General Chemistry - MCAT Chemical and Physical Foundations of Biological Systems

The sum of oxidation numbers of each element in this compound must add to the total charge of -2. Hydrogen always has an oxidation number of +1, and oxygen always has an oxidation number of -2.

So, in total for oxygen we have (4) * (-2) = -8, and for hydrogen we have (4) * (+1) = +4.

Combining the oxygen and hydrogen, (+4) + (-8) = -4, so we need an additional +2 to achieve the total charge of -2. Zn must provide this balancing charge, and have an oxidation number of +2.

To find the molecular weight (mass) of a molecule, simply add up the atomic weights of each atom within the molecule. The units used will be amu (atomic mass units) for molecular weight and g/mol for molar mass.

In order to find the mass percentage of an atom in a molecule, start by finding the total mass of one mole of the molecule. Glucose has 180 grams per mol.

Next, we determine the mass of the carbon atoms in one mole of the molecule. One carbon mole has a mass of 12 grams. Multiplied by the six carbon atoms in glucose gives a mass of 72 grams.

Finally, we divide the mass of carbon by the mass of the molecule.

So, 40% of glucose's mass is made up of carbon.

To find the empirical formula, use the mass percentage of each element to find mole ratios based on a hypothetical sample of .

We see that there is a 1:1 mole ratio for carbon to hydrogen, making the empirical formula .

The next step will be to find molecular formula by dividing the molar mass of the compound by the molar mass of the empirical formula.

Multiply the subscripts of the empirical formula for each element by six to get the molecular formula: .

The molecular formula is the same as the empirical formula if it cannot be reduced by any whole number. Any formula containing a single atom of any given element must be an empirical formula as well. The formula for DEET is . Since this contains a single atom each of oxygen and nitrogen, it cannot be further reduced and must be an empirical formula as well.

The molecular formula for ribose is , which can be reduced by a factor of five. The empirical formula for ribose (and most other carbohydrates) is .

The other two options require us to calculate pass percentages based on the given molecular formulas.

Although we can look at the formulas for chlorophyll and ethyl butyrate to deduce that oxygen makes up a larger percentage of the latter, we can double check mathematically. In order to find which compound contains oxygen in a larger percentage, divide the molar mass of the oxygen in the compound by the molar mass of the entire compound.

For ethyl butyrate:

For chlorophyll:

We see that it is in fact true that there is a larger percentage of oxygen in ethyl buyrate.

Next, find out if hydrogen and carbon make up the largest mass percentage of ribose by using the same method:

This amounts to 46.7% of the molecular mass, meaning that oxygen must account for the remaining 53.3%. Oxygen thus makes up a greater mass percentage of ribose than hydrogen and carbon combined.

Given the percentages by mass of the compound, we can convert the percentages to grams by considering a 100-gram sample of the compound. The next step involves dividing each mass by the element's molar mass.

By dividing each value by the smallest molar quantity, we can determine the ratio of elements in the compound.

As a result, the empirical formula for the compound is .

The empirical formula for carbohydrates is , where is the number of carbon atoms. The question states that molecule A has 6 carbons; therefore, and the molecular formula for molecule A is . The molecular weight (MW) of this compound is calculated using the molecular weight of each atom. The MW of each atom is obtained from the periodic table and is multiplied by the number atoms.

So, the molecular weight of molecule A is . Since it is also a carbohydrate, molecule B will have the same empirical formula as molecule A. Molecule B has half the molecular weight of molecule A; therefore, molecule B must have half the atoms as molecule A. The molecular formula of molecule B is . Note that MW of is .

The MW contribution of oxygen to each compound is calculated as follows.

%MW of oxygen for molecule A =

%MW of oxygen for molecule B =

Therefore, oxygen contributes to 53% of MW in both compounds.

The reaction given is the complete oxidation of glucose. Reactants are glucose and oxygen (from inhalation) and products are carbon dioxide and water, which are released through exhalation and excretion.

The complete balanced reaction is:

When balancing reactions, it is generally easiest to leave oxygen and hydrogen alone until the end. First, balance the carbon. Next, balance the hydrogen since it is only found in one reactant molecule and one product molecule. Finally, balance the oxygen.

Electrolyte solutions are formed when a compound creates ions once in solution. Carbon dioxide will not create ions in solution, so it is not a good electrolyte.

(This is the acid dissociation for acetic acid.)

A saturated solution is a solution that is at equilibrium. The ionization product is equal to the Ksp in a saturated solution, and the rate of the forward reaction is equal to the rate of the reverse reaction. In an unsaturated solution, the solute continues to dissolve until equilibrium is reached. In a supersaturated solution, salt will "crash out," or precipitate, from solution.

An electrolyte is a substance that creates ions when in solution. These compounds conduct electricity very well by donating ions to an aqueous solution. The worst electrolyte will be the compound that creates the smallest amount of ions in solution. Silver chloride is relatively insoluble in solution, meaning that it will make the smallest amount of ions out of the given options.

Van't Hoff factor can also be useful in determining electrolyte effectiveness, but is irrelevant in this particular question.

Boiling point elevation is a colligative property, meaning that it depends on the relative number of solute particles in solution. The answer choice with the largest number of moles of particles will show the greatest boiling point elevation. The equation for boiling point elevation is:

Molality is equal to moles of solute per kilogram of solvent, meaning that it will be proportional to the moles of solute added. Each solute is added to equal amounts of water, allowing us to keep this value constant. Similarly, will be constant for all of the solutions. Overall, boiling point elevation will be proportional to the moles of solute multiplied by the van't Hoff factor.

Using this proportion, we can find the solute that will most impact the boiling point of water.

Since sodium chloride results in the greatest moles of ions in solution, it will yield the greatest boiling point elevation.

Raoult's law can be written as .

When the formation of a solution has a positive enthalpy, it is considered to be endothermic. An endothermic reaction will result in the solution's vapor pressure being higher than predicted by Raoult's law. This is because an endothermic reaction results in weaker intermolecular bonds, which increases the vapor pressure.

In fission a large atom, with low binding energy per nucleon, splits into two approximately equal smaller atoms, which are more stable due to greater binding energy per nucleon. All the other properties mentioned are accurate.

The first two statements are true. Statement III is false becuse radioactive decays tend to produce nuclei with higher binding energy per nucleon than the parent nucleus, since nuclei with higher binding energy are held together more tightly and thus are more stable.

Potassium (K) is orignially in the electron configuration of \[Ar\]4s1. To obtain a +1 charge it loses an electron, resulting in a configuration of \[Ar\].

Bond distance is related to the type of bond between the two atoms. The more bonds between two atoms, the shorter the distance. In other words, a triple bond is shorter than a double bond, and a double bond is shorter than a single bond. Ethyne is the only option that has a triple bond between the carbons, so it has the shortest bond distance.

First note that 15g is 1/8 of the original 120-gram mass, so all but 1/8 of these radium nuclei have decayed. By definition, ½ of the nuclei decay during one half-life. So the number of nuclei remaining after n half-lives is . In this case, three half-lives have elapsed since, . The answer is 3 * 1600 = 4800 years.

Like any system, energy is absorbed to break bonds and released when bonds are formed. Energy is absorbed to get U-235 to an unstable state. It then breaks apart, and new species form. These new species release more energy than was originally absorbed when they form their bonds, thus leading to a net release of energy from the system.