dat booster gen chem test 2
Iodine-131 has a half-life of 8 days. How long will it take for a 36 g sample to decay to 4.5 g? A. 8 days B. 16 days C. 24 days D. 32 days E. 40 days
24 days In radioactive decay, the half-life of a radioisotope is the time required for the amount of the isotope to decrease by half. With each subsequent half-life interval that passes, the amount of the isotope that remains is again decreased by half. The fraction of the initial amount remaining becomes exponentially smaller and eventually approaches zero. Since we are starting out with 36 grams, we will see how many times we can half it until we reach 4.5 g: 36 → 18 → 9 → 4.5 We halved it 3 times. In other words, it took three half-lives to reach 4.5 g. Since one half-life takes 8 days, 3 half-lives take 24 days.
Which of the following interactions is the weakest? A. Hydrogen bonds B. Covalent bonds C. Hydrophobic interactions D. Ionic bonds E. Van der Waals forces
Van der Waals forces (Intermolecular forces are the forces between two neighboring molecules. They are weak relative to intramolecular forces - the forces which hold a molecule together, such as covalent bonds and ionic bonds. Therefore, in terms of bond strength, intermolecular forces are weaker than both covalent and ionic bonds Within intermolecular forces, individual interactions are ranked from weakest to strongest as shown below. Van der Waals interactions are the weakest of all intermolecular forces, partly because they are the cause of temporary dipole movements in the atoms.) Vander Waal forces < Dipole-Induced Dipole < Dipole-Dipole < Hydrogen bonds < Ion-dipole
What is the E ° for the voltaic cell composed of the half cells shown below? Br2 + 2 e- → 2 Br- (E°= +1.09 v) Ag+ + e- → Ag (E°= + 0.80 v) A. E°cell = - 0.29 v B. E°cell = + 0.29 v C. E°cell = - 1.89 v D. E°cell = + 1.89 v E. E°cell = - 0.58 v
E°cell = + 0.29 v To determine the E°cell for a voltaic cell, we must keep in mind that they undergo spontaneous reactions, as shown in the image below. In order for a reaction to be spontaneous, the E°cell must be positive (also, delta G will be negative and K > 1). In the information provided, the "E" represents the reduction potentials. Therefore, one reaction must be reversed in order to have one reduction and one oxidation reaction. Since it's a spontaneous cell, we want the largest E cell value. To get that, we reduce Br2 (+1.09 v E°) and oxidize Ag (-0.8 v E°). To determine E°cell, we add the oxidation and reduction halves as shown below. E°cell = +1.09 v + (-0.8 v) E°cell = +0.29 v
Which of the following is the correct molecular geometry of NH3? A. Bent B. Trigonal planar C. Tetrahedral D. Trigonal pyramidal E. Octahedral
Trigonal pyramidal STEP 1: To determine the correct molecular geometry, we must begin by drawing the VESPR diagram of ammonia, NH3. STEP 2: By drawing out the VESPR diagram, we can determine that ammonia is sp3hybridized and has a tetrahedral electron geometry and a trigonal pyramidal for its molecular geometry.
Each of the following are intramolecular forces EXCEPT one. Which one is the EXCEPTION? A. Polar covalent bonds B. Metallic bonds C. Ionic bonds D. Van der Waals forces E. Non-polar covalent bonds
Van der Waals forces All of the options listed are intramolecular forces except for Van der Waals forces which is an intermolecular force. Other intermolecular forces include Hydrogen Bonding and the various kinds of dipole interactions mentioned previously. Within intermolecular forces, hydrogen bonding (H bonded to N, O and F) is the strongest, followed by Dipole-Dipole forces, and finally, London dispersion forces, both of which are considered to be Van der Waals forces. A hydrogen bond is an electrostatic force of attraction between a hydrogen atom which is covalently bound to a more electronegativeatom or group, like nitrogen, oxygen, or fluorine and another electronegative atom bearing a lone pair of electrons.
Bromine exists as two stable isotopes: 50.69% 79Br and 49.31% 81Br. Whereas, hydrogen exists as 99.98% 1H and 0.02% 2H in nature. What are the possible peaks for the isotopes of HBr on the mass spectrum? A. Peaks for mass: 80 (1H and 79Br), 81 (2H and 79Br), 82 (1H and 81Br), and 83 (2H and 81Br) B. Peaks for mass: 80 (2H and 79Br), 81 (1H and 79Br), 82 (1H and 81Br), and 83 (2H and 81Br) C. Peaks for mass: 80 (1H and 79Br), 81 (2H and 79Br), 82 (2H and 81Br), and 83 (1H and 81Br) D. Peaks for mass: 80 (2H and 79Br), 81 (2H and 79Br), 82 (1H and 81Br), and 83 (1H and 81Br)
Peaks for mass: 80 (1H and 79Br), 81 (2H and 79Br), 82 (1H and 81Br), and 83 (2H and 81Br) This is a tricky general chemistry question. Upon first glance, this question may look like an organic chemistry question that involves an analytical understanding of mass spectrometry, but in actuality, it tests your understanding of isotopic mass. Essentially, the question can be rephrased as: "What are the various isotopic combinations of atomic mass that could display as distinctive peaks on the mass spectrum?". This type of question has shown up on the DAT recently under the chemistry section, so it is important to understand how to solve it. Given in the question, 1H has an atomic mass of 1, 2H has an atomic mass of 2, 79Br has an atomic mass of 79, and lastly, 81Br has an atomic mass of 81. The 4 combinations of these isotopes yield: 1 + 79 = 80 2 + 79 = 81 1 + 81 = 82 2 + 81 = 83 These numbers all correspond to the peaks shown in Option A. Therefore, the answer is Peaks for mass: 80 (1H and 79Br), 81 (2H and 79Br), 82 (1H and 81Br), and 83 (2H and 81Br).
It takes 512 g of diesel with a density 0.832 g•mL-1 to fill a certain container. At constant temperature and with a density of 0.64 g•mL-1, what mass of petroleum ether would be required to completely fill the container? A. (0.832) ---------- (512)(0.64) B. (0.832)(512) --------------- (0.64) C. (0.64) ---------- (512)(0.832) D. (0.64)(512) -------------- (0.832) E. (0.64) (0.832) ----------------- (512)
STEP 1: To solve this question, we must first rearrange the density equation to find the volume occupied by diesel: Density = mass/volume → Volume = mass/density Volume occupied by diesel = 512 g diesel/ 0.832 g•mL-1 STEP 2: Next, we use the same density equation to find the mass (in grams) of petroleum required to fill up the same volume of the container. So we can solve using this equation: Grams of petroleum = density of petroleum * volume of canister Grams of petroleum = 0.64 * volume occupied STEP 3: From step 1, we know the volume occupied was 512 g/0.832, so we get the following answer: Grams of petroleum = (0.64*512)/0.832
According to the following reaction, how many grams of AlCl3 are formed when 50 g of aluminum reacts with excess chlorine gas? 2 Al +3 Cl2 → 2 AlCl3 A. (132) ---------- (50) (27) B. (132)(50) ------------ (27) C. (27) -------- (50)(132) D. (27)(50) ------------ (132)
(132)(50) ------------ (27) Since it says that Cl2 is in excess, we know that Al would be the limiting factor/reagent. Remember, the limiting reagent is what determines the extent to which the reaction would proceed. Looking at the periodic table, we see that Al has a molar mass of ~27 g•mol. STEP 1: Determine the # of moles of Al: moles of Al = mass/molar mass moles of Al = (50 g Al) / (27 g•mol) STEP 2: Next, we determine how many moles of AlCl3 are present relative to the number of aluminum moles: moles of AlCl3 = (50/27 mol Al) * (2 AlCl3/2 Al) moles of AlCl3 = 50/27 mol STEP 3: The molar mass of AlCl3 is ~132 g•mol-1, and now you simply insert the known values and rearrange the equation to get the mass of AlCl3: moles of AlCl3 = mass/molar mass (50/27) = mass/(132 g•mol) mass = (50*132)/27 grams
Which of the following solutions has the highest boiling point? A. 0.10 m K2SO4 B. 0.10 m NaCl C. 0.10 m NaNO3 D. 0.10 m KNO3 E. 0.10 m KCl
0.10 m K2SO4 Boiling-point elevation describes the phenomenon where the boiling point of a liquid will be higher when another compound is added. Specifically, this happens whenever a non-volatile solute, such as a salt, is added to a pure solvent, such as water. Therefore, a solution will have a higher boiling point than a pure solvent. In fact, boiling point is a colligative property, meaning it will only be affected by the effective concentration of solutes, regardless of the type of solute present. We obtain boiling point elevation by the change in "iKm" equation, as shown below: ΔTb = iKbm Where ΔTb = change of temperature, i= the van't Hoff constant (i.e # of ions per molecule formed after dissolving),Kb = boiling point elevation constant, and m = molality of solution (mol solute/Kg of solvent) Going back to the question, all of the given solutes have the same molality and are all assumed to dissolve in the same solvent. Therefore, the only factor that can elevate the boiling point is the van 't Hoff constant. It is very important to know this for the DAT! Since the solute with the highest van 't Hoff constant has the highest boiling point, Option A. 0.10 m K2SO4 is the correct answer (it has a van 't Hoff constant of 3 since it is a salt which dissociates into twoK and one SO4).
Which of the following numbers possess the most significant digits? A. 10.20 B. 0.000761 C. 1,000.0 D. 812.0 E. 0.905
1,000.0 Significant digits are an important topic that have shown up on the DAT in the past. To solve this question, you must know the following significant digit rules: Non-zero digits are always significant Any zeros between two significant digits are significant A final zero or trailing zeros in the decimal portion ONLY are significant From the given option choices, Option C. 1,000.0 possesses the most significant digits (5 significant digits); according to the second rule above, the zeros in the middle are significant and according to the third rule, the last zero is also significant. Therefore, this is the correct answer.
The alpha decay of iridium-168 results in emission of an alpha particle and what other nuclei? 168 Ir →4 α+? 77 2 A. 172 Re 75 B. 164 Re 75 C. 172 Au 79 D. 172 Au 75 E. 164 Au 79
164 Re 75 Radioactive decay is the process by which an unstable nucleus becomes stable. There are many different types of decay, as shown in the image below. In particular, alpha decay results in the emission of an alpha particle which leads to: Decrease in atomic mass by 4 (lose 2 protons + 2 neutrons) Decrease in the atomic number by 2 We can subtract the values from the original atomic mass and weight of iridium-168 to get the atomic mass and number of the other emitted nuclei: Atomic mass = 168 - 4 = 164 Atomic number = 77 - 2 = 75
A 20g block of ice is at -10°C temperature. What is the total energy required to heat the ice 50°C above boiling temperature? (heat of fusion of water = 334 J/g, heat of vapourization of water = 2257 J/g, specific heat of ice = 2.09 J/g·°C, specific heat of water = 4.18 J/g·°C, specific heat of steam = 2.09 J/g·°C) A. 24 200 J B. 36 400 J C. 58 400 J D. 62 600 J E. 72 100 J
62 600 J The equation we will be using to calculate the heat energy required to raise the temperature within the same state is: q = mcΔT Where q = heat energy, m = mass, c = specific heat, and ΔT = change in temperature The equation we will be using to calculate the heat energy required to cause a change of phase is: q = mΔHf OR q = mΔHv Where q = heat energy, m = mass, ΔHf = heat of fusion (used to convert solid to liquid), and ΔHv = heat of vapourization (used to convert liquid to gas) STEP 1: Calculate the energy required to first heat the -10°C ice to 0°C ice. For this step, we will use the specific heat of ice = 2.09 J/g·°C: q = mc(ice)ΔTq = (20g) x (2.09 J/g·°C) x (10°C)q = 418 J = ~400 J STEP 2: Calculate the heat required to convert 0°C ice to 0°C water. For this equation, we will use the ΔHf = heat of fusion: q = mΔHfq = (20g) x (334 J/g)q = 6680 J = ~6700 J STEP 3: Calculate the energy required to now heat the 0°C water to 100°C water. For this step, we will use the specific heat of water = 4.18 J/g·°C: q = mc(water)ΔTq = (20g) x (4.18 J/g·°C) x (100°C)q = 8360 J = ~8400 J STEP 4: Calculate the heat required to convert 100°C water to 100°C steam. For this equation, we will use the ΔHv = heat of vapourization: q = mΔHvq = (20g) x (2257 J/g)q = 45140 J = ~45000 J STEP 5: Calculate the energy required to now heat the 100°C steam to 150°C steam. For this step, we will use the specific heat of steam = 2.09 J/g·°C: q = mc(steam)ΔTq = (20g) x (2.09 J/g·°C) x (50°C)q = 2090 J = ~2100 J STEP 6: The last step is to calculate the total heat energy: Heat Total = Heat Step 1 + Heat Step 2 + Heat Step 3 + Heat Step 4 + Heat Step 5 Heat Total = 400 J + 6700 J + 8400 J + 45000 J + 2100 J Heat Total = 62600 J Therefore, the heat required to convert 20 grams of -10 °C ice into 150 °C steam is 62600 J (or 62.6 kJ).
Given the ΔH for the reaction data set, calculate the enthalpy change for the following reaction. H2SO4(l) → SO3(g) + H2O(g) H2S(g) + 2 O2(g) → SO3(g) + H2O(l) ΔH = -200 kJ H2S(g) + 2 O2(g) → H2SO4(l) ΔH = -230 kJ H2O(l) → H2O(g) ΔH = 45 kJ A. 430 kJ B. 380 kJ C. 245 kJ D. 150 kJ E. 75 kJ
75 kJ To determine the enthalpy change for this reaction, we must use Hess's Law to solve this question. Note: We cannot consider the elements or reactants that cancel out, such as O2, or elements that are common in multiple reactions like liquid water. STEP 1: We know that H2SO4 will be on the left side of the reaction and that it is only present in the 2nd equation on the right. Thus, we reverse that reaction and get a 230 KJ for delta H: H2SO4(l) → H2S(g) + 2 O2(g) ΔH = +230 kJ STEP 2: Next, we note that SO3 is only common to and on the right side the first reaction. Also, in the overall reaction, it is on the right side with the same coefficient, and thus we do not change anything: H2S(g) + 2 O2(g) → SO3(g) + H2O(l) ΔH = -200 kJ STEP 3: The same logic can be applied for the 3rd reaction with respect to H2O in the gas state, where it's on the same side of the reaction and has the same coefficient as the overall: H2O(l) → H2O(g) ΔH = 45 kJ STEP 4: Lastly, we add the three steps up: H2SO4(l) → H2S(g) + 2 O2(g) ΔH = +230 kJH2S(g) + 2 O2(g) → SO3(g) + H2O(l) ΔH = -200 kJH2O(l) → H2O(g) ΔH = 45 kJ Therefore, the total delta H becomes 230 + (-200) + 45 = +75 kJ.
Which of the following elements is a metalloid? A. Aluminum B. Titanium C. Antimony D. Gallium E. Tin
Antimony In the periodic table, elements can be broadly classified as metals, metalloids, and nonmetals. Metalloids share some characteristics of metals and some characteristics of nonmetals. Physically, metalloids are typically solid, metallic in lustre, more brittle than metal, and semi-conductive (allow for the average transmission of heat). Chemically, they mostly resemble the behaviour of non-metals, and form alloys with metals. The 8 metalloids are: Boron (B), Silicon (Si), Germanium (Ge), Arsenic (As), Antimony (Sb), Tellurium (Te), Polonium (Po), and Astatine (At). They are found along the stair-step (or zig-zag) line on the periodic table, as shown in the image below.
Which of the following elements has the highest ionization energy? A. Si B. P C. S D. Cl E. Ar
Ar Ionization energy is the input of energy required to remove an electron. Ionization energy INCREASES as we go to the top and right of the periodic table, as shown in the image below. As a result, Helium has the highest ionization energy, while Francium has the lowest. This phenomenon can be explained using the net effective nuclear charge (Zeff) and atomic size. Zeff is the attractive force of the positively charged nucleus felt by an electron in a specific shell. Zeff is dependent on the number of protons and the number of electrons between the nucleus and the electron in the given shell. The electrons in between cause electron shielding, which means that electrons on the lower shells reduce the nucleus' attractive forces experienced by the valence electrons. As a result, the lower we go on the periodic table, the greater the number of lower electron shells that reduce the attractive forces holding the valence electrons, and so the easier it would be to remove an electron. However, as we go to the right of the periodic table, the number of electrons in the lower shells stays constant, but the number of protons increases and thus the positive charge increases. Thus, the Zeff increases across the table and so does the ionization energy. Furthermore, as we go up and to the right of the periodic table, the smaller the atomic radii will be. As a result, the outer electrons will be closer to the nuclei and will experience a greater attractive force. Thus, the smaller the atom, the more energy required to remove an electron from the outer shell. The opposite applies as we go down the periodic table where we increase in atomic size.
Which of the following compounds contains chlorine with an oxidation number of +3? A. HClO B. HClO2 C. HClO3 D. ClO2 E. HClO4
HClO2 In a redox reaction, oxidation and reduction always occur together. An atom is reduced by gaining one or more electrons, but only if another atom loses the electrons. In order to track the gain or loss of electrons by atoms in reactions, an oxidation number is assessed for each atom in a compound relative to the number of electrons in a neutral atom of that element. To solve this question, we would have to calculate the oxidation number of chlorine in each compound given. As an example, we will calculate the oxidation number of chlorine for Option B. HClO2 as an example. We know the oxidation of the entire compound is = 0. Next, we need to calculate the oxidation number for oxygen, hydrogen, chlorine. For oxygen, recall that the valency of oxygen is -2, we therefore multiply it by 2 since there are 2 oxygen atoms. For hydrogen, it would be +1. And lastly, for chlorine, we must isolate for it: (H × 1) + (Cl × 1) + (O × 2) = Oxidation # of compound (+1 × 1) + (Cl × 1) + (-2 × 2) = 0 (+1) + (Cl) + (-4) = 0 Cl = +3
Identify the acids in the following reaction: HF + HCO3- ↔ F- + H2CO3 A. HF and F- B. HF and HCO3- C. HF and H2CO3 D. F- and HCO3- E. F- and H2CO3
HF and H2CO3 Acids are entities that either form/donate protons (Arrhenius and Bronsted-Lowry) or accept electrons (Lewis). A generic formula for acids is: HA ↔ H+ + A- Bases on the other hand are those entities that either form hydroxides in water (Arrhenius), accepts protons (Bronsted-Lowry) or donates electrons (Lewis). A generic formula for bases is A- +H+ ↔ HA For the reaction in the question, HF donates an H+ to HCO3- to form F- and H2CO3. As a result: HF and H2CO3 (which obtained the H+) are acids HCO3- and F- (which lost H+) are bases
The following reaction is an exothermic all gas reaction known as the Haber process. Which of the following will occur if the pressure in the reaction chamber were to increase? N2(g) + 3 H2(g) → 2 NH3(g) + heat A. Increase in NH3 concentrations B. Increase in N2 concentrations C. Decrease in NH3 concentrations D. Increase in H2 concentrations E. Drop in chamber temperature
Increase in NH3 concentrations Le Chatelier's principle can be used to predict the effect of a change in conditions on some chemical equilibria. The principle states that when any system at equilibrium is subjected to change in concentration, temperature, volume, or pressure, the system changes to a new equilibrium to counteract the applied change. Changing the concentration of a chemical will shift the equilibrium to the side that would reduce that change in concentration. Changing the temperature in the equilibrium can be understood by incorporating heat as either a reactant or a product, and assuming that an increase in temperature increases the heat content of a system. Changing the volume or pressure of the system changes the partial pressures of the products and reactants and can affect the equilibrium concentrations. When pressure increases due to a decrease in volume, the side of the equilibrium with fewer moles is more favorable, however, when pressure decreases due to an increase in volume, the side with more moles is more favorable. Getting back to the question, when we increase the pressure, we shift in the direction to neutralize this offset. As a result, we move towards the side with the fewer moles of gas, in this case, to the right. As a result, we lose heat (exothermic process) and form ammonia.
Which of the following is a neutral salt? A. NaF B. NaCl C. NaH2PO4 D. NH4Cl
NaCl Salts may be defined as the product of a neutralization reaction of an acid and a base. The neutralization of any strong acid with a strong base always results in a neutral salt product. However, salts can also be acidic or basic, as well as neutral. To determine the acid-base properties of salt, we must consider the strengths of the reactant acid and base that make up the salt. For example, the neutralization of a strong acid with a weak base results in an acidic salt. In comparison, the neutralization of a weak acid with a strong base results in a basic salt. Based on this, Option B.NaCl is the only neutral salt since both cation and anion are derived from strong conjugates. DAT Pro Tip: Most group 1 and 2 metals form strong bases. As a result, their conjugates are neutral. Therefore, the cation half containing Na will be neutral.
According to the following redox reaction, identify the compounds that are reduced: CH4 + 2 O2 → CO2 + 2 H2O A. CH4 B. O2 C. CO2 D. CH4 and H2O E. CH4 and CO2
O2 Reduction is the gain of electrons or a decrease in the oxidation state of an atom by a molecule, ion, or another atom. An example of a reduction is when iron reacts with oxygen, forming a chemical called rust. In that example, the iron is oxidized and the oxygen is reduced. Getting back to the question, we know that whenever an oxygen element is added, the molecule becomes oxidized, and whenever hydrogen is added, it gets reduced. In this case: CH4 becomes CO2 and thus it is oxidized (due to the addition of oxygen). O2 becomes H2O and thus is reduced (due to the addition of hydrogen).
The nuclear binding energy of sodium-21 is 7.77 MeV and the binding energy of sodium-23 is 8.11 MeV. Which statement is true about these two isotopes? A. Sodium-21 is more stable B. Sodium-21 has greater mass C. Sodium-23 has a greater mass defect D. Sodium-23 is less likely to lose an electron E. Sodium-23 has greater intermolecular force
Sodium-23 has a greater mass defect Nuclear binding energy is the energy required to disassemble the nucleus of an atom into its separate nucleons (protons and neutrons). The greater the binding energy, the more stable the nucleus is and the less likely it is to decay by losing a proton or neutron. The equation to calculate nuclear binding energy is: E = mc2 Where E is the nuclear binding energy, m is the mass defect, and c is the speed of light in a vacuum. Mass defect is the difference in the mass of the nuclei and the total mass of its constituent nucleons. For example, the mass of helium on the periodic table is 4.00 amu. However, the mass of 2 protons + 2 neutrons is 4.03 amu; therefore, there is a mass defect of 0.03 amu. Based on the equation, nuclear binding energy is proportional to the mass defect; the greater the mass defect, the greater the nuclear binding energy. Nuclear binding energy has no relation to intermolecular nor intramolecular forces, nor electron affinity.
Which of the following properties determine the strength of an acid? A. Stability of conjugate base B. Stability of conjugate acid C. Number of covalent bonds D. Number of ionic bonds E. Volume of an acid present
Stability of conjugate base Acid strength refers to the tendency of an acid (HA) to dissociate into a proton, H+, and an anion, A−. Examples of strong acids are hydrochloric acid (HCl), perchloric acid (HClO4), nitric acid (HNO3) and sulfuric acid (H2SO4). A weak acid is only partially dissociated, with both the undissociated acid and its dissociation products being present in equilibrium with each other. Acetic acid (CH3COOH) is an example of a weak acid. HA ⇌ H+ + A− DAT Pro Tip: The stronger the acid, the weaker is its conjugate. In other words, the greater the likelihood of losing a proton/accepting an electron depends on how weak (or stable) the conjugate is. Therefore, all strong acids and bases have very stable conjugates.
A student uses an uncalibrated pH meter on a solution with a pH of 7.25. They obtain the values 6.25, 6.24, 6.25. What conclusions can be made about these measurements? A. The measurements are precise but not accurate B. The measurements are accurate but not precise C. The measurements are both accurate and precise D. The measurements are neither accurate nor precise
The measurements are precise but not accurate To solve this problem involving accuracy vs. precision, let's first define the two properties: Accuracy describes the overall closeness of experimental values to the true literature value. If a set of data is accurate, that means the experimental values are grouped around the true value. Precision describes the overall consistency of experimental values. If a set of data is precise, that means the experimental values are grouped closely to each other, but not necessarily around the true value. Accuracy and precision are independent of one another. In this case, the student's measurements are all very close to each other, indicating high precision. However, the measurements are not close to the true value of the solution, which is pH 7.25. Therefore, these measurements are not accurate.
Which of the following statements best describes Boyle's law? A. Volume of a gas is directly proportional to pressure B. Volume of a gas is directly proportional to its temperature C. Volume of a gas is inversely proportional to temperature D. Volume of a gas is inversely proportional to pressure E. Pressure is inversely proportional to temperature
Volume of a gas is inversely proportional to pressure Boyle's law states that at equal temperature and moles of gas, pressure is inversely proportional to volume. According to the kinetic molecular theory of gases, pressure results from the number of collisions that gas particles make with the walls of a container. Decreasing volume while maintaining the same temperature and the same number of gas particles, increases the number of collisions with the walls of the container and increases the pressure. Consequently, pressure and volume are inversely proportional, and every gas has a particular pressure-volume constant at a given temperature. Boyle's law describes this inverse relationship between volume(V) and pressure (P) for a fixed number of molecules of an ideal gas at constant temperature, where k is a constant: P∝1/V PV = k
An experiment is conducted and the titration curve was recorded as shown below. What was titrated to create this curve? A. Weak acid by adding strong base B. Weak base by adding strong acid C. Weak base by adding weak acid D. Strong base by adding weak acid
Weak base by adding strong acid In a titration experiment, a measured amount of an acid or a base solution of a known concentration is slowly added to another solution containing a base or an acid of an unknown concentration. During this experiment, the acid or base being titrated is fully neutralized at the equivalence point. When this occurs, the pH changes rapidly when passing through the equivalence point is seen as a nearly vertical segment on a titration curve. The equivalence point is located in the middle of this vertical segment. Going back to the question, we know that since the pH at the start of the solution is 10, as shown in the image below, the solution was initially a base. At the end of the curve, the pH is 2 and therefore the solution turned acidic. The only way to turn a basic solution acidic is by adding a strong acid to an already weak base.
To electroplate a piece of copper metal with zinc, what reaction must occur at the cathode? A. Cu2+ + 2e- → Cu B. Cu → Cu2+ + 2e- C. Zn → Zn2+ + 2e- D. Zn2+ + 2e- → Zn
Zn2+ + 2e- → Zn Electroplating is the process of plating one metal with another metal using an electrolytic cell. The piece of metal which is being plated is made into the cathode because reduction always occurs at the cathode. The electrolyte contains positive cations of the metal that is being used to plate, which are reduced at the cathode. The anode thus contains a block of metal that is being used to plate. As electrons are transferred from the anode to the cathode, the positive Zn2+ ions are attracted to the negative cathode and reduction occurs, producing Zn that deposits on the copper metal. The reduced cations are then replaced by the anode. The electrolytic cell image below depicts the process of electroplating as a piece of metal is electroplated with copper. In this question, the piece of metal to be electroplated is a piece of copper metal (labeled 'Me' in the image below) and the anode consists of a sacrificial piece of zinc instead of the copper that is depicted below. Based on this, we can see that the following reactions are occurring: The reaction that occurs at the anode: Zn → Zn2+ + 2e-. The oxidation of Zn (i.e. corrosion) releases Zn2+ ions into the solution. The reaction that occurs at the cathode: Zn2+ + 2e- → Zn. The Zn2+ ions in the solution migrate to the negative cathode where they regain electrons. In comparison, a Galvanic cell generally consists of two different metals immersed in an electrolyte. Sometimes it can include individual half-cells with different metals and their ions in solution connected by a salt bridge. Oxidation occurs at the anode in both galvanic and electrolytic cells, which means that electrons flow from the anode to the cathode.
Which of the following is the correct electron configuration for Cu? A. [Ar] 4s13d10 B. [Ar] 4s13d9 C. [Ar] 4s23d10 D. [Ar] 4s23d104s1 E. [Ar] 4s13d104s1
[Ar] 4s13d10 Electron configuration is the distribution of electrons of an atom or molecule in atomic or molecular orbitals. The first three orbitals are the s, p and d orbitals which hold 2, 6, and 10 electrons respectively. When we try to solve this question, at first glance the electron configuration of copper appears to be: [Ar] 4s23d9 However, there are certain elements that have special exceptions, copper being one of them since it is a transition metal. Transition metals such as copper need an electron to fill their d-orbitals; they get this electron from the full s-orbital. This occurs because elements with either half or full sub-orbitals are more stable than if they were empty orbitals. Therefore, we take the original electron configuration of [Ar] 4s23d9, and we remove one electron from the 4s and add it to the 3d-orbital, to get: [Ar] 4s13d10 This effectively results in a partial occupancy of the s-orbital and a full d-orbital. The same rule can be applied to the 4th group which are the transition metals with 4 electrons in their d-orbitals.
Which of the following represent the solubility product constant for magnesium hydroxide? Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH-(aq) A. [Mg2+][OH-] B. [Mg2+][OH-]2 C. 2[Mg2+][OH-] D. 2[Mg2+][OH-]2 E. [Mg2+]2[OH-]
[Mg2+][OH-]2 STEP 1: To begin, we can write the given reaction in this format: a A(s) ⇌ c C(aq) + d D(aq) STEP 2: To tackle this question, you must remember that the values for the Q and K constants (i.e Ksp, Kp, Kseq, Ka, Kb, etc) are obtained by making a ratio of products over reactants. Also, remember that solids and liquids have activity indexes of 1 and thus are not included in the solubility constant. Additionally, the coefficients in the reactions become the exponent of the element under question, and not a coefficient. Thus, we get: Ksp = [C]c[D]d STEP 3: The next step is to simply substitute the reaction products, and the Ksp value for the above chemical equation equals to: Ksp = [Mg2+][OH-]2
Transition metals have partially filled A. s-subshells. B. p-subshells. C. d-subshells. D. f-subshells. E. g-subshells.
d-subshells. A transition metal is any element in the d-block of the periodic table, which includes groups 3 to 12 on the periodic table. The f-block and actinide series are also considered transition metals. The definition of a transition metal that you should know for the DAT is that transition metals have a partially filled d-subshell. There are a number of properties of transition metals that are not found in other elements, because of the partially filled d-shell. These include Formation of compounds whose color is due to d-d electronic transitions Formation of compounds in many oxidation states Formation of many paramagnetic compounds due to the presence of unpaired delectrons
The transition from gas to solid phase is known as A. fusion. B. freezing. C. condensation. D. sublimation. E. deposition.
deposition. Substances on Earth can exist in one of three phases: solid, liquid or gas. There are six changes of phase that you should be familiar with: freezing, melting, condensation, vaporization, sublimation, and deposition. Freezing: the substance changes from a liquid to a solid. Melting: the substance changes back from the solid to the liquid. Condensation: the substance changes from a gas to a liquid. Vaporization: the substance changes from a liquid to a gas. Sublimation: the substance changes directly from a solid to a gas without going through the liquid phase. Deposition: is a process where a gas changes phase and turns directly in solid without passing through the liquid phase. It's also the opposite of sublimation. Here is a quick summary of all the different types of thermodynamic processes involving phase changes that you must know for the DAT
