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Nimfa-mama [501]
3 years ago
8

32 P P32 is a radioactive isotope with a half-life of 14.3 days. If you currently have 65.1 65.1 g of 32 P P32 , how much 32 P P

32 was present 5.00 5.00 days ago?
Chemistry
1 answer:
kirza4 [7]3 years ago
3 0

Answer:

Original Mass before 5 days = 82.97 gm

Explanation:

The problem can be solved using simple formula:

How much mass remains = (1 / 2^n) x Original Mass ------ (1)

Since,

How much mass remains = 65.1 gm

n = Number of half-lives during the interval = 5 days / 14.3 days = 0.35

Original Mass = ?

Now, equation (1) implies:

⇒ 65.1 = ( 1 / 2^0.35 ) x Original Mass

⇒ 65.1 = (0.785) x Original Mass

⇒ Original Mass = 82. 97 gm

Alternatively, another formula is:

N(t) = No (1/2)^ (t / t1/2) ------- (2)

where,

Nt = Mass Remains

No = Initial Mass

t = Time

t1/2 = Half Life

Equation (2) implies:

⇒ 65.1 = No (1/2) ^ (5/14.3)

Solving above equation we get,

No = 82.97 gm

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Explanation:

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Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O

According to question, solid calcium hydroxde is added to 38.0 mL of 0.410 M of HCl solution, so the chemical equation can be written as:

Ca(OH)_2(s)+2HCl(aq)\rightarrow CaCl_2(aq)+2H_2O(l)

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