General Chemistry - MCAT Physical

An oxidizing agent is one that oxidizes another element and, becomes reduced in the process. Oxidizing and reducing agents are found in the reactants of the equation, eliminating Al3+ and Cr(s) as answer choices. Cr3+ gains three electrons to become Cr(s), which has an oxidation number of zero. Cr3+ becomes reduced, while Al(s) is oxidized, therefore, Cr3+ is the oxidizing agent. Remember "OIL RIG": Oxidation Is Loss of electrons and Reduction Is Gain of electrons.

Good insulators are usually non-metals, in which the electrons are not able to freely move. Insulators, unlike conductors, do not carry current. This is because charges cannot move freely in an insulating material, making this choice the correct answer.

Sound would travel slowest through an insulating material, as there is less ability to compress and propagate the sound wave. Copper is an example of a good conductor, and is a poor insulator.

Halogens are in the group next to the noble gasses. They have seven valence electrons, and therefore have a high electronegativity. The addition of only a single electron (production of an anion) generates a full valence octet.

Their diameters vary within the group. The diameter can be very small, like fluorine, or large, like iodine. They do not conduct electricity well, as they are non-metals.

Transition elements have multiple oxidation states because of the d-orbitals they possess. This allows them to lose or gain electrons in a variety of ways, often leading to the standard characteristics of metals, such as electrical conductivity. Lanthanides and actinides are less commonly tested, but also have the ability to form multiple oxidation states due to their large and variable orbitals. Ytterbium is one of the lanthanides, and has Yb(II) and Yb(III) oxidation states.

The correct answer is cobalt, since it is the only metal among the answer choices. Metals have all the properties described (malleability, ductility, and conductivity). Sulfur, boron, and silicon do not exhibit these properties to the same extent.

The halogens are the second to last column in the periodic table, meaning that they have an affinity for a single additional electron. Halogens would be most likely to react with alkali metals, which contain only one loosely bound electron in the valence shell. Alkali metals have very low ionization energy, readily losing an electron, while halogens have very high electronegativity, readily gaining an electron. This interaction allows the alkali metals to form ionic bonds with the halogens.

Sodium iodide is given by the formula NaI, and has a molecular weight of 150g/mol. Molarity is found by dividing the moles of solute by liters of solvent.

We find the moles of sodium iodide by using the mass (300g) and molecular weight.

We know our volume is 1L, so now we can solve for the molarity.

Solutions can occur in all three phases of matter. They can also occur between two compounds in a single phase. For example, brass is a solution of two metals: zinc and copper.

The properties described fit well with the noble gases. Alkali and alkaline earth elements are solid at room temperature, meaning that they have a high boiling point. Halogens can be gaseous at room temperature, but are very reactive. Noble gases have low boiling points and rarely act in spontaneous reactions. Their properties are due to their full valence shell, which is the source of their stability. Changes to their electron configuration (such as removing an electron) require large amounts of energy.

The mass number is the total number of nucleons (protons and neutrons) in an atom's nucleus. An atom that undergoes alpha decay will lose a helium nucleus. This decreases its mass number by four. All other forms of radioactive decay listed alter the atomic number of the atom, but not the mass number.

Beta emission involves the conversion of a neutron to a proton and an electron, but only expels the electron.

Electron capture involves the conversion of a proton and an electron into a neutron.

Positron emission involves the conversion of a proton to a neutron and a "positron," or positively charged electron.

Le Chatelier's principle emphasizes three main stressors: changes in temperature, pressure, and concentration of reactants and products. If any of these three stressors are added, the Keq will shift in a way that counters this added stress. Temperature and concentration changes will affect any reaction, while pressure changes will only have a marked effect on reactions involving at least one gaseous compound.

Alkali metals are much less dense than other metals due to their large radii, which results from having a single loosely bound valence electron. Some of the alkali metals have such low densities that they can float on water.

Atomic radius can be determined using the periodic trends. Atomic radius increases to the left of a period and down a group of the periodic table. Electronegativity, in contrast, increases to the right of a period and up a group of the periodic table. Relating the two, we can see that the greater the atomic radius, the weaker its electronegativity because the electrons are farther away from the nucleus and are unable to feel the attractive force of the protons in the nucleus.

Remember that electronegativity increases as you approach the top right corner of the periodic table. Since oxygen is the farthest right and the highest up on the perioidic table out of these choices, we conclude that it has the highest electronegativity.

This question requires use of the simple titration equation M1V1 = M2V2. The key is to identify that sulfuric acid has two equivalents of acidic hydrogens while NaOH has only one hydroxide equivalent. All wrong answer choices result from making this mistake or other calculation errors.

Remember that equilibrium constants are not affected by the concentrations of the reactants and products. Since an acid is being added to the solution, the pH of the solution will be affected. This means that the pOH will be affected as well.

There are a couple options to solve this problem. First, we could simply remember that anything which combines with oxygen in a redox reaction is oxidized, and the associated oxygen is reduced.

Another option is to think about the oxidation numbers of each element (an element whose oxidation number increases is oxidized, and one whose oxidation number decreases is reduced). If we choose this method, recall that free elements (Fe and O2 in this case) have oxidation number 0. Also, oxygen in compounds has oxidation number -2. Oxygen's oxidation number decreases from 0 to -2, meaning that oxygen is reduced. Since oxidation and reduction occur at the same time in different elements of the redox reaction, iron's oxidation number must increase, so iron is oxidized.

This problem requires us to determine the oxidation number of carbon as a reactant and as a product. The oxidation number of carbon as a reactant is –4, because it is attached to four hydrogens each with a charge of +1. The overall charge on the molecule is zero, the carbon must cancel out the charges contributed by hydrogen.

Carbon as a product has an oxidation number of +4, because it is attached to two oxygens, each with an oxidation number of –2. Again, we know that the molecule is neutral, and carbon must balance the charges from oxygen

Since the oxidation number of carbon went from –4 to +4, we conclude that carbon has been oxidized in the reaction. Any atom that loses electrons is oxidized, while any atom to gain electrons is reduced.

The atom that is oxidized will have a higher oxidation number as a reactant than as a product. Since reactant phosphorus (P4) is in elemental form, it has an oxidation number of 0. As a product, phosphorus (PO43-) has an oxidation number of +5.

Total charge = (P) + 4(O) = –3

–3 = (P) + 4(–2) = (P) + (–8)

p = +5

Because it is now more positive, we say that the phosphorus has been oxidized.