u> </u><u>after</u><u> </u><u>9</u><u> </u><u>hours</u><u>.</u>

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Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

  • In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
  • Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

<u><em>In our problem:</em></u>

P1 = ??? <em>(is needed to be calculated) </em>and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.


2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

<u><em>In our problem:</em></u>

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? <em>(is needed to be calculated) </em>

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.


3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

<u><em>In our problem:</em></u>

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? <em>(is needed to be calculated) </em> and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.


4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; <em><u>101.325 kPa = 1.0 atm</u></em>, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.


5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

<u><em>In our problem:</em></u>

P1 = 2200.0 mmHg and T1 = ??? <em>(is needed to be calculated) </em>.

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

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