Graham's law of effusion is written as

Defining helium as compound 1 in the equation above, we can substitute

Square both sides:

The molecular masses of the possible compounds are

(A) CO₂ = 44, (B) CH₄ = 16, (C) C₅H₁₂ = 72, (D) C₈H₁₈ = 114, is obviously the sample in this experiment.

Work can be determined from PV data. The equation is

work (w) = -P Δ V.

We can use P₁V₁ = P₂V₂ to calculate the final volume in the problem above:

The vapor pressure of a mixture is found to be

This electron configuration has 20 electrons in the lowest possible energy levels according to the Aufbau ordering. Element 20 is calcium.

To calculate the oxidation number of an element in a polyatomic ion, the charges of the atoms must add up to the charge on the ion.

Molarity is a ratio of units (moles per liter); therefore, the setup of the calculation must start with a ratio, such as another molarity.

Setup:

Expand molarities to mol/L and insert conversion factors:

The amount of matter is equal on both sides of the reaction. None of the other options are supported by the diagram.

An intermediate is defined as a substance that is neither a reactant nor a product. Intermediates are often difficult to detect. Catalysts fit the above description, but they are substances that are added to the reaction mixture and can be isolated afterward.

By convention, liquids are not written as part of the equilibrium law. Liquids always have the same number of molecules per liter, which is a constant, combined into the value of the K. The only nonsolid substances in this reaction are O₂(g) and CO₂(g). In this reaction, CO₂ to the third power appears in the numerator because it is a product, and O₂ to the fifth power is in the denominator.

Reactions with high activation energies that do not proceed at a measurable rate are considered to be under kinetic control-that is, their rate of progress is based on kinetics instead of thermodynamics.

The reaction is 2 HI(g) ↔ H₂(g) + I₂(g), and the equilibrium law is

When HI reacts, 2x moles of HI form x moles of H₂ and x moles of I₂. The equilibrium law is

Because x = [I₂], that is the concentration of I₂ vapor expected. We keep only two significant figures because K_c had only two significant figures.

Hydrogen bonds can form when a compound contains a fluorine, an oxygen, or a nitrogen atom with a hydrogen bonded to it. In this question, this criterion is fulfilled only by CH₃NH₂.

The Gibbs free energy equation is ΔG° = ΔH° - T ΔS°. To determine ΔG° at a temperature other than 298 K we need ΔH° and ΔS° so that the above equation can be solved at a temperature different from 298.

This response is true because of the large distance between gas particles. The other four responses are not necessarily true about any given gas sample.

This pair differs by only one H⁺ between their two formulas.

Alkenes have a double bond. The double bond consists of one sigma and one pi bond connecting a pair of carbon atoms. The simplest alkene, ethane, has one pi bond and five sigma bonds. Geometrically, they are triangular planar around the carbon atoms (sp² hybrid). Because of that geometry, the simplest alkene molecule does not have isomers or optical activity.

The neutron, with no charge, has tremendous penetrating ability. Many can pass through the earth without stopping.

With a half-life of 36 minutes, there will be 120/36 = 3.33 half-lives. Therefore there are more than 3 and less than 4 half-lives. Three half-lives leaves 1/8 of the material, and 4 half-lives leaves 1/16 of the material. 1/8 of the initial 10 grams is a little more than 1 gram, while 1/16 of 10 grams is a little more than 1/2 gram. The only answer that falls in that range is 1.00 gram.

Use the integrated rate equation: ln(g₀/g_t) = kt, also ln(2) = kt_1/2.

Therefore k = (0.693)/36 min = 0.0193 min^-1.

Then

The most important factor in entropy change is whether or not there is a change in volume as denoted by the change in the number of gas molecules in the chemical equation (Δn_g). For the reactions given, the Δn_g is: (A) 0, (B) -2, (C) 0, (D) -1. Reaction (B) has the greatest decrease in the moles of gas and should have the largest decrease in entropy.

From the integrated rate law we can derive that ln(2) = kt_1/2, where t_1/2 is the half-life.

Converting 34 minutes to seconds yields 34 × 60 = 2040 seconds. The rate constant is calculated as

We round the final answer to two significant figures because 34 minutes has only two significant figures.

Molarity is the moles of solute in each liter of solution; molarity = mol solute/L solution

Hydrogen bonding is the strongest intermolecular attractive force and is listed first. Induced dipoles are the weakest attractive forces and last for a short period of time.

ONLY temperature changes will result in changes in the numerical value of the equilibrium constant.

The symbol represents an isotope of chromium. Since there is no charge, it is not an ion that may have gained or lost electrons. Therefore the numbers of electrons and protons are equal (24 of each), which eliminates response (C). Response (B) does not have 24 protons, and response (D) is the remaining one where the protons and neutrons add to 52.

Mathematically, we start with = mol C₃H₈ = 6.2 g C₃H₈. We then use the conversion factor 1 mol C₃H₈ = 44 g C₃H₈ to convert g to mol.

The overall rate law is always equal to the rate law of the slowest elementary step. The rate law of any elementary step can be determined using the coefficients of the reactants in that step.

To determine the molar mass, we need to calculate the moles of the acid and divide it into the mass of the acid used in this experiment. Since each acid molecule has one proton (it is monoprotic), determining the moles of protons gives us the moles of the acid.

moles of protons = (0.0263 L) (0.122 M KOH) = 3.21 × 10^-3 mol

molar mass = (0.682 gram)/(3.21 × 10^-3 mol) = 212 g mol^-1