This equation has two polyatomic ions. If you recognize that right away, the question is not as difficult. Treat the polyatomic ions as one species. The balanced equation is
The sum of the coefficients is then 1 + 3 + 2 + 3 = 9.
Here you are basically given the correct chemical formula for the product. The balanced reaction is
The coefficient of the indium (III) bromide is 2.
The balanced equation is
The sum of the coefficients is thus 1 + 3 + 2 + 3 = 9.
You must know your solubility rules to get the correct answer. Aluminum sulfate is soluble, as is sodium hydroxide. The ions participating are Al3+, , Na+, and OH-. It is a double displacement, and the key is knowing that sodium sulfate is soluble, while aluminum hydroxide is written molecularly since it is a weak base. The balanced net ionic equation is
This is also a double displacement reaction. The balanced molecular equation is
6. T, T
(Fill in CE.) The first statement is true: the conjugate base of a weak acid is a strong base. Remember that conjugate acid-base pairs are acids and bases that differ only in the presence or absence of a proton. The second statement is also true: a strong Brønsted-Lowry acid will not be a strong Brønsted-Lowry base. It is an explanation for the first statement, so you would fill in the CE oval.
This question asks you to find the compound with the greatest [H3O+], which means it’s asking you to find the compound with the lowest pH or determine which one is the most acidic. As you look down the list, you can see only one compound that is on our list of the six strong acids, and that’s choice B, HNO3.
If the temperature is 25ºC, then Kw = 1.010-14, so A is true. If the hydroxide concentration is equal to 1.010-6 M, then the pOH = 6, so E is true. (You know that 1.010-6 is the same as plain old 10-6. The log of 10-6 is -6. Simply use the exponent when a number, any number, is written as 10power, so the “negative” of the log is equal to -(-6) or simply 6.) Therefore the pH is equal to 14 - 6, which equals 8. If the pH = 8, then the solution is basic, so C is true. Furthermore, when the pH = 8, the [H+] = 1.010-8, so D is true. Only B is not true.
Recall that aqueous solutions of oxides and hydroxides of 1A and 2A metals form strong bases. Now review what you learned about strong bases: the strong bases are hydroxides (—OH), and oxides of 1A and 2A metals (except Mg and Be), H-, and CH3-.
Perchloric acid is the strongest acid in this list. Remember our discussion of how the increased number of oxygens makes the proton more likely to dissociate. Here’s that list of strong acids again: the hydrohalic acids (HCl, HBr, HI), nitric acid (HNO3), sulfuric acid (H2SO4), and perchloric acid (HClO4).
In this problem, it is probably easiest to determine the charge on each nitrogen first. The oxygen in each case is -2. In choice A, NO, the nitrogen has a charge of +2. In choice B, N2O, the charge on nitrogen is +1. In choice C, NO2, and in choice D, N2O4, the nitrogen has a charge of +4. In choice E, , the nitrogen has a charge of +5.
12. T, F
(Do not fill in CE.) Statement I is a true statement: voltaic cells harness the energy released during redox reactions, and this energy is used to perform work. However, statement II is false. The flow of electrons from anode to cathode takes place through a wire, not the salt bridge. The salt bridge is present in order to retain electrical neutrality in the cells. Since the second statement is not an explanation or reason for the first statement, you would not fill in the CE oval.
13. T, T
(Do not fill in CE.) Look at the statements one at a time. The first statement is true— electrolytic cells require the input of energy. These cells are used to separate ores, to electroplate, and as car batteries. The second statement is also true: whereas voltaic cells have two containers, one at the anode and one at the cathode, electrolytic cells have just one (although they still have an anode and a cathode). Since the second statement is not an explanation for the first, you would not fill in the CE oval.
The anode and cathode are not labeled in the diagram, so you must first remember that electrons travel from the anode to the cathode—according to the arrow showing electron movement in the diagram, this means that the chamber on the left must be the anode. This narrows the answer choices down to A or D since these are the only two ions in the anode half-cell. Now, since oxidation occurs at the anode, the metal will lose electrons, and you can determine that choice A shows the appropriate half-reaction.
The first step to solving a problem like this is to assign charges to each element and then decide exactly what’s happening before you look at the answer choices. On the reactants side, lithium and oxygen both are assigned charges of zero. In lithium oxide, Li has a +1 charge and O has a -2 charge. Li has been oxidized: it lost 2e- and is the reducing agent. O has been reduced: it gained 2e- and is the oxidizing agent. The only statement that matches is choice C.