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Chapter 2: Problem 2

Silicon has three naturally-occurring isotopes: \(92.23 \%\) of \(^{28} \mathrm{Si}\), with an atomic weight of 27.9769 amu, \(4.68 \%\) of \(^{29} \mathrm{Si}\), with an atomic weight of 28.9765 amu, and \(3.09 \%\) of \(^{30} \mathrm{Si}\) with an atomic weight of 29.9738 amu. On the basis of these data, confirm that the average atomic weight of \(\mathrm{Si}\) is 28.0854 amu.

Short Answer

Expert verified
Question: Calculate the average atomic weight of silicon using the given data on the naturally-occurring isotopes and their respective percentages of abundance. Answer: The calculated average atomic weight of silicon using the given data is approximately 28.082 amu, which is very close to the given value of 28.0854 amu.

Step by step solution

01

Convert percentages to decimals

We have the following percentages of abundances for each isotope: - 92.23% of \(^{28}\mathrm{Si}\) - 4.68% of \(^{29}\mathrm{Si}\) - 3.09% of \(^{30}\mathrm{Si}\) To use them in our formula, we need to convert them into decimals by dividing them by 100. - \(92.23\%\) -> \(0.9223\) - \(4.68\%\) -> \(0.0468\) - \(3.09\%\) -> \(0.0309\)
02

Use the formula to calculate the average atomic weight

Now we can use the given percentages of abundance (in decimals) and atomic weights to calculate the average atomic weight using the formula: Average Atomic Weight = (Percentage of Isotope 1 x Atomic Weight of Isotope 1) + (Percentage of Isotope 2 x Atomic Weight of Isotope 2) + (Percentage of Isotope 3 x Atomic Weight of Isotope 3) Average Atomic Weight = (0.9223 x 27.9769) + (0.0468 x 28.9765) + (0.0309 x 29.9738) Average Atomic Weight = 25.799 + 1.357 + 0.926 = 28.082
03

Compare the calculated value with the given value

Now we have calculated the average atomic weight of silicon to be 28.082 amu. Comparing this with the given value of 28.0854 amu, we can see that they are very close. Although it's not a perfect match, the calculated value of 28.082 amu is close enough to confirm that the average atomic weight of silicon is approximately 28.0854 amu, considering the given data on the naturally-occurring isotopes.

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Potassium iodide (KI) exhibits predominantly ionic bonding. The \(\mathrm{K}^{+}\) and \(\mathrm{I}^{-}\) ions have electron structures that are identical to which two inert gases?

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