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Marina CMI [18]
2 years ago
9

David kept 100 grams of radioactive sodium in a container. He observed the amount of sodium left in the container after regular

intervals of time and recorded them in the table shown below.
Time (hours) Amount of sodium in container (in grams)
0 100
15 50
30 25
45 12.5

Based on the observations, which of these is most likely David's inference?
A.The half life period of radioactive sodium is 15 hours.
B.The half life period of radioactive sodium is 50 hours.
C.The amount of sodium left in the container after 60 hours will be 10.5 grams.
D.The amount of sodium left in the container after 60 hours will be 1.5 grams.
Chemistry
2 answers:
Anna [14]2 years ago
5 0
First we have to understand what a half life is. A half life is a measure of time when the amount of a certain object is 50% of the original amount. Hence the answer for this is letter A. The initial amount is 100. fifty percent of 100 is 50 and that happens after 15 hours. Hence, 15 hours is the half life period. 
Fudgin [204]2 years ago
5 0

Answer: A. The half life period of radioactive sodium is 15 hours.

Explanation:

Half life is the time taken by a reactant to reduce to its original concentration. It is designated by the symbol t_\frac{1}{2}.

Given:

At 0 hours the amount of sodium in container is 100 grams

After 15 hours , the amount of sodium in container is 50 grams which is half of the original amount of 100 grams.

It will take 15 hours for half of a 100 g sample to decay to half of the amount i.e. 50 grams.

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7 0
2 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
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Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

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3 years ago
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Answer:

{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 =  \frac{11}{m _{r}}  \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\  \\ { \bf{vapour \: density = 2 \times m _{r}}} \\  = 2 \times 14.85 \\  = 29.7 \: { \tt{g {dm}^{ - 3} }}

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3 years ago
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