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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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If 842 grams of sodium hydroxide reacts with 750.0 grams of aluminum, how many grams of aluminum hydroxide should theoretically
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548.55 grams of aluminum hydroxide should theoretically form.

Explanation:

Balanced equation for the reaction:

3 NaOH + Al ⇒ Al(OH)3 +3 Na

DATA GIVEN:

mass of NaOH = 842 grams, atomic mass =39.9 grams/mole

mass of Al = 750 grams, atomic mass = 26.9 grams/mole

aluminum hydroxide theoretical yield = ?

Moles of NaOH reacted

number of moles = \frac{mass}{atomic mass of 1 mole}

putting the values in the equation

NaOH = \frac{842}{39.9}

           = 21.1 MOLES OF NaOH

Al = \frac{750}{26.9}

   = 27.8 moles

from the equation

 from 3 moles of NaOH 1 mole of Al(OH)3 is produced

21.1 moles of NaOH will react to give x moles of Al(OH)3

\frac{1}{3} = \frac{x}{21.1}

7.03 moles of Al(OH)3 is formed.

and

1 mole of Al(OH)3 is formed from 1 mole of Al in the reaction

so, 27.8 Moles will react to give give 27.8 moles of Al(OH)3 limiting reagent of the given reaction is NaOH

mass of Al(OH)3 =7.03 x 78 (atomic mass of Al(OH)3)

          = 548.55 grams

theoretical  yield from the given data is 548.55 grams

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