Gases
Thermodynamics
Kinetics
Stoichiometry
Atomic Structure
100
An ideal gas contained in 5.0 liter chamber at a temperature of 37ºC. If the gas exerts a pressure of 2.0 atm on the walls of the chamber, which of the following expressions is equal to the number of moles of the gas? a. (2.0)(5.0)/(0.082)(37) mol b. (2.0)(0.082)/(5.0)(37) mol c. (2.0)(0.082)/(5.0)(310) mol d. (2.0)(310)/(0.082)(5.0) mol e. (2.0)(5.0)/(0.082)(310) mol
e. Use the ideal gas law and solve for n. n = PV/RT Substitute in the values given in the problem
100
Which of the following terms must be negative, in reactions that are spontaneous (feasible)? (A) Delta H (B) Delta S (C) Delta G (D) Temperature (E) Activation Energy\
C. Delta G is negative in spontaneous reactions
100
When a catalyst is added to a system at equilibrium, a decrease occurs in the A. activation energy B. heat of reaction C. potential energy of the reactants D. potential energy of the products
A A catalyst speeds up a chemical reaction by lowering (decreasing) the activation energy.
100
The simplest formula for a hydrocarbon that is 20.0 percent hydrogen by mass is a. CH b. CH2 c. CH3 d. C2H2 e. C2H3
C Again, it is important to know that a hydrocarbon is any compound made of just hydrogen and carbon. Probably the easiest way to work this problem would be to use the answers and calculate the mass ratio for the mass of H to the total mass of compound: CH is 1/13, CH2 is 3/14, CH3 is 3/15, C2H2 is 2/26, and C2H3 is 3/27. The CH3 compound: 3/15 should pop out as ⅕ which is of course 20%
100
Fluorine is the most active nonmetal. What explains this high attraction for electrons? A. Fluorine is the most negative ion of any element. B. All of the fluorine electrons are in the outer shell. C. The fluorine nucleus is larger than any other nonmetal. D. Fluorine has atomic and pass numbers which are odd numbers. E. The fluorine atom has little shielding of its nucleus from its outer electrons.
E. Fluorine has 7 electrons in its outermost shell, one short of the ultimate objective, an octet of electrons. The outer shell of electrons (has 2) of fluorine is not far away from the nucleus, which decreases the shielding of the outer electrons from the nucleus.
200
A gaseous mixture of oxygen and nitrogen is maintained at a constant temperature. Which of the following MUST be true regarding the two gases? a. Their average kinetic energies will be the same. b. Their average molecular speeds will be the same. c. Their partial pressures will be the same. d. Their total masses will be the same e. Their densities will be the same
a. If substances are at the same temperature, they must have the same amount of kinetic energy.
200
To convert L atm to the metric unit of joules, we need to know A. Avogardro's constant and Planck's constant B. the universal gas law constant in units of L atm/molK C. the universal gas law constant in units of J/ molK D. both B and C E. A, B, and C
D. Both forms of the universal gas constant are needed. The ratio of the two forms of this constant has units of L atm/J is used as the factor label for the conversion.
200
Which statement explains why the speed of some chemical reactions is increased when the surface area of the reactant is increased? A. This change increases the density of the reactant particles. B. This change increases the concentration of the reactant. C. This change exposes more reactant particles to a possible collision. D. This change alters the electrical conductivity of the reactant particles.
C Reaction rate is affected by nature and concentration of reactants, temperature, surface area and a catalyst. Increasing surface area exposes more particles to contact with reactants, increasing the number of particle collisions. Imagine a cube of chocolate 3 feet by 4 feet and 2 inches thick. Take the same block and make it into a 24 candy bars. A class of chemistry students could eat the bars faster than the single chocolate block.
200
How many grams of H2O will be formed when 32.0 g H2 is allowed to react with 16.0 g O2 according to 2 H2 + O2 → 2 H2O a. 9.00 g b. 16.0 g c. 18.0 g d. 32.0 g e. 36.0 g
C Again, this is a limiting reactant problem, 32 g of H2 is 16 moles and is far in excess of the 0.5 moles of O2 available. Thus the O2 limits the reaction. Set up the dimensional analysis and look for easy factors. 16gO2 x (1 mol/32g) x (2 mol H2O/1 mol O2) x (18g/1mol)=18g H2O
200
Which does not make a good electron donor in the formation of a coordinate covalent bond? A. Cl- B. H20 C. PF3 D. CH4 E. NH3
D. Methane (CH4) is not a good electron donor in the formation of a coordinate covalent bond because it doesn't have a lone pair of electrons to donate to the bond.
300
A sample of 18.0 g of aluminum metal is added to excess hydrochloric acid. The volume of hydrogen gas produced at 0.0ºC and 1 atm pressure is approximately a. 67 L b. 45 L c. 22 L d. 11 L e. 7 L
c. Write a balanced equation: 2Al + 6HCl → 2AlCl3 + 3H2 Convert the 18 g of Al to moles: 18g/27(g/mol)= 2/3mol 2/3mol × 3H2/2Al× 22.4L/1mol =~ 22L
300
Which of the following describes a system that CANNOT be spontaneous? A. delta H is positive, and delta S is negative B. delta H is positive, and delta S is positive C. delta H is negative, and delta S is negative D. delta H is negative and delta S is positive. E. delta H is 0, and delta is positive.
A. Delta G must be positive. So, delta H must be positive and delta S must be negative.
300
For a first order reaction of half-life 75 min, what is the rate constant in min-1? A. (0.693)/75 B. (0.693)/1.25 C. (0.693)(75) D. 75/(0.693) E. 0.693
A The equation for the half-life of a first order reaction is t1/2=0.693/k. Substituting, 75 min=0.693/k or k=(0.693)/75 min-1
300
What is the total mass of products formed when 16 grams of CH4 is burned with excess oxygen? a. 32 g b. 36 g c. 44 g d. 62 g e. 80 g
E Balance: CH4 + 2O2 → CO2 + 2H2O Then do some stoichiometry using “easy math” 16 g of methane (MM = 16) is 1 mole and 1 mole of methane will produce 1 mole of CO2 = 44 g, and 2 moles of H2O which is 36 g for a total of 80 g
300
The boron atom in the boron tribromide has sp2 hybridization. What are the angles between the bromine atoms in BBr3? A. All angles are 90 degrees B. Two angles are 90 degrees and the other is 120 C. All angles are 120 D. Two angles are 120 and the other is 180 E. All angles are 180
C. The bromine atoms will be evenly spaced around the central boron atom. The three bromine atoms form a triangle with the boron atom at the center. All four atoms are in a single plane. (study the chart!)
400
A sealed flask at 20ºC contains 1 molecule of carbon dioxide, CO2 for every 3 atoms of helium, He. If the total pressure is 800 mmHg, the partial pressure of helium is a. 200 mmHg b. 300 mmHg c. 400 mmHg d. 600 mmHg e. 800 mmHg
d. The helium must be ¾ of the total pressure
400
In expanding from 3.00 to 6.00 liters at a constant pressure of 2.00 atmospheres, a gas absorbs 00.0 calories (24.14 calories=1 liter atm). The change in energy, delta E, for the gas is A. -600. calories B. -100. calories C. -44.8 calories D. 44.8 calories E. 100. calories
C The first law of thermodynamics states that delta E=q+w. Since w=-P x delta V, the equation can be stated as delta E=delta H-PxdeltaV delta E= 100.0 calories-(2.00 atmospheres x 3 liters24.14cal/L atm)= -44.8 calories
400
For irreversible chemical reactions, the rate will be affected by changes in all these factors except: A. temperature B. concentration of reactants C. presence of catalyst D. concentration of products E. surface area of solid reactant
D The frequency of collisions of reactant molecules largely determines reaction rate. Any factor that changes the frequency of collisions will affect the rate of reaction because collisions of product molecules do not aid the forward reaction. Changing temperature, reactant concentrations, and the surface are of a solid reactant, all change the frequency of collisions. A catalyst increases the rate of reaction by changing the reaction mechanism, eliminating the slowest step, and lowering the activation energy of the overall process.
400
Balance the following equation with the SMALLEST WHOLE NUMBER COEFFICIENTS possible. Select the number that is the sum of the coefficients in the balanced equation: ___KClO3 → ___KCl + ___O2 a. 3 b. 5 c. 6 d. 7 e. 8
D Balance: 2KClO3 → 2KCl + 3O2
400
What is the wavelength of light that has a frequency of 4.00x10^-14 s^-1 (The speed of light is 3.00x10^8m s^-1) A. 7.5 nm B. 1333 nm C. 750. nm D. 1.33 cm^-1 E. 1.2x10^23 m
C
500
Consider the reaction below, in which 6 atm of ammonium nitrite is added to an evacuated flask with a catalyst and then heated. NH4NO2(g) → N2(g) + 2 H2O(g) At equilibrium the total pressure is 14 atm. Calculate the partial pressure of the water vapor at equilibrium a. 2.0 atm b. 4.0 atm c. 6.0 atm d. 8.0 atm e. 10.0 atm
d. Use the ICE chart to solve this problem. The volume is constant, and we can assume that at equilibrium (the end) that the temperature will return to the starting temperature, thus T is constant, which allows us to substitute with pressure values. NH4NO2 → N2 + 2 H2O I 6 0 0 C -x + x + 2x E 6−x + x + 2x We know that the pressure at the end must add up to 14. Thus (6 − x) + x + x = 14 and solve for x = 4 but the question asks for the pressure of water vapor, which is 2x, so 8 atm
500
A gas which initially occupies a volume of 6.00 liters at 4.00 atm is allowed to expand to a volume of 14.00 liters at a pressure of 1.00 atm. Calculate the valoue of work, w, done by the gas surroundings A. -8.00 L atm B. -7.00 L atm C. 6.00 L atm D. 7.00 L atm E. 8.00 L atm
A w=-PdeltaV=-(1.00 atm)(8.00 liters)=-8.00 L atm Because the gas was expanding, w is negative (work was being done by the system).
500
The rate law of a certain reaction is rate=k[X][Y]. The units of k, with time measured in seconds, is: A. s-1 B. M-1 s-2 C. M-2 s-1 D. M-1 E. M-1 s-1
E The sum of the exponents of the rate law is 2, so this is a second-order reaction. Second-order reactions are typified by one concentration unit and one time unit for a total of two units.
500
11.2 g of metal carbonate, containing an unknown metal, M, were heated to give the metal oxide and 4.4 g CO2. MCO3(s) + heat → MO(s) + CO2(g) What is the identity of the metal M? a. Mg b. Pb c. Ca d. Ba e. Cr
E This problem requires a bit more clever thought and understanding. Work the problem “backwards” starting with the information that the carbon dioxide product tells you. Since 44g CO2 x (1 mol/44g CO2)=0.1 mol CO2, the balanced equation tells us that 0.1 mol of CO2 must come from 0.1 mole of MCO3. Further, we know that 0.1 mole of MCO3 contains 0.1 mole of M and also 0.1 mole of CO3. Since CO3 has a mass of 60 g/mol, we can calculate that the 0.1 mole of the CO3 has a mass of 6 g and the remainder of the compound (whose mass we were told was 11.2 g), the M part, must weigh 5.2 g. But remember there is 0.1 mole of M in this compound, thus (5.2g/0.1mol)=(52g/1mol) and so you are looking for an element that has a molar mass of 52, which is Cr.
500
Which element has an electronic configuration that does NOT follow the Aufbau principle? A. Fe B. Mg C. Al D. Ag E. Ni
D
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