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Chapter 2: Problem 30

Fill in the gaps in the following table, assuming each column represents a neutral atom. $$ \begin{array}{l|l|l|l|l|l} \hline \text { Symbol } & { }^{89} \mathrm{Y} & & & & \\ \text { Protons } & & 78 & & & 89 \\ \text { Neutrons } & & & 123 & & \\ \text { Electrons } & & & 81 & 50 & \\ \text { Mass no. } & & 195 & & 119 & 227 \\ \hline \end{array} $$

Short Answer

Expert verified
The completed table is as follows: \( \begin{array}{l|l|l|l|l|l} \hline \text { Symbol } & { }^{89} \mathrm{Y} & \mathrm{Pt} & \mathrm{Tl} & \mathrm{Sn} & \\ \text { Protons } & 89 & 78 & 81 & 50 & 89 \\ \text { Neutrons } & 111 & 117 & 123 & 69 & 138 \\ \text { Electrons } & 89 & 78 & 81 & 50 & 89 \\ \text { Mass no. } & 200 & 195 & 204 & 119 & 227 \\ \hline \end{array} \)

Step by step solution

01

Finding the element's symbol for 78 protons

According to the periodic table, the element with the atomic number (number of protons) 78 is platinum (Pt).
02

Finding the number of electrons for a neutral ^{89}Y atom

A neutral atom has the same number of electrons as protons. In the case of Yttrium (Y), it has 89 protons, so it must have 89 electrons.
03

Finding the number of neutrons for a mass number of 195 and 78 protons

We can find the number of neutrons using the formula A = Z + N. In this case, the mass number (A) is 195 and the number of protons (Z) is 78, which gives us N = 195 - 78 = 117 neutrons.
04

Finding the element's symbol for 81 electrons

A neutral atom has equal numbers of protons and electrons. Therefore, an atom with 81 electrons has 81 protons. The element with 81 protons is Thallium (Tl).
05

Finding the number of protons for an atom with 50 electrons

A neutral atom has equal numbers of protons and electrons. Therefore, an atom with 50 electrons has 50 protons. The element with 50 protons is Tin (Sn).
06

Finding the number of neutrons for an atom with 89 protons and mass number 227

We can find the number of neutrons using the formula A = Z + N. In this case, the mass number (A) is 227 and the number of protons (Z) is 89, which gives us N = 227 - 89 = 138 neutrons. The completed table is as follows: $$ \begin{array}{l|l|l|l|l|l} \hline \text { Symbol } & { }^{89} \mathrm{Y} & \mathrm{Pt} & \mathrm{Tl} & \mathrm{Sn} & \\\ \text { Protons } & 89 & 78 & 81 & 50 & 89 \\\ \text { Neutrons } & 111 & 117 & 123 & 69 & 138 \\\ \text { Electrons } & 89 & 78 & 81 & 50 & 89 \\\ \text { Mass no. } & 200 & 195 & 204 & 119 & 227 \\\ \hline \end{array} $$

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Most popular questions from this chapter

Predict the chemical formula for the ionic compound formed by (a) \(\mathrm{Ca}^{2+}\) and \(\mathrm{Br}^{-}\), (b) \(\mathrm{K}^{+}\)and \(\mathrm{CO}_{3}^{2-}\), (c) \(\mathrm{Al}^{3+}\) and \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), (d) \(\mathrm{NH}_{4}^{+}\)and \(\mathrm{SO}_{4}^{2-}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{PO}_{4}{ }^{3-}\).

A cube of gold that is \(1.00 \mathrm{~cm}\) on a side has a mass of \(19.3 \mathrm{~g}\). A single gold atom has a mass of \(197.0\) amu. (a) How many gold atoms are in the cube? (b) From the information given, estimate the diameter in ? of a single gold atom. (c) What assumptions did you make in arriving at your answer for part (b)?

The natural abundance of \({ }^{3} \mathrm{He}\) is \(0.000137 \%\). (a) How many protons, neutrons, and electrons are in an atom of \({ }^{3} \mathrm{He}\) ? (b) Based on the sum of the masses of their subatomic particles, which is expected to be more massive, an atom of \({ }^{3} \mathrm{He}\) or an atom of \({ }^{3} \mathrm{H}\) (which is also called tritium)? (c) Based on your answer to part (b), what would need to be the precision of a mass spectrometer that is able to differentiate between peaks that are due to \({ }^{3} \mathrm{He}^{+}\)and \({ }^{3} \mathrm{H}^{+}\)?

Give the name or chemical formula, as appropriate, for each of the following acids: (a) \(\mathrm{HBrO}_{3}\), (b) \(\mathrm{HBr}\), (c) \(\mathrm{H}_{3} \mathrm{PO}_{4}\), (d) hypochlorous acid, (e) iodic acid, (f) sulfurous acid.

The element lead ( \(\mathrm{Pb}\) ) consists of four naturally occurring isotopes with atomic masses \(203.97302,205.97444,206.97587\), and \(207.97663\) amu. The relative abundances of these four isotopes are \(1.4,24.1,22.1\), and \(52.4 \%\) respectively. From these data, calculate the atomic weight of lead.
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