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Levart [38]
1 year ago
14

The activity of 14C in a 50 -kg human body is estimated to be 2.6 kBq . What is this activity in microcuries?

Chemistry
1 answer:
cupoosta [38]1 year ago
6 0

To make the conversion we must take into account the relationship between kBq and microcuries. Both are units that represent radioactivity activity. The relationship is:

1 kilobecquerel (kBq) = 0.027027027 microcurie

Therefore, 2.6kBq will be equal to:

\begin{gathered} microcuries=givenkBq\times\frac{0.027027027microcurie}{1kBq} \\ m\imaginaryI crocur\imaginaryI es=2.6kBq\times\frac{0.027027027m\imaginaryI crocur\imaginaryI e}{1kBq} \\ microcuries=0.070 \end{gathered}

Answer= That activity in microcuries will be 0.070 microcuries

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Calculate 18.9 minus 2.15. Round the answer to the correct number of significant figures.
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If gas pressure is held constant and the temperature is decreased, gas volume will _____
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- How many grams are in 4.6 x 1016 molecules of phosphorus trichloride (PCI)?
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Answer:

0.0000101 grams.

Explanation:

First of all, use Avogadro's number to convert from molecules to moles.

Avogadro's number is 6.02 x 10^23.

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Now find the molar mass of PCl3

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Cl = 35.5 amu

They are three chlorine atoms and one phosphorus atom. So you do this.

31.0(1) + 35.5(3) = 137.5 g/mol

Now use dimensional analysis to show your work

4.6 x 10^16 grams of PCl3 * 1 mol/ 6.02 x 10^23 * 137.5 g/mol / 1 mol

The grams and the moles cancel out.

So now do [(4.6 x 10^16) / (6.02 x 10^23)] x 137.5

and you will get 0.000010507

4.6 has the fewest digits which is 2.

So use sig figs and the answer rounds to the nearest millionths.

0.000010507 rounds to 0.0000101

So the final answer is 0.0000101 grams(don't forget the units)

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