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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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6 0
3 years ago
Read 2 more answers
Which of these substances has the lowest pH? 0.5 M HBr, pOH = 13.5 0.05 M HCl, pOH = 12.7 0.005 M KOH, pOH = 2.3
vagabundo [1.1K]
<h3><u>Answer;</u></h3>

0.5 M HBr, pOH = 13.5 ; Has the lowest pH

<h3><u>Explanation;</u></h3>

From the question;

pH = -Log [OH]

or pH = 14 - pOH

Therefore;

For 0.5 M HBr

[H+] = 0.5 M

pH = - Log [0.5]

     = 0.30

For;  pOH = 13.5

pH = 14 - pOH

     = 14 -13.5

     = 0.5

For; 0.05 M HCl

pH = - log [H+]

[H+] = 0.05

pH = - Log [0.05]

     = 1.30

For; pOH = 12.7

pH = 14 -pOH

     = 14 -12.7

     = 1.30

For;  0.005 M KOH,

pOH = - log [OH]

[OH-] = 0.005

pOH = - Log 0.005

        = 2.30

pH = 14 - 2.30

     = 11.7

For; pOH = 2.3

   pH = 14 -pOH

         = 14- 2.3

         = 11.7

6 0
3 years ago
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