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Chapter 1: Problem 14

Which of the following nuclei has 3 more neutrons than protons? (Remember: The number before the symbol indicates atomic mass.) (A) 11 \(\mathrm{B}\) (B) 37 \(\mathrm{Cl}\) (C) 24 \(\mathrm{Mg}\) (D) 70 \(\mathrm{Ga}\)

Short Answer

Expert verified
The nucleus which has 3 more neutrons than protons is \(37Cl\) - option (B).

Step by step solution

01

Calculate Neutrons in each Nucleus

For each of the nuclei given in the options, subtract the number of protons from the total atomic mass to get the number of neutrons. The atomic number (which is also the number of protons) of each element is: \(B\) - 5, \(Cl\) - 17, \(Mg\) - 12, \(Ga\) - 31. So number of neutrons for each are: \(11B\) - 6, \(37Cl\) - 20, \(24Mg\) - 12, \(70Ga\) - 39.
02

Calculate Difference between Neutrons and Protons

Subtract the number of protons from the calculated number of neutrons for each element to confirm which has exactly 3 more neutrons than protons. The difference for each are: \(11B\) - 1, \(37Cl\) - 3, \(24Mg\) - 0, \(70Ga\) - 8.
03

Choose the Correct Nucleus

The nucleus which has exactly 3 more neutrons than protons is \(37Cl\). Therefore, the correct choice is option (B).

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Most popular questions from this chapter

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) What is the enthalpy change for reaction 1\(?\) (A) \(-148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (B) \(-56 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (C) \(-18 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) (D) \(+148 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\)

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) What is the expression for the equilibrium constant, \(K_{\mathrm{c}} ?\) (A) \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}\) (B) \(\frac{2\left[\mathrm{SO}_{3}\right]}{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}\) (C) \(\frac{\left[\mathrm{O}_{2}\right]\left[\mathrm{SO}_{2}\right]^{2}}{\left[\mathrm{SO}_{3}\right]^{2}}\) (D) \(\frac{\left[\mathrm{O}_{2}\right] 2\left[\mathrm{SO}_{2}\right]}{2\left[\mathrm{SO}_{3}\right]}\)

\(\begin{array}{ll}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-390 \mathrm{kJ} / \mathrm{mol}} \\\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=-290 \mathrm{kJ} / \mathrm{mol}} \\ {2 \mathrm{C}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)} & {\Delta H^{\circ}=+230 \mathrm{kJ} / \mathrm{mol}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { (A) }-1,300 \mathrm{kJ} \\ \text { (B) }-1,070 \mathrm{kJ} \\ \text { (C) }-840 \mathrm{kJ} \\ \text { (D) }-780 \mathrm{kJ} \end{aligned} $$

Which substance would have the highest boiling point? (A) Ethanol, because it is the most asymmetrical (B) Acetone, because of the double bond (C) Ethylene glycol, because it has the most hydrogen bonding (D) All three substances would have very similar boiling points because their molar masses are similar.
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