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Chapter 1: Problem 17
A 22.0 gram sample of an unknown gas occupies 11.2 liters at standard temperature and pressure. Which of the following could be the identity of the gas? (A) \(\mathrm{CO}_{2}\) (B) \(\mathrm{SO}_{3}\) (C) \(\mathrm{O}_{2}\) (D) He
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150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . What are the concentrations of \(\mathrm{Sr}^{2+}\) and \(\mathrm{F}^{-}\) in the beaker? (A) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (B) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\) (C) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (D) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} \mathrm{M}\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\)
Questions 54-56 refer to the following. GRAPH CAN'T COPY Between propane and ethene, which will likely have the higher boiling point and why? (A) Propane, because it has a greater molar mass (B) Propane, because it has a more polarizable electron cloud (C) Ethene, because of the double bond (D) Ethene, because it is smaller in size
The value of \(K_{a}\) for \(\mathrm{HSO}_{4}^{-}\) is \(1 \times 10^{-2} .\) What is the value of \(K_{b}\) for \(\mathrm{SO}_{4}^{2-} ?\) (A) \(K_{b}=1 \times 10^{-12}\) (B) \(K_{b}=1 \times 10^{-8}\) (C) \(K_{b}=1 \times 10^{-2}\) (D) \(K_{b}=1 \times 10^{2}\)
$$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The reaction above came to equilibrium at a temperature of \(100^{\circ} \mathrm{C} .\) At equilibrium the partial pressure due to NOBr was 4 atmospheres, the partial pressure due to NO was 4 atmospheres, and the partial pressure due to \(\mathrm{Br}_{2}\) was 2 atmospheres. What is the equilibrium constant, \(K_{p},\) for this reaction at \(100^{\circ} \mathrm{C}\) ? (A) \(\frac{1}{4}\) (B) \(\frac{1}{2}\) (C) 1 (D) 2
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 \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ If, instead of copper, a nickel bar were to be used, could nickel be plated onto the zinc penny effectively? Why or why not? (A) Yes, nickel’s SRP is greater than that of zinc, which is all that is required for nickel to be reduced at the cathode (B) Yes, nickel is able to take electrons from the \(\mathrm{H}^{+}\) ions in solution, allowing it to be reduced (C) No, nickel's SRP is lower than that of \(\mathrm{H}^{+}\) ions, which means the only product being produced at the cathode would be hydrogen gas (D) No, nickel's SRP is negative, meaning it cannot be reduced in an electrolytic cell
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