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Dafna1 [17]
1 year ago
6

131i has a half-life of 8.04 days. assuming you start with a 1.53 mg sample of 131i, how many mg will remain after 13.0 days ___

_______?
a.0.835
b.0.268
c.0.422
d.0.440
e.0.499
Chemistry
1 answer:
inn [45]1 year ago
8 0
For this problem we can use half-life formula and radioactive decay formula.

Half-life formula,
t1/2 = ln 2 / λ

where, t1/2 is half-life and λ is radioactive decay constant.
t1/2 = 8.04 days

Hence,         
8.04 days    = ln 2 / λ                         
λ   = ln 2 / 8.04 days

Radioactive decay law,
Nt = No e∧(-λt)

where, Nt is amount of compound at t time, No is amount of compound at  t = 0 time, t is time taken to decay and λ is radioactive decay constant.

Nt = ?
No = 1.53 mg
λ   = ln 2 / 8.04 days = 0.693 / 8.04 days
t    = 13.0 days 

By substituting,
Nt = 1.53 mg e∧((-0.693/8.04 days) x 13.0 days))
Nt = 0.4989 mg = 0.0.499 mg

Hence, mass of remaining sample after 13.0 days = 0.499 mg

The answer is "e"

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Why does 4.03/0.0000035 = 1.2 x 106, instead of a different number of significant figures?
jasenka [17]

Explanations:- As per the significant figures rule, In multiplication and division, we go with least number of sig figs.

4.03 has three sig figs where as 0.0000035 has two sig figs only, The zeros in this number are not sig figs as they are just holding the place values. As the least number of sig figs here is two, the answer needs to be reported with two sig figs only.

\frac{4.03}{0.0000035}=1.2*10^6



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Who helps recycle nutrients through an ecosystem? A. producers and decomposers B. producers and consumers C. consumers and decom
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On a trip to the natural history museum you find two minerals that are similar in color. You can see from their chemical formula
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Surface currents are mainly caused by prevailing winds. What is the best synonym for "prevailing?"
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6 0
2 years ago
Ancient Egyptians used a variety of lead compounds as white pigments in their cosmetics, including PbS, PbCO3, PbCl(OH), and Pb2
ser-zykov [4K]

Answer : The compound contains the highest percentage of lead (by mass) is, PbS.

Explanation :

To calculate the percentage of lead in sample, we use the equation:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100

<u>For PbS :</u>

Mass of PbS = 239.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{239.3g}\times 100=86.58\%

The percentage of lead in the PbS is 86.58 %.

<u>For PbCO_3 :</u>

Mass of PbCO_3 = 267.2 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{267.2g}\times 100=77.55\%

The percentage of lead in the PbCO_3 is 77.55 %.

<u>For PbCl(OH) :</u>

Mass of PbCl(OH) = 259.7 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{259.7g}\times 100=79.78\%

The percentage of lead in the PbCl(OH) is 79.78 %.

<u>For Pb_2Cl_2CO_3 :</u>

Mass of Pb_2Cl_2CO3_ = 545.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{545.3g}\times 100=37.99\%

The percentage of lead in the Pb_2Cl_2CO3_ is 37.99 %.

Hence, from this we conclude that, the compound PbS contains the highest percentage of lead (by mass).

6 0
1 year ago
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