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Chapter 1: Problem 21

How many moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) must be added to 500 milliliters of water to create a solution that has a 2 -molar concentration of the Na' ion? (Assume the volume of the solution does not change). (A) 0.5 \(\mathrm{mol}\) (B) 1 \(\mathrm{mol}\) (C) 2 \(\mathrm{mol}\) (D) 5 \(\mathrm{mol}\)

Short Answer

Expert verified
The number of moles of \(Na_2SO_4\) that should be added is 0.5 moles. So the correct answer is (A) 0.5 moles.

Step by step solution

01

Convert volume to liters

Before proceeding, convert the volume of the solution from milliliters to liters. 1 L = 1000 mL, so 500 mL = 0.5 L.
02

Determine molar concentration

We want a 2 -molar concentration of \(Na^+\) ions. However, each \(Na_2SO_4\) molecule contains two \(Na^+\) ions, so the molar concentration we need for \(Na_2SO_4\) is actually 2 mol of \(Na^+\) per 1 mol of \(Na_2SO_4\). Thus, we need a 1-molar solution of \(Na_2SO_4\).
03

Calculate the moles of \(Na_2SO_4\)

We use the formula for molarity, which is \(M = n / V\), where \(M\) is Molarity, \(n\) is number of moles and \(V\) is volume in liters. Solving for \(n\) gives us \(n = M*V\). Substituting our values into the equation gives us \(n = 1*0.5 = 0.5\) moles.

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Most popular questions from this chapter

A sample of a hydrate of \(\mathrm{CuSO}_{4}\) with a mass of 250 grams was heated until all the water was removed. The sample was then weighed and found to have a mass of 160 grams. What is the formula for the hydrate? (A) \(\mathrm{CuSO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\) (B) \(\mathrm{CuSO}_{4} \cdot \mathrm{TH}_{2} \mathrm{O}\) (C) \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (D) \(\mathrm{CuSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

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

Use the following information to answer questions 12-15. When heated in a closed container in the presence of a catalyst, potassium chlorate decomposes into potassium chloride and oxygen gas via the following reaction: \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) If 12.25 g of potassium chlorate decomposes, how many grams of oxygen gas will be generated? (A) 1.60 g (B) 3.20 g (C) 4.80 g (D) 18.37 g

The following reaction is found to be at equilibrium at 25°C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) The value for \(K_{\mathrm{c}}\) at \(25^{\circ} \mathrm{C}\) is \(8.1 .\) What must happen in order for the reaction to reach equilibrium if the initial concentrations of all three species was 2.0 \(M\) ? (A) The rate of the forward reactions would increase, and \(\left[\mathrm{SO}_{3}\right]\) would decrease. (B) The rate of the reverse reaction would increase, and \(\left[\mathrm{SO}_{2}\right]\) would decrease. (C) Both the rate of the forward and reverse reactions would increase, and the value for the equilibrium constant would also increase. (D) No change would occur in either the rate of reaction or the concentrations of any of the species.

Which gas has the strongest IMFs? (A) He (B) Ne (C) NO (D) All gases have identical IMFs.
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