ari gchem test 5
The step that determines the overall speed of the reaction is called the A. rate-determining step. B. elementary step. C. activated complex step. D. intermediate step. E. catalytic step.
A In mechanisms, the entire reaction cannot proceed faster than the slowest step. This step is called the rate-determining step. Intermediates are species that are found during the mechanism, but are canceled out in the balanced equation. The activated complex is the species with the highest amount of energy, found at the tip of the activation energy peak.
A Lewis acid can be described as a species that A. can accept an electron pair. B. can donate an electron pair. C. can accept a proton. D. can donate a proton. E. increases the concentration of [H+].
A The Lewis definition of acids revolves around electrons pairs. Those species that can accept an electron pair (ex. H+) are deemed acids, and those species that can donate an electron pair (ex. OH-) are bases. Answer choices [C] and [D] are Bronsted-Lowry definitions of acids and bases and revolve around protons. Molecules that can accept a proton are deemed bases, and molecules that can donate a proton are deemed acids.
Which of the following can act as both an oxidizing agent and a reducing agent? A. NO3-1 B. Co C. F-1 D. MnO4-1 E. H2O2
An oxidizing agent becomes reduced. A reducing agent becomes oxidized. This is just one of those weird things about chemistry, you have to remember that things become backwards in electrochemistry if the word agent is placed after it. To be able to be oxidized, the species cannot be at its maximum oxidation state. Similarly, to be reduced, the species cannot be at its minimum oxidation state. H2O2 can act as either an oxidizing agent or a reducing agent. This is because the oxygen in H2O2 is at a -1 oxidation state. The oxygen can become further reduced into a -2 oxidation state, such as in H2O, or it can become oxidized to a 0 oxidation state in O2. All the other species are at either their minimum or maximum oxidation state. NO3-1 N = +5 Nitrogen is at its maximum oxidation state and cannot be oxidized anymore and cannot act as a reducing agent. However, it can still act as an oxidizing agent. Co Co = 0 Cobalt cannot be reduced because it is at its minimum oxidation state. Co-1 does not normally exist. However, it can act as a reducing agent and become oxidized into Co2+. F-1 F = -1 Fluorine is at its minimum oxidation state and cannot be reduced anymore. MnO4-1 Mn = +7 Manganese is at its maximum oxidation state and cannot be oxidized any further.
Parallax error may occur when: A. an experiment does not account for changes in altitude. B. a measurement is not properly converted between the metric and customary systems. C. a meniscus in a buret is not observed directly straight on. D. an object under water is observed at an angle from above the surface. E. the concentration of an acid to be used in a titration is not first checked against a standard.
C All options are common laboratory errors, but parallax error may occur when a pointer on a meter or surface of a liquid in a buret or pipette is not observed from directly straight on. If they are observed from above or below then the liquid surface or pointer will not be seen to line up with the scale. It will result in a positive mistake if the meniscus is viewed from below, and a negative mistake if the meniscus is viewed from above.
How would the bond energy and bond length of a carbon-carbon bond in C2H6 compare to C2H4? A. The bond energy for C2H4 would be greater, but the bond length would be shorter. B. The bond energy and bond length for C2H4 would be greater than C2H6. C. The bond energy for C2H4 would be smaller, but the bond length would be greater. D. The bond energy and bond length for C2H4 would be smaller than C2H6. E. The bond energy and bond length for C2H4 would be the same in C2H6.
C2H6 has a single bond between the two carbons, while C2H4 has a double bond between the two carbons. Double bonds have more energy between them and are shorter than normal single bonds. Triple bonds are even shorter and their bond energies are even higher. E= 1/lamda
Electron affinity measures the: A. tendency to attract electrons towards the atom and form a bond. B. chemical reactivity of the valence electrons of the atom. C. amount of energy absorbed when an electron is added to an atom. D. ease with which an atom gains an electron. E. average distance between the electrons and nucleus.
D Electron affinity is the measured change in energy (how much energy is released) when an atom gains an electron. This can be restated as the ease with which a neutral atom gains an electron. The tendency to attract electrons towards the atom and form bonds is known as electronegativity, and this value is estimated based on ionization energies and electron affinities. The amount of energy it takes to remove an electron from an atom is known as the ionization energy. Electron affinity generally increases as we move across the periods from left to right. Depending on the textbook you read, it may or may not say that moving down the groups will decrease electron affinity. Since there are conflicting statements, you won't be tested on the periodic trends of electron affinities down a group on the DAT. Generally though, electron affinity trends follow electronegativity (fluorine is the most electronegative element).
Why should one always slowly add acid to a beaker of water rather than water to a beaker of acid? A. To ensure the acid does not react with impurities in beaker B. To prevent the water from sinking beneath the acid and remain unmixed C. To maximize the ionization of the acid being diluted D. To ensure there is enough water to absorb the heat released E. To solvate the conjugate base to prevent a reaction
D This is an important rule in chemistry safety. You always add acid to water, never the other way around. If you add water to acid, so much heat will be released that the water cannot contain it and the reaction may violently boil and splash. Adding acid to water ensures that the large volume of water will absorb the heat produced and prevents splashing.
decreasing pressure while temperature kept constant
If we decrease pressure while keeping temperature constant, three possibilities can happen. We can go from: Solid to gas: Sublimation Liquid to solid: Crystallization Liquid to gas: Vaporization If we decrease the pressure at a constant temperature, there is no way we can go from a solid to a liquid (because the liquid will always come first), so therefore fusion is not possible for the phase diagram of water, because water has a negative solid-liquid line slope.
ZnCl2 is electrolyzed with a current of 7.00 amperes for 3 minutes. Which of the following is equal to the maximum amount of Zn(s) that is produced in grams? (1 faraday = 96,500 coulombs).
In quantitative electrolytic cell calculations, remember the following formulas. grams produced = A. S. MW /n.F moles produced = A. S / n. F (amps) = amperes, or the current (seconds) = the amount of time the reaction took place in seconds (MW) = molecular weight of the product (n) = moles of electrons in the reaction (F) = Faraday constant In this question, we have to convert the 3 minutes into 180 seconds. Next, we use the atomic weight of Zn(s) because it is our product. Lastly, in this reaction Zn goes from a +2 to a 0 oxidation state, indicating that 2 electrons were moved in the reaction, therefore we will use 2 moles of electrons in the equation. Just plug in our numbers and we're done!
A 1.0 L sample of an aqueous solution contains 0.20 mol of NaCl and 0.10 mol of MgCl2. What is the minimum amount of moles of AgNO3 that must be added to the solution to precipitate all of the Cl- ions as AgCl(s)?
In this problem, we assume that all of the Cl- dissociates in NaCl and MgCl2. Since NaCl only has one chlorine atom, when it dissociates it will produce 0.20 mol of Cl-. Since MgCl2 has 2 chlorines, it will produce 0.20 mol of Cl- (0.10 mol x 2 mol of Cl-). In total, we will have 0.40 mol of Cl- in the solution. Since there is only one silver atom in AgNO3, we will need 0.40 mol of Ag to react with all 0.40 mol of Cl-.
Water molecules in an aqueous solution will have the strongest interactions with ions with which of the following characteristics?
Large charge and small size. Let's first examine charge. A polar water molecule will be more likely to interact with an ion with more charge. This is a stretch of the analogy "like dissolves like". The high polarity molecule will enjoy the company of a highly charged ion, and this will result in a greater interaction between the two. For size, think about the charged ion and water molecule as two tiny magnets. When these magnets are spaced far apart, they barely have an effect on each other. As you bring the magnets closer together, you begin to feel their force of interaction. The same principle applies to a water molecule and a charged ion in an aqueous solution. We need to minimize the distance between the water molecule and charged ion, and we can do that by choosing an ion with a smaller size. F= K(q1q₂)/d² Where F is the force of interaction, q1 and q2 are charges and d is the distance. Maximizing the charge would increase the force. Minimizing the distance between the two particles would maximize force as well.
An exothermic reaction at equilibrium will proceed at a: A. slower rate while its Kc increases at higher temperatures. B. slower rate while its Kc decreases at higher temperatures. C. faster rate while its Kc increases at higher temperatures. D. faster rate while its Kc decreases at higher temperatures.
Let's solve this by making up an example exothermic reaction (shown below). The general concept of an exothermic reaction is that heat is released, or heat is a product: A + B → C + D + heat Kc = [C] [D]/ [A] [B] From Le Chatlier's principle, increasing the temperature would increase the heat, so therefore the reaction would shift to the left (towards the reactants, [A] and [B]). If the concentrations of [A] and [B] increase, then the overall Kc of the reaction would decrease at higher temperatures, eliminating answer choices A and C. Lastly, increasing the temperature will increase the rates of both the forward and the reverse reactions, so the reaction will proceed at a faster rate. Thus we arrive at answer choice D.
If the Ka of acetic acid is 1.8 × 10⁻⁵, what is the pKb of acetate
The "p" in pKa and pH stands for "-log". Here are some useful equations to know for this problem. (Ka)(Kb) = Kw = 10⁻¹⁴ pKa + pKb = 14 pKb = 14 - pKa pKb = 14 - (-log(Ka)) pKb = 14 - (-log(1.8 × 10-5)) The two negatives cancel to form our final equation. pKb = 14 + log (1.8 × 10-5)
Which of the following is true for the energy diagram of a reversible exothermic reaction? A. The forward reaction proceeds slower than the reverse reaction. B. The enthalpy of the products is greater than the enthalpy of the reactants. C. The reverse reaction has a higher activation energy than the forward reaction. D. The reverse reaction has a lower activation energy than the forward reaction. E. The forward reaction has a higher activation energy than the reverse reaction.
The energy diagram for an exothermic reaction is shown above. As we can see, the activation energy for the reverse reaction is larger than the forward reaction. A larger activation energy means the reaction would proceed slower, so therefore choice [A] is incorrect (the forward reaction is faster). The enthalpy of the reactants is larger than the products, which is why exothermic reactions have a negative ΔH, so choice [B] is incorrect. Choices [D] and [E] can be eliminated as well by looking at the energy diagram for an exothermic reaction.
Two separate mixtures are created. One using water and KCl and the other using water and CaCl2. Assuming the molality is equal in both mixtures, which of the following is true?
The equation to calculate the change in boiling point is: ∆Tb = m × Kb × i Since the molality is the same for both mixtures, this can be disregarded. Both compounds are ionic and since the solvent is water, the Kb is also the same for both equations, which for water is 0.512. i factor: The difference between the two compounds is the number of the ions that each will produce with KCl producing 2 ions and CaCl2 producing 3 ions. Therefore, the i or the Van't Hoff factor is the only input that varies. Thus, CaCl2 will have the higher boiling point and the lower freezing point.
10 kJ of heat was applied to a 150 g sample of copper at 50°C. The specific heat of copper is 0.39 J·g-1·°C-1. What is the final temperature of the copper after the heat is applied in °C?
The equation we have to use here is: q = mCpΔT q = heat m = mass Cp = specific heat ΔT = change in temperature It is important to make sure all the units agree with the constant, the specific heat. In this case, we must convert the 10 kJ to 10,000 J. We don't convert the copper into moles because the specific heat's unit for mass is grams. We can break the ΔT down into (Tf - Ti). Ti = 50°C, as stated in the problem. Q= mcΔT = Q/mc = T₂ - T₁ --> (Q/mc) +50
2 A(g) + B(s) ⇌ 3 C(g) The above reaction is at equilibrium at STP and the molar concentrations found are [A] = 2.0 M, B = 2.0 mol, and [C] = 4.0 M. What is the equilibrium constant for this reaction at STP?
The equilibrium constant of this reaction can be solved by: Kp = [C]³/[A]² = 64/4 = 16 Note that B is not included because B is a solid. Solids are not included in equilibrium constant equations. Only gases and aqueous states of matter are used to solve equilibrium constant equations. Next, we just plug in the values of the concentrations and simplify until we arrive at an answer.
The ΔH for which of the following reactions is the standard enthalpy of formation (ΔH°f) for sodium chloride? A. Na+ (aq) + Cl- (aq) --> NaCl (aq) B. Na (s) + 1/2 Cl2 (g) --> NaCl (s) C. Na (g) + 1/2 Cl2 (g) --> NaCl (g) D. 2Na (s) + Cl2 (g) --> 2NaCl (s) E. 2 Na (g) + Cl2 (g) --> 2NaCl (s)
The ΔH°f for a compound is the enthalpy chance for the formation of one mole of the compound from reactants in their most stable forms at standard conditions (298K and 1 atm). Sodium is solid and chlorine is a gas at standard conditions. To form one mole of the compound we must use 1 mole of Na (s) and 1/2 a mole of Cl2 (g).
100 mL of 0.50 M CaCl2(aq) is mixed with 100 mL of 0.25 M H2SO4(aq). What is the final concentration of [Ca2+] in this solution? CaSO4 is insoluble in solution.
This is a very tricky problem. First we have to examine the reaction that is occurring when we mix these two solutions: Ca2+ + SO4-2 → CaSO4 Since Ca2+ and SO42- are in a 1:1 ratio in the above balanced equation, then we have to see how much calcium will react with the sulfate to produce insoluble CaSO4. (100 mL)(0.25 M) = 0.025 mol SO42- (100 mL)(0.50 M) = 0.050 mol Ca2+ 0.050 mol Ca2+ - 0.025 mol SO42- = 0.025 mol Ca2+ remaining free in the solution. Now that we have the mols of Ca2+ remaining, we divide by the final volume (100 mL + 100 mL) = 200 mL to find the concentration: M= mol of calcium /Total volume --> 0.025 mol [Ca2+] / 0.200 L = 0.125 M [Ca2+]
Which of the following would be strongly attracted to magnetic fields? A. He B. H2 C. Zn D. O2 E. N2
This question is asking about the properties of diamagnetic and paramagnetic atoms and molecules. A paramagnetic atom or molecule has at least one unpaired electron which will result in a tiny magnetic field that can interact with a larger magnetic field. If all of the electrons are paired within their orbitals, then the molecule is said to be diamagnetic (di- for two) and will not interact with magnetic fields (some diamagnetic molecules may repel magnetic fields). To solve this question, we have to draw out the atomic or molecular orbital diagrams of each answer and determine which one has at least one unpaired electron (or is paramagnetic). MO diagram of He (diamagnetic): http://www.youtube.com/watch?v=c1wzgky77hs MO diagram of H2 (diamagnetic): http://youtu.be/DyL5pQc0Hsw?t=5m30s Zinc has 10 electrons in its 3d orbital and 2 electrons in its 4s orbital, so every electron is paired up. It is thus diagmagnetic. MO diagram of O2 (paramagnetic): http://youtu.be/dVWtvG9ztMQ?t=3m11s MO diagram of N2 (diamagnetic): http://youtu.be/1Bgjka4kLYo?t=17m40s A popular demonstration of this principle is pouring liquid nitrogen through a magnetic field and watching nothing spectacular happen, and then pouring liquid oxygen through a magnetic field and watching it stick to the magnets. http://youtu.be/KcGEev8qulA?t=40s
Which of the following are the correct quantum numbers for the 13th electron in chlorine's electron configuration? A. 3, 0, -1, 1/2 B. 3, -1, 1, 1/2 C. 3, 1, -1, 1/2 D. 4, 2, -1, 1/2 E. 4, 0, 2, 1/2
To assign quantum numbers the following rules must be applied: n = energy level of the electron l = this identifies the shape of the orbital that the electron occupies. S orbital = 0, p orbital = 1, d orbital = 2, f orbital = 3. ml = this helps to identify the suborbital that is being occupied. This runs from -l to +l. ms = -1/2 if the electron is spinning down; 1/2 if the electron is spinning up The 13th electron of chlorine's electron configuration is in the last suborbital of the p orbital, spinning upwards. Assigning the numbers as described results in the quantum numbers 3, 1, -1, 1/2. Quantum Number Principal Azimuthal Magnetic Spin Symbol n l ml ms Possible values 1,2,3,4... 0,1,2,3... (n-1) +/- (1/2 s=0; p=1; d=2; f=3 -l,...0...+l Characteristic Size Shape Orientation Spin
Which of the following is the ground state electron configuration for Cr2+? A. [Ar] 4s23d2 B. [Ar] 4s13d3 C. [Ar] 4s23d8 D. [Ar] 4s03d4 E. [Ar] 4s13d2
Watch out for exceptions of filling d before s orbitals. First draw the ground state with first filling d orbita then one electron in s then take 2 electrons out; one from higher s orbital and one from d orbital. We have a special case here, the ground state electron configuration for normal Cr is [Ar] 4s13d5. This leads to both the 4s and the 3d subshells being half-filled, which they prefer. Other commonly tested exceptions in electron configuration in addition to Chromium are: Copper, Silver, Gold, and Molybdenum. When an atom becomes ionized, we remove electrons from the highest level (n value) first. Therefore, Cr will lose 1 electron from the 4s orbital before it loses 1 electron from the 3d orbital. This leaves us with an electron configuration of: [Ar] 4s⁰3d⁴
How much water should be ADDED to 10 mL of 3.00 M HCl(aq) in order to dilute it to a 2.00 M solution of HCl(aq)?
We can use the equation M1V1 = M2V2 to solve this problem. We are given the following values: M1 = 3.00 M HCl(aq) V1 = 10 mL M2 = 2.00 M HCl(aq) We'll use the above equation to solve for V2: (3.00)(10) = (2.00)(V2) V2 = 15 mL The question asks how much water should be ADDED to the original solution, so we subtract by the original amount, V1. 15 - 10 = 5 mL,
ideal gas
like ideal Gal with low Pressure but high temperature. large tits and low penis. No intermolecular interactions and collisions are elastic. (bounce off one another with same KE). The high temperatures contribute to kinetic energy to allow the molecules to overcome intermolecular forces. The low pressure spaces out the molecules far enough that the individual size of the molecules becomes insignificant compared to the entire volume of the gas.