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wel
3 years ago
5

One of the radioactive isotopes used in chemical and medical research is sulfur-35, which has a half-life of 87 days. How long w

ould it take for 0.25 g to remain of a 1.00 g sample of sulfur-35
Chemistry
1 answer:
ASHA 777 [7]3 years ago
7 0

Answer:

Time taken = 174 days

Explanation:

Half life is the time take for a subsrtance taken to decay to half of it's origial or initial concentration.

In this probel, the haf life is 87 days, this means that after evry 87 days, the concentration or mass of sulfur-35 decreases by half.

If the starting mass is 1.00g, then we have;

1.00g --> 0.5g (First Half life)

0.5g --> 0.25g (Second half life)

This means that sulphur-335 would undergo two half lives for 0.25g to remain.

Total time taken = Number of half lives  * Half life

Time taken = 2 * 87

Time taken = 174 days

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Answer:

2.25×10¯³ mm.

Explanation:

From the question given above, we obtained the following information:

Diameter in micrometer = 2.25 μm

Diameter in millimetre (mm) =?

Next we shall convert 2.25 μm to metre (m). This can be obtained as follow:

1 μm = 1×10¯⁶ m

Therefore,

2.25 μm = 2.25 μm / 1 μm × 1×10¯⁶ m

2.25 μm = 2.25×10¯⁶ m

Finally, we shall convert 2.25×10¯⁶ m to millimetre (mm) as follow:

1 m = 1000 mm

Therefore,

2.25×10¯⁶ m = 2.25×10¯⁶ m /1 m × 1000 mm

2.25×10¯⁶ m = 2.25×10¯³ mm

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3 years ago
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Answer:

1. V2.

2. 299K.

3. 451K

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Explanation:

1. The data obtained from the question include:

Initial volume (V1) = 0.25mL

Initial temperature (T1) = 26°C

Final temperature (T2) = 178°C

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2. Conversion from celsius to Kelvin temperature.

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Final temperature (T2) = 451K

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