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Alja [10]
4 years ago
11

The half-life of phosphorus-32 is 14.30 days. how many milligrams of a 20.00 mg sample of phosphorus-32 will remain after 85.80

days?
Chemistry
1 answer:
ZanzabumX [31]4 years ago
7 0
The amount of sample that is left after a certain period of time, given the half-life, h, can be calculated through the equation.

             A(t) = A(o) (1/2)^(t/d)

where t is the certain period of time. Substituting the known values,

             A(t) = (20 mg)(1/2)^(85.80/14.30)

Solving,

           A(t) = 0.3125 mg

Hence, the answer is 0.3125 mg. 
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Read 2 more answers
10. If 200. L of ammonia (NH3) are produced in the reaction between nitrogen and hydrogen at STP, how many
erma4kov [3.2K]

Answer:

100L of N2

Explanation:

First let us obtain the number of mole of NH3 that occupied 200L at stp. This can be achieved by doing the following:

1mole of NH3 occupied 22.4L at stp.

Therefore, Xmol mole of NH3 will occupy 200L i.e

Xmol of NH3 = 200/22.4 = 8.93moles.

Now let us generate a balanced equation for the reaction.

N2 + 3H2 —> 2NH3

From the equation above,

1mole of N2 produced 2moles of NH3.

Therefore, Xmol of N2 will produce 8.93moles of NH3 i.e

Xmol of N2 = 8.93/2 = 4.47moles

Now let us convert 4.47moles to N2 to Litres. This is illustrated below:

1mole of N2 occupied 22.4L at stp

Therefore, 4.47moles of N2 will occupy = 4.47 x 22.4 = 100L

Therefore, 100L of N2 is used in the reaction

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