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Chapter 2: Problem 36

Identify the isotope \(X\) that has one more neutron than protons and a mass number equal to nine times the charge on the ion \(X^{3+}\)

Short Answer

Expert verified
There seems to be an error in the problem as the conditions do not provide a valid result.

Step by step solution

01

Define the conditions

Firstly, let's establish the conditions outlined in the question. \(X\) is an isotope, meaning it has a certain number of protons and a different number of neutrons. We know that \(X\) has one more neutron than proton, so we can state that the number of neutrons = number of protons + 1. Furthermore, the mass number of an isotope \(X\) is the sum of protons and neutrons, and it's given as nine times the charge on \(X^{3+}\). The ion \(X^{3+}\) indicates that there are 3 fewer electrons than protons, hence its atomic number (proton number) is 3.
02

Determine the proton and neutron number

From the above analysis, the number of protons is equal to 3 (atomic number). Since the number of neutrons is one more than the number of protons (from the problem statement), the number of neutrons is equal to 3 + 1 = 4.
03

Compute the mass number

Next, we calculate the mass number of \(X\) which is the sum of protons (3) and neutrons (4). This comes out to be 7.
04

Check the mass number condition

Finally, we need to verify if the mass number (7) is equal to nine times the charge on \(X^{3+}\). Since the charge on \(X^{3+}\) is of +3, nine times this charge is 27. This does not equal to our calculated mass number (7) which indicates a mistake in our calculation.

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Most popular questions from this chapter

\begin{tabular}{l} There are four naturally occurring isotopes of \\ \hline \end{tabular} chromium. Their masses and percent natural abundances are \(49.9461 \mathrm{u}, 4.35 \% ; 51.9405 \mathrm{u}, 83.79 \% ; 52.9407 \mathrm{u}\) \(9.50 \% ;\) and \(53.9389 \mathrm{u}, 2.36 \% .\) Calculate the weightedaverage atomic mass of chromium.

One oxide of rubidium has \(0.187 \mathrm{g}\) O per gram of Rb. A possible O:Rb mass ratio for a second oxide of rubidium is (a) \(16: 85.5 ;\) (b) \(8: 42.7 ;\) (c) \(1: 2.674 ;\) (d) any of these.

Samples of pure carbon weighing \(3.62,5.91,\) and \(7.07 \mathrm{g}\) were burned in an excess of air. The masses of carbon dioxide obtained (the sole product in each case) were \(13.26,21.66,\) and \(25.91 \mathrm{g},\) respectively. (a) Do these data establish that carbon dioxide has a fixed composition? (b) What is the composition of carbon dioxide, expressed in \% C and \% O, by mass?

Deuterium, \(^{2} \mathrm{H}(2.0140 \mathrm{u}),\) is sometimes used to replace the principal hydrogen isotope \(^{1} \mathrm{H}\) in chemical studies. The percent natural abundance of deuterium is 0.015\%. If it can be done with 100\% efficiency, what mass of naturally occurring hydrogen gas would have to be processed to obtain a sample containing \(2.50 \times 10^{21}^{2} \mathrm{H}\) atoms?

Hydrogen and chlorine atoms react to form simple diatomic molecules in a 1: 1 ratio, that is, \(\mathrm{HCl}\). The natural abundances of the chlorine isotopes are \(75.77 \%^{35} \mathrm{Cl}\) and \(24.23 \%^{37} \mathrm{Cl} .\) The natural abundances of \(^{2} \mathrm{H}\) and \(^{3} \mathrm{H}\) are \(0.015 \%\) and less than \(0.001 \%,\) respectively. (a) How many different HCl molecules are possible, and what are their mass numbers (that is, the sum of the mass numbers of the H and Cl atoms)? (b) Which is the most abundant of the possible HCl molecules? Which is the second most abundant?
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