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natta225 [31]
3 years ago
7

As part of her nuclear stress test, Simone starts to walk on the treadmill. When she reaches the maximum level, Paul injects a r

adioactive dye with201Tl with an activity of 72 MBq . The radiation emitted from areas of her heart is detected by a scanner and produces images of her heart muscle. After Simone rests for 3 h, Paul injects more dye with 201Tl and she is placed under the scanner again. A second set of images of her heart muscle at rest is taken. When Simone's doctor reviews her scans, he assures her that she had normal blood flow to her heart muscle, both at rest and under stress.
Part A
If the half-life of 201Tl is 3.0 days, what is its activity, in megabecquerels after 3.0 days?
Express the activity to two significant figures and include the appropriate units.

Part B
What is its activity in megabecquerels after 6.0 days?
Express the activity to two significant figures and include the appropriate units.
Chemistry
1 answer:
Naya [18.7K]3 years ago
3 0

Answer:

Part A: 36 MBq; Part B: 18 MBq

Explanation:

The half-life is the time it takes for half the substance to disappear.

The activity decreases by half every half-life

A =Ao(½)^n, where n is the number of half-lives.

Part A

3.0 da = 1 half-life

A =  Ao(½) = ½ × 72 MBq = 36 MBq

Part B

6.0 da = 2 half-lives

A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq

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<u>Answer:</u> The percentage yield of water is 9.5 %

<u>Explanation:</u>

We are given:

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By Stoichiometry of the reaction:

2 moles of hydrogen gas reacts with 1 mole of oxygen gas

So, 14 moles of hydrogen gas will react with = \frac{1}{2}\times 14=7mol of oxygen gas

As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrogen gas produces 2 moles of water

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To calculate the percentage yield of water, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of water = 1.33 moles

Theoretical yield of water = 14 moles

Putting values in above equation, we get:

\%\text{ yield of water}=\frac{1.33mol}{14mol}\times 100\\\\\% \text{yield of water}=9.5\%

Hence, the percent yield of the water is 9.5 %.

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