Gen chem #5

The following mixtures (assume volumes are additive) will result to a buffer solution EXCEPT one. Which one is the EXCEPTION? A. 100 mL 0.10 M CH3COOH + 100 mL 0.10 NaCH3COOB. 100 mL 0.40 M HCN + 100 mL 0.20 M KOHC. 100 mL 0.050 M HCl + 100 mL 0.075 M NaOHD. 100 mL 0.125 M NH4Cl + 100 mL 0.100 M NH3E. 100 mL 0.100 M HCl + 100 mL 0.20 M NH3

A buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid. C. 100 mL 0.050 M HCl + 100 mL 0.075 M NaOH This will not result in a buffer because there is no source of a weak acid and its conjugate base or a weak base and its conjugate acid. HCl is a strong acid and NaOH is a strong base. Reaction of HCl and NaOH forms H2O and NaCl. [C] is the answer. A. 100 mL 0.10 M CH3COOH + 100 mL 0.10 NaCH3COO This will result in a buffer because a weak acid and its conjugate base (CH3COOH and CH3COO-, respectively) are present in the solution. Eliminate [A]. B. 100 mL 0.40 M HCN + 100 mL 0.20 M KOH This will result in a buffer solution. HCN will react with KOH in a 1:1 molar ratio to form H2O and KCN: HCN + KOH → H2O + KCN Because the number of moles of HCN is twice that of KOH, there will be HCN leftover after this reaction. In the process, KCN is formed, which dissociates into CN-, the conjugate base of HCN. Eliminate [B] D. 100 mL 0.125 M NH4Cl + 100 mL 0.100 M NH3 This will result in a buffer solution because a weak base and its conjugate acid is present (NH3 and NH4+, respectively). Eliminate [D]. E. 100 mL 0.100 M HCl + 100 mL 0.20 M NH3 This will result in a buffer solution. This is similar to mixture [B], but in this case a weak base is present initially (and in excess) and will react with a strong acid to form its conjugate weak acid. Eliminate [E].

Which of the following would be strongly attracted to magnetic fields? A. HeB. H2C. ZnD. O2E. N2

A paramagnetic atom or molecule has at least one unpaired electron which can interact with a magnetic field. If all electrons are paired, then the molecule is diamagnetic and will not interact with a magnetic field. To solve, draw each atomic or molecular orbital diagram to determine which answer choice has at least one unpaired electron. D. O2 MO diagram: http://youtu.be/dVWtvG9ztMQ?t=3m11s O2 is paramagnetic. [D] is the answer. A. He MO diagram: http://www.youtube.com/watch?v=c1wzgky77hs He is diamagnetic. Eliminate [A]. B. H2 MO diagram: http://youtu.be/DyL5pQc0Hsw?t=5m30s H2 is diamagnetic. Eliminate [B]. C. Zn Zn has 10 electrons in its 3d orbital and 2 electrons in its 4s orbital, so every electron is paired. It is diamagnetic. Eliminate [C]. E. N2 MO diagram: http://youtu.be/1Bgjka4kLYo?t=17m40s N2 is diamagnetic. Eliminate [E]. DAT Pro-tip: Diamagnetism and paramagnetism questions usually reference single atoms on the DAT. However, it is possible that they are asked about simple compounds. For the sake of efficiency, memorize that N2 is diamagnetic and O2 is paramagnetic. These are the two most common examples.

Which of the following acts as a reducing agent? A. Cl-B. IO3-C. F2D. Zn2+E. Cl2

A reducing agent is oxidized in a chemical reaction. Conversely, an oxidizing agent is reduced. The reducing agent must have be element that will likely give up electrons (or become oxidized). Cl- is a great candidate to act as a reducing agent because it will change oxidation states from -1 to 0. [A] is the answer. IO3- looks like a possibility, but the oxidation state on iodine is +5, and it isn't likely to give up more electrons. [B] is incorrect. F2 isn't likely to give up more electrons, and neither is Zn2+ or Cl2. [C], [D], and [E] are incorrect. FAQs - frequently asked questions Q: Why doesn't one of the O's in IO3- just donate electrons?Generally, oxygen is a GREAT oxidizing agent rather than a reducing due to its tendency to attract/steal electrons. It is more more energetically favorable for O to hold onto its electrons. That being the case, I would be the first to donate electrons as the reducing agent rather than O in IO3-, so the reducing potential of the compound is dependent on I! Since the oxidation number of I is 5+ (a large positive number), it will want to hold on to its few electrons and NOT act as a reducing agent.

Suppose a one-step reaction has forward and reverse activation energies that are equal to each other, as shown in the reaction coordinate diagram below. Which of the following accurately describes the reaction? A. Isenthalpic B. Adiabatic C. Isentropic D. Catalytic E. Isobaric

A. Isenthalpic The change in enthalpy ∆H is defined as the difference between the energy levels of reactants and products in the reaction coordinate diagram. ∆H = Eproducts - Ereactants If the forward and reverse activation energies of the reaction are equal, Eproducts = Ereactants and therefore ∆H = 0. Because enthalpy remains constant, the reaction can be described as isenthalpic. [A] is correct. B. Adiabatic An adiabatic reaction is one in which heat energy remains constant (∆q = 0). Eliminate [B]. C. Isentropic An isentropic reaction is one in which entropy remains constant (∆S = 0). Eliminate [C]. D. Catalytic Recall that catalysts speed up a reaction by providing an alternate pathway between reactants and products that innately has a lower activation energy. Eliminate [D]. E. Isobaric An isobaric reaction is one in which pressure remains constant (∆P = 0). Eliminate [E]. Topic:Chemical Kinetics

The preferred colligative property for measuring the molecular weight of polymers is osmotic pressure. Which of the following best explains why osmotic pressure is preferred to measure the molecular weight of polymers? A. Small changes are easier to measure in osmotic pressure than in other colligative properties. B. Polymers cannot pass through the semi-permeable membrane, but simple salts can. C. Osmotic pressure measures the concentration of solute in molarity instead of molality. D. The changing temperature of the solution needed for measuring other colligative properties changes the number of solute particles in solution. E. Other colligative properties are more sensitive to solute concentration than osmotic pressure.

A. Small changes are easier to measure in osmotic pressure than in other colligative properties. Colligative properties depend on the concentration of a solute in a solvent, and do not depend on the identity of the actual solute. Osmotic pressure is the pressure that must be exerted on one side of a U tube to prevent the osmosis of water across a semi-permeable membrane. If the osmotic pressure changes slightly, the height of the solution column will raise or lower. Small changes are easier to detect in osmotic pressure than in other colligative properties. [A] is the answer. B. Polymers cannot pass through the semi-permeable membrane, but simple salts can. Semi-permeable membranes are capable of filtering simple salts. This statement is incorrect and does not answer the question. [B] is incorrect. C. Osmotic pressure measures the concentration of solute in molarity instead of molality. While the formula for osmotic pressure does use molarity and not molality, this doesn't answer the question that was asked. [C] is incorrect. D. The changing temperature of the solution needed for measuring other colligative properties changes the number of solute particles in solution. While temperature change might influence the volume of the solution, the number of particles of solute will not change. [D] is incorrect. E. Other colligative properties are more sensitive to solute concentration than osmotic pressure. While it is true that in theory all colligative property changes are equally sensitive to changes in solute concentration, other colligative properties are not more sensitive than osmotic pressure. [E] is incorrect.

Which of the following can act as both an oxidizing agent and a reducing agent? A. NO3-1B. CoC. F-1D. MnO4-1E. H2O2

An oxidizing agent gets reduced. A reducing agent gets oxidized. To be able to get oxidized, the species cannot be at its maximum oxidation state. To get reduced, the species cannot be at its minimum oxidation state. H2O2 can act as either an oxidizing agent or a reducing agent. The oxygen in H2O2 is at a -1 oxidation state. The oxygen can become reduced into a -2 oxidation state, like in H2O, or it can become oxidized into a 0 oxidation state, like in O2. [E] is the answer.

Calcium has a larger atomic radius than magnesium because of the A. difference in the number of neutrons in their nucleus. B. increase in ionization energy. C. lanthanide contraction. D. increase in effective nuclear charge. E. increase in electron shielding.

Atomic size generally gets larger as you go down and towards the left on the periodic table, and smaller as you go up and towards the right. DAT Pro-tip: Occasionally, the DAT will ask about the fundamental concepts behind atomic size. Many of the periodic trends, including atomic size, can be derived from the two concepts of effective nuclear charge and electron shielding. 1. Going across the periodic table towards the right, elements have more protons, while the number of valence shells is held constant. So, the electrons experience a stronger effective nuclear charge. The increasing nuclear charge is responsible for the decrease in atomic size across the periodic table. 2. The electron shielding effect reduces the effect of a full nuclear charge. As more core electrons are added, they begin to "shield" the valence electrons from the increasingly positive nucleus. This allows the outer electrons to move farther away from the nucleus, increasing atomic size. The increasing electron shielding effect is responsible for the increase in atomic size moving down the periodic table. Calcium and magnesium are in the same column, so the size difference is due to the increase in electron shielding. [E] is the answer.

An extracted pure compound appears silvery white, exhibits high conductivity to heat and electricity, yet has a very low melting point. Which of the following elements is a possible match for this compound? A. PlatinumB. SodiumC. GermaniumD. SeleniumE. Iodine

B. Sodium The compound described matches that of an alkali metal, the first group in the periodic table. The alkali metals display metallic properties due to their low first ionization energies, but they have low melting points due to their weak inter-atomic bonding. Sodium fits the given characteristics. [B] is the answer. A. Platinum Platinum is a metal, but it also has a high melting point. Eliminate [A]. C. Germanium Germanium is a metalloid and has a high melting point as well, and is only conductive under special circumstances (a semi-conductor). Eliminate [C]. D. Selenium Selenium is used in Head and Shoulders shampoo, and is a nonmetal, so it doesn't have a high conductivity. Eliminate [D]. E. Iodine Iodine does not conduct electricity. Eliminate [E].

The central atom of a molecule has 4 pairs of electron, and the molecule is V-shaped (bent). Which of these statements is always true regarding this molecule? A. The central atom is sp2 hybridized. B. There are two bonding orbitals on the central atom. C. The bond angle around the central atom is 120o. D. The molecule is miscible with hexane. E. The molecule can participate in hydrogen bonding.

B. There are two bonding orbitals on the central atom. In order for a molecule with a central atom with four pairs of electrons to be V-shaped, the central atom must have two pairs of bonding electrons and two pairs of lone pairs. [B] is the answer. A. The central atom is sp2 hybridized. The central atom is sp3 hybridized. Eliminate [A]. C. The bond angle around the central atom is 120o. The bond angle is slightly less than 109.5°. Eliminate [C]. D. The molecule is miscible with hexane. The molecule is polar. Hexane is nonpolar, and the polar, bent molecule will not be miscible with hexane. Eliminate [D]. E. The molecule can participate in hydrogen bonding. There is not enough information given to prove that the molecule can participate in hydrogen bonding. A H bonded to a N, O, or F is necessary for hydrogen bonding to occur. Eliminate [E].

A student needs to measure 9.50 mL of a liquid reagent. Which of the following glassware should the student use to get the highest precision? A. 10 mL beakerB. 10 mL Erlenmeyer flaskC. 10 mL graduated cylinderD. 10 mL test tubeE. 10 mL Buchner flask

C. 10 mL graduated cylinder A 10-mL graduated cylinder gives the highest precision for measuring volumes of small liquid. It has a flat base, ensuring the fluid's surface can be made level for measurement. [C] is the answer. A. 10 mL beaker A beaker may have measurement labels on its side, but beakers are mainly for containing and mixing different solutions (larger volumes), and are not precise. Eliminate [A]. B. 10 mL Erlenmeyer flask An Erlenmeyer flask is designed for stirring solutions. It is not designed for accurate measurements. Eliminate [B]. D. 10 mL test tube Test tubes can have measuring labels, but the base is not flat and it is not designed for taking accurate measurements. Eliminate [D]. E. 10 mL Buchner flask A Buchner flask can attach to a vacuum pump and is used for the separation of a solid from a liquid. Eliminate [E].

All of the following are assumptions of an ideal gas EXCEPT one. Which one is the EXCEPTION? A. All gas molecules are in constant, rapid, and random motion. B. There are no attractive or repulsive forces between the gases. C. The average kinetic energy is proportional to the number of gas molecules. D. The volume of the molecules is negligibly small compared to the volume of the gas. E. All collisions within the gas are perfectly elastic.

C. The average kinetic energy is proportional to the number of gas molecules. The average kinetic energy of the gas molecules depends on the temperature of the system. This statement is false. [C] is the answer. A. All gas molecules are in constant, rapid, and random motion. This is true. Eliminate [A]. B. There are no attractive or repulsive forces between the gases. There are no intermolecular forces occurring between gas molecules. So, they are assumed to be non-polar. Eliminate [B]. D. The volume of the molecules is negligibly small compared to the volume of the gas. This is true. Eliminate [D]. E. All collisions within the gas are perfectly elastic. This is true. Eliminate [E].

A chemist adds 26.5 g of ammonium chloride to 10 g of sodium hydroxide, which follows the reaction below. Assuming the reaction goes to completion, how much of which reactant is left in excess? NH4Cl + NaOH → NH3 + H2O + NaCl A. 0.5 mols, ammonium chloride B. 0.25 mols, ammonium chloride C. 0.5 mols, sodium hydroxide D. 0.25 mols, sodium hydroxide E. 1 mol, ammonium chloride

Calculate the number of mols of NH4Cl present: Calculate the number of mols of NaOH present: The stoichiometric ratios are all 1:1 in this balanced equation: NH4Cl + NaOH → NH3 + H2O + NaCl NaOH is the limiting reagent because all 0.25 mols of it will react with 0.25 mols of NH4Cl to produce 0.25 mols of NH3. This will leave an excess of 0.25 mols of NH4Cl. [B] is the answer.

Which of the following are the correct quantum numbers for the 13th electron in chlorine's electron configuration? A. 3, 0, -1, 1/2B. 3, -1, 1, 1/2C. 3, 1, -1, 1/2D. 4, 2, -1, 1/2E. 4, 0, 2, 1/2

Chlorine's electron configuration: 1s22s22p63s23p5 The 13th electron of chlorine's electron configuration is in the first orbital of the p shell. Assign the quantum numbers: n = 3 l = 1 ml = -1, 0, or 1 ms = 1/2, or -1/2 [C] is the answer.

In which instance is a gas most likely to behave ideally? A. At high temperatures, because the molecules will always be far apart. B. At high pressures, because the molecules will be close to each other relative to the size of the molecules. C. When molecules are nonpolar, because the molecules are more likely to interact. D. At high temperatures and low pressures, because the molecules are far apart and not likely to interact. E. At low temperatures, because the molecules are less likely to collide with each other.

D. At high temperatures and low pressures, because the molecules are far apart and not likely to interact. A gas becomes more ideal as the pressure decreases and the temperature increases. In an ideal gas, the gas molecules do not interact with each other through intermolecular forces and all collisions are elastic. High temperatures increase kinetic energy, allowing molecules to overcome intermolecular forces. Low pressures space out molecules far enough apart that the size of the molecules themselves is insignificant compared to the volume of the gas. [D] is the answer. A. At high temperatures, because the molecules will always be far apart. Pressure affects the spacing between the molecules, not temperature. Eliminate [A]. B. At high pressures, because the molecules will be close to each other relative to the size of the molecules. Gases behave more ideally when the molecules are far apart. Eliminate [B]. C. When molecules are nonpolar, because the molecules are more likely to interact. Nonpolar molecules are less likely to interact. Eliminate [C]. E. At low temperatures, because the molecules are less likely to collide with each other. Gases behave more ideally at high temperatures. Eliminate [E].

Which statement explains why HI is a stronger acid than HBr, which is a stronger acid than HCl? A. Iodine is the smallest element out of the 3 and thus creates the strongest acid. B. The acid strength is increased as electronegativity increases in the binding element. C. The number of available electrons in iodine for binding is greater than that of Cl or Br. D. The bond strength between H-I is lower than that of H-Cl or H-Br. E. Acid dissociation decreases as the shielding effect increases.

D. The bond strength between H-I is lower than that of H-Cl or H-Br. The strength of an acid is usually determined by how much the acid dissociates. Thus, the weaker the bond holding the proton to the molecule, the stronger the acid is. The bond strength is largely determined by the difference in electronegativity. Fluorine is the most electronegative elements, and thus is not as strong of an acid as HCl or HI, which have weaker bonds, and thus a more readily available proton. [D] is the answer. A. Iodine is the smallest element out of the 3 and thus creates the strongest acid. Iodine is the largest element out of the three, not the smallest. [A] is incorrect. B. The acid strength is increased as electronegativity increases in the binding element. This is a partially true statement, but it does not answer the question because acid strength is decreased as electronegativity increases in the halogens, as seen in HF. [B] is incorrect. C. The number of available electrons in iodine for binding is greater than that of Cl or Br. All of the halides have 7 available valence electrons. [C] is incorrect. E. Acid dissociation decreases as the shielding effect increases. The shielding effect increases as you go down a column, and the acid strength also increases in the halides. [E] is incorrect.

Which is the best definition of a boiling point? A. The temperature at which the vapor pressure of the liquid equals one atmosphere of pressure B. The temperature at which the liquid vaporizes C. The temperature at which the surrounding pressure becomes greater than the vapor pressure of the liquid D. The temperature at which the vapor pressure of the liquid equals the surrounding pressure E. The pressure at which a liquid can no longer vaporize

D. The temperature at which the vapor pressure of the liquid equals the surrounding pressure The boiling point of a liquid is reached when the vapor pressure of the liquid equals the surrounding environment's pressure. This is why liquids boil faster at higher altitudes. There is less pressure, so the vapor pressure of the liquid can be raised to a lesser value in order to equal the surrounding pressure. [D] is the answer. A. The temperature at which the vapor pressure of the liquid equals one atmosphere of pressure This answer choice would be correct if the question asked for the definition of the normal boiling point, which is when the vapor pressure of the liquid equals sea level pressure, or 1 atm. Eliminate [A]. B. The temperature at which the liquid vaporizes This answer choice is not specific enough and is not the best answer choice. For example, a liquid can vaporize from evaporation, when it is not boiling. Eliminate [B]. C. The temperature at which the surrounding pressure becomes greater than the vapor pressure of the liquid This occurs in the liquid phase. Eliminate [C]. E. The pressure at which a liquid can no longer vaporize It would be impossible for a substance to boil under these conditions. Eliminate [E]. Topic:Liquids and Solids

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. ease with which an atom gains an electron. Electron affinity is the amount of energy release when an atom gains an electron. This can be restated as the ease with which a neutral atom gains an electron. [D] is the answer. A. tendency to attract electrons towards the atom and form a bond. This is electronegativity, which is estimated based on ionization energies and electron affinities. Eliminate [A]. B. chemical reactivity of the valence electrons of the atom. The reactivity of valence electrons depends on their distance from the nucleus. Eliminate [B] C. amount of energy absorbed when an electron is added to an atom. Energy is typically released when an atom gains an electron. Eliminate [C]. E. average distance between the electrons and nucleus. This is the atomic radius. Eliminate [E].

How many electrons are transferred in this half-reaction?Mn2+ → MnO4- A. 2B. 3C. 5D. 7E. 14

DAT Pro-tip: On the DAT, you won't be expected to fully balance out entire redox equations, simply because it takes too much time. Instead, you'll be asked to do a part of a balance redox equation. Find the oxidation states of manganese on the reactant side and on the product side, to solve for the number of electrons that were transferred in this half-reaction. The oxidation state of manganese in the reactant is +2. The oxidation state of manganese in the product is a little trickier. Setup an algebraic equation: Mn + (standard state of oxygen)(number of oxygen atoms) = overall charge of compound x + (-2)(4) = -1 x = 7 The oxidation state of Mn on the product side is +7. The difference between Mn2+ and Mn7+ is 5 electrons.

N2(g) + 3H2(g) ⇌ 2NH3(g) ; ΔH° = -92.22 kJ•mol-1 Which change would decrease the Keq in the above reaction? A. Adding an iron catalyst B. Decreasing the volume of the vessel C. Injecting H2 gas D. Cooling ammonia gas into a liquid E. Increasing the temperature

DAT Pro-tip: The trick to this question is to realize the only thing that can change the Keq is changing the temperature, thus Choice [E] must be the correct answer. All the other answers will not affect the equilibrium or only affect the reaction quotient, Q. The easiest way to think about the ΔHo of a reaction is to add it into the chemical equation. A negative ΔHo means the reaction is exothermic, so energy is given off, or it is part of the products. The chemical reaction can be rewritten as: N2(g) + 3H2(g) ⇌ 2NH3(g) + energy Keq is the concentration of products divided by the concentration of reactants: Keq= Product/Reactant Decreasing the Keq requires increasing the amount of reactants, so look for the answer choice that will will shift the equilibrium towards the reactants, or towards the left. E. Increasing the temperature Increasing the temperature will shift the equilibrium towards the left, or towards the reactants, and thus decrease the Keq. [E] is the answer. A. Adding an iron catalyst Adding a catalyst does not change the Keq. [A] is incorrect. B. Decreasing the volume of the vessel Decreasing the volume of the vessel will increase the pressure, and the reaction will shift towards the side that has less moles of gas to reach equilibrium, or in this case towards the products, thus increasing Q. [B] is incorrect. C. Injecting H2 gas Injecting H2 gas will push the equilibrium towards the products, increasing Q. [C] is incorrect. D. Cooling ammonia gas into a liquid Lastly, cooling ammonia gas into liquid will pull the reaction towards the right to produce more products, increasing Q. [D] is incorrect.

Acetic acid (Ka = 1.8 x 10-5) is titrated with KOH. Which pH indicator should you use to signal the end point of the titration? A. Thymol blue; pH = 2 - 2.5 B. Methyl yellow; pH = 3 - 4 C. Bromocresol purple; pH = 5 - 7 D. Cresol red; pH = 7.5 - 9 E. Alizarine Yellow; pH = 10.5 - 12

DAT Pro-tip: This looks like an intimidating problem and that you have to solve for the pH, but no calculation is required. Instead, an understanding of titration techniques is required. For the titration of a weak acid with a strong base, the equivalence point will be a weak base. For the pH indicator to properly signal the end point (or the "equivalence point"), the indicator's pH range must be similar to the pH at the equivalence point. Only cresol red's pH indicator is in the weak base range. [D] is the answer.

An unknown gas has a density of 1.79 g•L-1 at 273K and 1 atm. Identify the gas. A. HeB. NeC. ArD. KrE. Xe

DAT Pro-tip: This may look like a challenging problem, but you still don't need a calculator. In fact, you don't even need the formula for gas density (you could use it, but it'd be difficult without a calculator). All that you need to know is that there is 1 mol in 22.4L at STP (273K and 1 atm). Commit 22.4L in 1 mol of gas at STP to memory, it'll help you out a lot on the DAT. From this information you can find the molar mass of the gas, which will reveal the identity of the gas: The only gas close to 40 g•mol-1 is argon, Ar, at 39.9g•mol-1.

A student measures the mass of a compound to be 9.01 g and its volume to be 11.0 mL. What is the density in g/mL? A. 0.8B. 0.82C. 0.819D. 0.8191E. 0.81909

DAT pro-tip: This problem isn't testing your calculation skills; it is testing your knowledge of significant figures. For multiplying and dividing, the number of significant figures is determined by the measurement with the fewest number of significant figures. For adding and subtracting, the answer must have the same number of decimal places as the measurement with the fewest number of decimal places. There are 3 significant figures in 9.01g and 3 significant figures in 11.0mL, so the answer must have 3 significant figures in this division problem. [C] is the answer.

What is the boiling point of a solution of 5.0 mol of glucose dissolved in 1000 mL water? (Kb for water = 0.512oC / m; assume density of water = 1g/mL) A. 2.6°CB. 97.4°CC. 100°CD. 102.6°CE. 105.2°C

Dissolving a solute in a solvent will cause a boiling point elevation. The boiling point of pure water is 100°C. DAT pro-tip: Answer choices [A], [B], and [C] can immediately be eliminated because their boiling point is at 100°C or below. They do not have the boiling point elevation expected from dissolving a solute in water. Use the boiling point elevation formula: i is the van't Hoff factor. Glucose does not dissociate in water, so its van't Hoff factor is one. Kb is the boiling point elevation constant, 0.512 oC/m. m is the molality, mol of solute/kg of solvent. The density of water is 1 g/mL. 1000 mL of water contains 1000 g of water, equal to 1 kg water. There are 5 mol of glucose in 1 kg solute, so the molality is 5 mol/kg. Plug in the values and solve: Because the boiling point of water is 100°C, the boiling point of the solution is 102.5°C, increased by 2.5°C. [D] is the answer. Frequently Asked Question: Why do we have to add 100 to get the correct answer? Answer: Remember that after you compute the calculation, you're figuring out the boiling point ELEVATION, not the boiling point. This is the rise in boiling point from the standard boiling point. The boiling point of water is 100 oC, so the boiling point elevation must be added to the original boiling point to get the new boiling point. Topic:Chemical Solutions

Which of the following liquids would have the highest volatility? A. A liquid with hydrogen bonds B. A liquid with ionic bonds C. A liquid with a high viscosity D. A metal in liquid form E. A liquid with London dispersion forces

E. A liquid with London dispersion forces Volatility is the ability of a liquid to evaporate. Liquids that readily evaporate have weak intermolecular attractions and can turn into gas at low temperatures. London dispersion forces are weak intermolecular forces that arise from instantaneous dipoles in molecules without dipoles. The weaker the intermolecular force, the higher the volatility. [E] is the answer. A. A liquid with hydrogen bonds Hydrogen bonds are stronger intermolecular forces. This liquid would not evaporate as easily. Eliminate [A]. B. A liquid with ionic bonds Ionic bonds are stronger than intermolecular forces. This liquid would not evaporate as easily. Eliminate [B]. C. A liquid with a high viscosity A liquid with a high viscosity has strong bonds and will not evaporate at low temperatures. Viscous liquids include substances like honey and molasses, neither of which are able to evaporate easily. Eliminate [C]. D. A metal in liquid form A liquid metal would not readily evaporate. Eliminate [

Which statement best explains why bromine is a liquid and iodine is a solid at STP? A. At a given temperature, bromine has a higher vapor pressure than iodine. B. Smaller atoms tend to be more polarizable and exhibit greater intermolecular forces. C. The dipole-dipole interactions in iodine are stronger than in bromine. D. Elements in the same column tend to have different chemical properties. E. London dispersion forces increase as molar mass increases.

E. London dispersion intermolecular forces increase as molar mass increases. As molecules get heavier, they exhibit a greater amount of London dispersion forces (nonpolar forces), and these forces become so powerful that iodine actually turns into a solid. [E] is the answer. A. At a given temperature, bromine has a higher vapor pressure than iodine. While this may be true because bromine is a liquid, it does not explain why bromine is a liquid or iodine is a solid. Vapor pressures are a result of intermolecular forces and internal energy, not a cause of physical states of matter. Eliminate [A]. B. Smaller atoms tend to be more polarizable and exhibit greater intermolecular forces. Larger molecules tend to be more polarizable and exhibit greater intermolecular forces. Eliminate [B]. C. The dipole-dipole interactions in iodine are stronger than in bromine. The interactions in iodine and bromine (I2 and Br2) are non-polar. Both of these molecules do not have dipole-dipole interactions. Eliminate [C]. D. Elements in the same column tend to have different chemical properties. Elements in the same column tend to have similar chemical properties. Even if this was true, it doesn't address the question at hand. Eliminate [D].

Three elements in the same period are listed in order of decreasing atomic radius. Which of the following is an appropriate explanation for the non-metal in the list having the smallest atomic radius? A. Non-metals have an expanded octet. B. Non-metals have higher electronegativities. C. Non-metals have fewer electron orbitals. D. Non-metals have higher ionization energies. E. Non-metals have a higher effective nuclear charge.

E. Non-metals have a higher effective nuclear charge. These elements are arranged in the same period of the periodic table. A period is a horizontal row of the periodic table. Rows on the periodic table are very diverse as they may contain metals, metalloids, and non-metals, as you move across the periodic table from left to right. In the same period, across the periodic table, elements' number of protons increases. The negative electrons are more attracted to the positive nucleus resulting in a smaller atomic size. The higher the effective nuclear charge, the closer electrons are drawn to the nucleus. [E] is the answer. A. Non-metals have an expanded octet. Expanded octets have no relationship with atomic size. Eliminate [A]. B. Non-metals have higher electronegativities. The number of protons determines electronegativity, but electronegativity does not explain the atomic radius trend. Eliminate [B]. C. Non-metals have fewer electron orbitals. Non-metals' number of electron orbitals varies. Eliminate [C]. D. Non-metals have higher ionization energies. Ionization energy is a result of atomic size. Eliminate [D].

Gas A has a molar mass of 4 g/mol and gas B has a molar mass of 36 g/mol. If the gases are at the same temperature, what will be the rate of effusion of gas B relative to gas A? A. Rate of gas B = 1⁄9 × rate of gas A B. Rate of gas B = 1⁄4 × rate of gas A C. Rate of gas B = 1⁄3 × rate of gas A D. Rate of gas B = 3 × rate of gas A E. Rate of gas B = 9 × rate of gas A

Effusion of a gas is the movement of a gas through a small opening, from one compartment into another. The relative rate of effusion can be calculated via Graham's law: Plug in the values: The rate of gas A is three times that of gas B. The rate of gas B is one third the rate of gas A. [C] is the answer.

For a given substance, arrange the following values from least to greatest in magnitude: I. Enthalpy of sublimation, ∆Hsub II. Enthalpy of fusion, ∆Hfus III. Enthalpy of vaporization, ∆Hvap A. I < III < II B. II < III < I C. II < I = III D. III = II < I E. III < II < I

Fusion = solid → liquid (melting) Vaporization = liquid → gas (boiling) Sublimation = solid → gas If you remember the relationship between these phase changes and their associated enthalpies, you should have realized that they all endothermic and therefore involve a positive change in heat (+∆H). In terms of arranging magnitudes, sublimation will have the highest ∆H. It involves the most dramatic phase change out of all of them because it denotes a transition that essentially skips over one of the phases. The question is, which one has a higher ∆H between fusion and vaporization? It is useful to note that the ∆Hvap for a given substance is always significantly higher than its ∆Hfus. Think about it - the transition from a liquid to a gas involves creating a lot more disorder and degree of separation between molecules than the change between a solid and a liquid. This difference can be seen in the image below: https://upload.wikimedia.org/wikipedia/commons/8/89/States_of_matter_En.svg Heat of vaporization therefore involves more energy, making the correct answer choice for this question [B]

What volume of 0.5M H2SO4 is needed to neutralize 250mL of 0.5M KOH? A. 75mLB. 125mLC. 250mLD. 333mLE. 500mL

H2SO4 has two moles of H+ ions (n1=2), while OH only has one mole of OH-(n2=1). Therefore, we must remember to use the normality of the acid in this dilution reaction. (n1)(M1)(V1) = (n2)(M2)(V2) (2)(0.5)(V1) = (1)(0.5)(250) V1 = 125 mL

In a given electrochemical cell, a spontaneous reaction occurs where metal A forms the cathode and metal B forms the anode. Based on the information provided, which of the following statements is/are correct? I. Oxidation occurs at the anode II. The reduction potential of metal A > the reduction potential of metal B III. This reaction represents an electrolytic cell A. I onlyB. II onlyC. I and II onlyD. I and III onlyE. I, II, and III

I. Oxidation occurs at the anode Use the mnemonic AN OX RED CAT: oxidation occurs at the anode, and reduction occurs at the cathode. I is true. Eliminate [B]. II. The reduction potential of metal A > the reduction potential of metal B This is a spontaneous reaction so the reduction potential of the cathode (reduction half cell) must be greater than the reduction potential of the anode (oxidation half cell). Metal A forms the cathode, and metal B forms the anode. The reduction potential of metal A is greater than the reduction potential of metal B. II is true. Eliminate [A] and [D]. III. This reaction represents an electrolytic cell A spontaneous electrochemical cell is a voltaic or galvanic cell. A non-spontaneous electrochemical cell is an electrolytic cell, which requires the input of energy. III is false because this is a spontaneous reaction. Eliminate [E]. Options I and II are true, and option III is false. [C] is the answer.

Consider the reaction below and the provided experimental rate data. If the rate law for this reaction is rate = k[NO2]2, what is the rate constant? 2NO2 + F2 → 2NO2F

It is possible for DAT questions to include more information than what is necessary to solve a problem. Even though it seems like this could be a complicated question, the correct answer can be found from just manipulating the provided rate law equation! Before we get started with any equations, it can be smart to quickly scan the answer choices. Pay attention to the rate law: rate = k[NO2]2. Notice how it is a second order reaction. We know this because in order to determine a rate law's overall reaction order, we just have to add up all the exponents. This rate law only has one exponent with a value of 2, and therefore we refer to it as being second order. Next, check out the table below. For a second rate equation, the units of the rate constant must be M-1s-1. This allows us to eliminate choices [A] and [E] right off the bat! Now let's work on determining the exact answer. To make things clear, they are giving us a rate law and asking us to find the rate constant k, rate = k[NO2]2 k = rate / [NO2]2 In order to go further, we need to figure out the rate, as well as the concentration of NO2. Luckily, this question gives us a nice table with some experimental rate data, precisely the information we need for a kinetics problem. At this point, it is as easy as selecting any of the experiments 1-3 and plugging in the NO2 concentration and the rate for that NO2 concentration. You don't even need to look at the [F2] column — F2 is not a part of the rate law and therefore totally irrelevant towards our aim of finding k. Here is an example where we insert the values from Experiment 1, k = rate / [NO2]2 k = 4.0 x 10-5 / (0.1)2 *don't forget the exponent on the [NO2]! k = 4.0 x 10-5 / (0.01) k = 4.0 x 10-5 / 1.0 x 10-2 k = 4.0 x 10-3

Which of the following is a characteristic of network covalent solids? A. Malleability B. Conductivity C. High melting point D. High luster E. Frail

Network covalent solids are solids that consist of long chains of covalent bonds. These solids have higher melting points and are harder than molecular solids because covalent bonds are stronger than the intermolecular forces that hold molecular solids together. Luster, conductivity, and malleability (being able to hammer a solid into sheets) are all characteristics of a metal. A network covalent solid does not contain metals (it only contains non-metals) so it does not have any metal characteristics. Eliminate [A], [B], and [D]. Frailty is not a characteristic of a network covalent solid or a metal. Eliminate [E]. [C] is the answer.

The following exothermic reaction was allowed to reach equilibrium: P4(s) + 10Cl2(g) ⇌ 4PCl5(g) + heat Which of the following changes would increase the amount of PCl5(g)? A. Increasing the volume of the container B. Decreasing the temperature C. Adding more P4(s) D. Adding a nonreactive gas such as He E. Adding a catalyst

P4(s) + 10Cl2(g) ⇌ 4PCl5(g) + heat B. Decreasing the temperature Decreasing the temperature of an exothermic reaction will cause the reaction to shift right, towards the heat product. [B] is the answer. A. Increasing the volume of the container This will decrease pressure. Because there are more gaseous molecules on the reactant side than on the product side, the reaction will shift left. Eliminate [A]. C. Adding more P4(s) This is a pure solid, which is not included in the equilibrium expression. Adding more P4 will not cause a shift in the reaction, or disturb equilibrium. Eliminate [C]. D. Adding a nonreactive gas such as He He is not in the equilibrium expression for this reaction. He is a nonreactive gas that has no effect on the equilibrium. Eliminate [D]. E. Adding a catalyst Adding a catalyst would speed up the reaction, causing it to reach equilibrium faster. However, a catalyst will not affect the position of equilibrium. Eliminate [E].

An equilibrium mixture for the reaction A⇌B has a Kc = 0.40 at 350K. Which of the following statements is true regarding a A⇌B mixture with a Qc = 20 at 350K? A. The reaction will proceed to the right, forming more product to reach equilibrium. B. Once the reaction reaches equilibrium, the concentration of B will be greater than A. C. The concentrations of A and B are equal, and the reaction has reached a state of equilibrium. D. To reach equilibrium, the concentration of A will increase while the concentration of B will decrease. E. Kc will increase and Qc will decrease, and eventually equalize at Kc = 10, Qc = 10.

Qc is the reaction quotient. Q measures the relative amounts of products and reactants at a specific time that usually is not at equilibrium; its value varies. Kc is the equilibrium constant. K does not change because it is a constant. Compare the magnitude of Q and K to identify which direction the reaction will proceed to reach equilibrium. Q is greater than K (20 > 0.40). A⇌B Q and K are both measured as products over reactants, [B]/[A]. To reach equilibrium the reaction will shift left, decreasing the value of Q to equal that of K. The concentration of A will increase, and the concentration of B will decrease. [D] is the answer.

An unknown solution has its initial pH measured, is titrated, and the resulting analysis of the titration is shown below. Which of the following could be the compound in the original solution?

The acid-base equivalence point is the point at which an acid or base being added through titration is stoichiometrically equal, or equivalent, to the number of moles of the base or acid it is being added to. In the titration given, the initial solution has a low pH, and as titrant is added, the pH rises. The initial solution is acidic, and the titrant is basic. Eliminate [B] and [E] because they are bases. A polyprotic acid titration has more than one equivalence point. In a diprotic acid, there are two equivalence points with two curves in the titration graph, as is shown in the question. The acid must be able to lose two protons. H2SO3 is the only diprotic acid given.

What is the sum of the coefficients of the complete balanced equation from combusting C2H6O? A. 6 B. 9 C. 12 D. 18 E. 24

The balanced equation for the combustion of ethanol (C2H6O) is: 1 C2H6O + 3 O2 → 2 CO2 + 3 H2O From the balanced equation, we find 1 + 3 + 2 + 3 = 9. [B] is the answer.

Which of the following combinations of elements would most likely have the smallest bond length? A. An atom with low ionization energy and another atom with high metallic character B. An atom with high ionization energy and another atom with high electron affinity C. An atom with high metallic character and another atom with high atomic radius D. An atom with low electron affinity and another atom with low ionization energy

The bond length between two atoms is equal to the sum of the atomic radii of the two atoms. The smaller the atomic radii, the smaller the bond length. Atomic radius is smallest at the top right of the periodic table. B. An atom with a high ionization energy, and another atom with a high electron affinity. If ionization energy is high and electron affinity is high, atomic radii would be small for both atoms. [B] is the answer. A. An atom with a low ionization energy, and another atom with a high metallic character. If ionization energy is low, and metallic character is high, atomic radii would be high for both atoms. Eliminate [A]. C. An atom with high metallic character and another atom with high atomic radius. If metallic character is high, and atomic radius is high, atomic radii would be high for both atoms. Eliminate [C]. D. An atom with low electron affinity and another atom with low ionization energy. If electron affinity is low, and ionization energy is low, atomic radii would be high for both atoms. Eliminate [D].

If the bond length of H2 is experimentally found to be 74pm and the bond length of Br2 is experimentally found to be 228pm, what would be the bond length of H-Br? A. 138pm B. 142pm C. 151pm D. 156pm E. 159pm

The bond length is determined by adding the atomic radii of both of elements in a bond. To determine the individual radii of the elements, the average can be taken of the two bond lengths. H2 has a bond length of 74pm, where each hydrogen has an atomic radius of 37pm. Br2 has a bond length of 228pm, where each bromine has an atomic radius of 114pm. The H-Br bond would have a length of 37pm + 114pm, or 151pm. [C] is the answer.

A student finds that the average number of cells over 10 samples is 2130, with a standard deviation of ±6. However, the professor determines the true average number of cells is 2850. The student's measurement was A. inaccurate and precise. B. accurate and imprecise. C. inaccurate and imprecise. D. accurate and precise.

The definition of accuracy is that the measurement taken is close to the true value, whereas the definition of precise is that the measurement produces consistent results. The student's measurement was inaccurate, because the measured value differed significantly from the true value. However, the measurements were precise because the standard deviation is low compared to the total number of cells. The variation of each measurement was low.

The valence electron in ground state lithium can be described by which set of quantum numbers? A. 2, 0, 0, +1/2 B. 2, 0, 1, +1/2 C. 2, 1, 1, +1/2 D. 3, 0, 0, -1/2 E. 3, 1, 0, +1/2

The electron configuration of ground state lithium is 1s22s1. The 2s1 describes the valence electron specifically, where n = 2 (second period). This is the principal quantum number. The electron is in the s orbital due to being in the first two columns, so l = 0. This is the azimuthal quantum number. According to the table below, ml must be 0, and ms can be +1/2 or -1/2. [A] is the answer.

The empirical formula of a compound is found to be CO3. What is the molecular formula if the molar mass of the compound is 180 g/mol? A. CO9 B. C2O6 C. C3O9 D. C4O12 E. C6O18

The empirical formula is the lowest whole number ratio of the elements. Solve by dividing the molar mass by the empirical formula's mass: CO3 = 60 g/mol 180 / 60 = 3x multiplier This indicates that the subscripts in the empirical formula must be multiplied by 3 to obtain the molecular formula: CO3 × 3 = C3O9 [C] is the answer.

Two separate solutions are created. One using water and KCl, and the other using water and CaCl2. Assuming the molality is equal in both solutions, which of the following is true? A. KCl will have the lower freezing point. B. KCl will have the higher boiling point. C. CaCl2 will have the lower boiling point. D. CaCl2 will have the higher boiling point. E. KCl and CaCl2 will have the same freezing point.

The equation to calculate the change in boiling point or freezing point is: ∆T = m × Kb × i The molality is the same for both solutions and can be ignored. Both ionic compounds are dissolved in water, so the Kb is the same for both equations, and can be ignored. The difference between the two compounds is the number of the ions that each solution will produce. KCl dissolves into 2 ions and CaCl2 dissolves into 3 ions. The Van't Hoff factor, i, is larger for CaCl2. CaCl2 will have a higher boiling point and lower freezing point. [D] is the answer. Topic:Chemical Solutions

H2O + CO2 → H2CO3 Two steps are needed to form the above reaction. If the first step must be reversed and halved and the second step must be doubled, which of the following would be the correct mathematical representation? A. B.C.D.E.

The first step of the reaction is reversed and halved, so the enthalpy of step one must be multiplied by -1/2. The second step had is doubled, so the enthalpy of step two must be multiplied by 2. The enthalpies of the two steps can be added together to find the total enthalpy of the reaction:

What is the final volume (in mL) when 300mL of 0.8 M KCl is concentrated to 0.9 M KCl?

The formula for concentrations and volumes is: M1V1 = M2V2 Plug in the values given: (0.8)(300) = (0.9)(V2) (0.8/0.9)(300) = V2 [B] is the answer. DAT pro-tip: If you're using the same units on both sides of the equation, then you don't need to worry about changing the units. Answers [C],[D], and [E] are all designed to distract you with the value 1000 in the solution, encouraging you to convert from mL to L.

A first order reaction has a half-life of 90 minutes. What is the rate constant in hour-1? A. 90 / (0.693)B. (0.693) / 1.5C. (0.693)(1.5)D. (0.693) / 90E. (0.693)(90)

The half-life for a first order reaction is: t1/2 = 0.693/k We are given time in minutes and the final rate constant should be in hours. There are 1.5 hours in 90 minutes. Substitute in the values given: k = (0.693)/1.5 [B] is the answer.

After 2.2 billion years, how much of a 100-gram sample of Uranium-235 would remain, assuming a half-life of 704 million years? A. B.C.D.E.

The half-life is the amount of time it takes for one half of a substance to decay. The equation is: Amount remaining = original amount x (1/2)t The number of half-lives is represented by t. The half-life of Uranium-235 is 704 million years. 2.2 billion years is equivalent to 2,200 million years. The number of half-lives that have passed is 2,200 / 704. Plug in the values: Amount remaining = 100(1/2)(2200/704) [A] is the answer.

A mixture of gas is composed of 0.5 mol Ar, 1.5 mol He, and 2.0 mol of N2. What is the partial pressure of N2 in the mixture if the total pressure is 0.95 atm? A. 0.10 atmB. 0.48 atmC. 0.95 atmD. 1.7 atmE. 3.8 atm

The partial pressure of a gas, P1, in a mixture is related to the mole fraction of the gas x1, and the total pressure of the mixture, PT, according to the equation: The mole fraction of N2 is equal to the moles of N2 divided by the total number of moles. There are four total moles of gases present. The mole fraction of N2 is: The partial pressure of N2 is: [B] is the answer.

Consider a hypothetical reaction 2A + B → 3C, with a rate that is second order with respect to B and zero order with respect to A. How would doubling the concentration of A and increasing the concentration of B by a factor of three affect the rate of this reaction? A. The rate will increase by a factor of four. B. The rate will increase by a factor of two. C. The rate will increase by a factor of six. D. The rate will increase by a factor of three. E. The rate will increase by a factor of nine.

The rate law is the concentration of the reactants with the reactant order equalling the power it is raised to, multiplied by one another. Reactant A is zero order and has an exponent of zero, so it will not affect the rate. Reactant B is second order, and has an exponent of two. The rate law for this reaction is therefore: Rate = k[B]2 Increasing the concentration of B by a factor of three will increase the rate by a factor of nine (32 = 9). Increasing the concentration of A won't affect the rate since the reaction is zero order in A. [E] is the answer.

What is the molar solubility of PbSO4 in 4.0 × 10-2 M Pb(NO3)2 solution? The Ksp for PbSO4 is 1.6 × 10-8 A. 4.0 × 10-7 M B. 6.4 × 10-10 M C. 3.2 × 10-6 M D. 1.6 × 10-8 M E. 8.0 × 10-10 M

This is a common ion problem. You should recognize that there are already Pb2+ ions in the solution, so PbSO4 will be even less soluble. Setup an ICE table to solve: PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq) PbSO4. Pb2+. SO42- Initial. 0.04. 0 Change. +x. +x Equilibrium (0.04 + x). x Next, setup the equilibrium expression: Ksp = [Pb2+][SO42-] (0.04 + x)(x) Assume that x is a very small number and that (0.04 + x) is approximately 0.04: 1.6 × 10-8 = 0.04x x = 4 × 10-7 = molar solubility [A] is the answer. DAT pro-tip: If you move the decimal places and just divide the integers, you see that 16/4 = 4, and there is only one answer choice with 4 as the stem, leading you to answer [A].

Sulfuric acid is stored in a protective container to prevent it from corroding the container. If 250mL of the 4.5M stock solution of sulfuric acid is to be transferred to an empty flask, what volume of water in liters must be added to the flask to reduce the sulfuric acid's molarity to 2.2M? A.B.C. D.E.

This is a dilution problem because the molarity is being decreased. The dilution equation is: M1V1 = M2V2 The question asks for the volume in liters not milliliters, so the initial volume must be converted from 250mL to 0.250L. Substitute in the given values: (4.5)(.250) = (2.2)(V2), (4.5)(0.250) / 2.2 = V2. This is the final volume, which is the initial volume plus the added water. The actual volume of water added is: final volume - initial volume = volume of water added ((4.5)(0.250) / 2.2) - 0.250 = liters of water added [D] is the answer.

Ammonia is formed according to the reaction below. A chemist mixes 21 grams of nitrogen gas and 18 grams of hydrogen gas in a 2.0 L vessel. How many grams of hydrogen gas will be consumed? N2(g) + 3H2(g) → 2NH3(g) A. 0.75 gramsB. 1.1 gramsC. 4.5 gramsD. 9.0 gramsE. 18 grams

This is a limiting reagent problem. Calculate how many moles of each reactant are present: According to the balanced chemical equation, one N2 is consumed for every three H2. Consuming 0.75 moles of N2 requires 2.25 moles of H2. There are 9 moles of H2 present, so H2 is in excess and N2 is the limiting reagent. N2(g) + 3H2(g) → 2NH3(g) If 2.25 moles of H2 are consumed, and H2 has a molar mass of 2 g/mol, 4.5 grams of hydrogen will be consumed in this reaction. [C] is the answer.

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 beakerB. To prevent the water from sinking beneath the acid and remain unmixedC. To maximize the ionization of the acid being dilutedD. To ensure there is enough water to absorb the heat releasedE. To solvate the conjugate base to prevent a reaction

This is an important rule in chemistry safety. Always add acid to water, never the other way around. If water is added 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. [D] is the answer. Topic:Laboratory Techniques

The pressure of a gas is increased from 100 mmHg to 120 mmHg. If the initial temperature of the gas was 300K, which of the following could be the final temperature of the gas? A. 310K B. 360K C. 400K D. 600K E. 610K

This problem includes pressure and temperature, which involves Gay-Lussac's law. No calculations are needed for this problem if you know the relationship between pressure and temperature. Pressure and temperature are directly related. If the pressure is increased, the temperature must also increase. The pressure increased by about 20%, so temperature must do the same, increasing to about 360K. The only answer choice close to this approximation is [B]. [B] is the answer. DAT Pro-tip: To approximate, take 10% of 300K, equal to 30K. Double this to get 20%, or 60K. Add this value to the original 300K, for a 20% increase. Topic:Gases

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)? A. 5 mLB. 10 mLC. 15 mLD. 30 mLE. 50 mL

Use the dilution equation to solve: M1V1 = M2V2 Plug in the given values and solve for V2: (3.00 M)(10 mL) = (2.00 M)(V2) V2 = 15 mL The question asks how much water should be added to the original solution, so subtract by the original amount, V1: 15 - 10 = 5 mL

If the Ka of acetic acid is 1.8 × 10-5, what is the pKb of acetate? A. 14 - log (1.8 × 10-5)B. 14 + log (1.8 × 10-5)C. - log (1.8 × 10-5)D. -14 + log (1.8 × 10-5)E. -14 - log (1.8 × 10-5)

Use the equation: pKa + pKb = 14 Isolate pKb: pKb = 14 - pKa Remember that: pKa = -log(Ka) Plug this in for pKa in the above equation: pKb = 14 - (-log(Ka)) Plug in the given Ka value: pKb = 14 - (-log(1.8 × 10-5)) Cancel out the two negatives: pKb = 14 + log (1.8 × 10-5) [B] is the answer. DAT Pro-tip: The "p" in pKa and pH stands for "-log".

15 moles of a N2 gas sample are placed in a 30L container at 33°C. What is the pressure (in atm) exerted by this gas sample? A.B. C.D.E.

Use the ideal gas law: PV = nRT R, the gas constant, is 0.0821, not 8.314, because the question asks for the pressure in atmospheres. Substitute the given values: P(30) = (15)(0.0821)(273 + 33) P(30) = (15)(0.0821)(306) P = ((15)(0.0821)(306)) / 30

Which of the following is the most likely formed compound between atoms X and Y? Atom X: 1s22s22p63s2 Atom Y: 1s22s22p4 A. X3YB. X2YC. X2Y3D. XYE. XY2

Use the periodic table. Based on the given electron configurations, X is magnesium and atom Y is oxygen. These atoms are most often found in with a +2 and -2 charge in compounds, respectively. Therefore, we only need one of each atom to form the likely compound MgO, or magnesium oxide.

When the following equation is balanced using the smallest whole numbers, what is the coefficient in front of H2(g) on the product side of the reaction? Al(s) + HCl(aq) → AlCl3(aq) + H2(g) A. 1B. 2C. 3D. 5E. 6

Balance the equation: 2 Al(s) + 6 HCl(aq) → 2AlCl3(aq) + 3 H2(g) H2 has a coefficient of 3. [C] is the answer.

The solvation of ionic compounds in water is due to which of the following? A. Reactions with H3O+ and OH- B. Diffusion C. Dipoles of water D. High specific heat of water E. Cohesive and adhesive forces.

C. Dipoles of water Water's structure allows it to dissolve compounds. In O-H bonds, electrons are more attracted to the electronegative oxygen than the less electronegative hydrogen. Oxygen takes on a slightly negative charge and the hydrogens take on a slightly positive charge. These charges attract cations and anions in an ionic compound, dissolving them. [C] is the answer. A. Reactions with H3O+ and OH- Reactions with acid and base in water would change the ionic compound, but this does not answer the question. [A] is incorrect. B. Diffusion This is a solution already and diffusion not the driving force behind the solvation of the ionic compound. [B] is incorrect. D. High specific heat of water The high specific heat of water is irrelevant to the solvation of an ionic compound. [D] is incorrect. E. Cohesive and adhesive forces Cohesive and adhesive forces, which dictate capillary action, are irrelevant to the solvation of an ionic compound. [E] is incorrect.

All of the following will affect the rate of an irreversible chemical reaction EXCEPT one. Which one is this EXCEPTION? A. Pressure B. Concentration of reactants C. Presence of a catalyst D. Surface area of reactant solid E. Concentration of products

A chemical reaction occurs when two or more reactant molecules hit each other with enough force and in the proper orientation. The more the molecules hit each other, the greater the chance for a reaction, and the faster a reaction can proceed. E. Concentration of products The concentration of products will not affect the rate of an irreversible reaction. Product molecules colliding into each other will not help the forward reaction to occur quicker. Choice E is the EXCEPTION. [E] is the answer. A. Pressure Increasing the pressure will increase collision frequency, resulting in a change in the rate of the reaction. Eliminate [A]. B. Concentration of reactants Increasing the concentration of the reactants will also affect the frequency of molecular collisions. Eliminate [B]. C. Presence of a catalyst A catalyst will lower the activation energy required and speed up a reaction. Eliminate [C]. D. Surface area of reactant solid More surface area of a reactant will give more opportunities for molecules to strike to cause a reaction. Eliminate [D].

A volumetric flask weighs 185 g when empty and 380 g when completely filled with "liquid A" (density of 2 g•mL-1). If the flask is filled with 160 g of "liquid B", what is the density of "liquid B" in g•mL-1? A. B.C.D.E.

A volumetric flask is a piece of lab equipment that measures volume very accurately. Density is mass divided by volume. The density of liquid A is 2 g•mL-1. If the flask weighs 185 g when empty and 380 g when filled with liquid A, then the mass of liquid A is 380 g - 185 g. Use these values to convert into units of 1/mL: The mass of liquid B that is added is 160 g. Using dimensional analysis, dividing the mass of liquid B by the volume of the flask will give us the density of liquid B, in g•mL-1:

Buried steel propane tanks are often attached to a zinc plate to minimize corrosion. Which of the following explains why the zinc plate will corrode instead of the steel tank? A. Zinc acts like an anode and is more easily oxidized than steel. B. Zinc is more easily reduced than steel. C. Steel acts like a cathode and is more easily oxidized than zinc. D. Steel is more easily oxidized than zinc. E. Steel acts like an anode and is more easily reduced than zinc.

A. Zinc acts like an anode and is more easily oxidized than steel. Remember the mnemonic "RED CAT", and "AN OX". Steel tanks can be protected from corrosion by acting as the cathode in an electrochemical cell. Zinc will be oxidized, and zinc will corrode instead of the steel tank. This is called cathodic protection. [A] is the answer. B. Zinc is more easily reduced than steel. Zinc is oxidized. Eliminate [B]. C. Steel acts like a cathode and is more easily oxidized than zinc. When steel is oxidized it is acting as an anode. Eliminate [C]. D. Steel is more easily oxidized than zinc. Zinc is more easily oxidized than steel. Eliminate [D]. E. Steel acts like an anode and is more easily reduced than zinc. Reduction does not take place at the anode. Eliminate [E].

Which of the following has the highest boiling point? A. NeB. ArC. KrD. XeE. Rn

All elements given are noble gases. These exist as monoatomic gases (one atom) and are nonpolar. The only intermolecular forces within noble gases are Van der Waals forces. Traveling down a group, or column, of the periodic table the atomic size increases. As the atoms become larger with a greater mass, the Van der Waals forces increase, and boiling point rises. Rn has the highest mass and the highest boiling point.

When the following half-reaction is balanced in acidic medium, how many hydrogens are added to the left hand side of the half reaction after balancing? Cr2O72-(aq) → Cr3+(aq) A. 2B. 4C. 7D. 8E. 14

Balance the redox reaction in acidic medium: Cr2O72-(aq) → Cr3+(aq) Balance all atoms other than H and O: Cr2O72-(aq) → 2Cr3+(aq) Balance O by adding H2O: Cr2O72-(aq) → 2Cr3+(aq) + 7H2O Balance H by adding H+: Cr2O72-(aq) + 14H+→ 2Cr3+(aq) + 7H2O There are 14H+ on the left side after balancing.

The ∆Hf for Br(g) is +193 kJ/mol. What is the bond dissociation energy of a Br-Br bond? A. +193 kJ / mol B. -193 kJ / mol C. +386 kJ / mol D. -386 kJ / mol E. +96.5 kJ / mol

Bond dissociation energies are always positive because we have to put in energy to break the bond. In other words, the bond has to absorb energy, meaning bond breaking is an endothermic process, thus the answer must be positive. Eliminate [B] and [D]. The ∆Hf for Br(g) is: (1/2) Br2(l) -> Br(g) ; ∆Hf = +193 kJ / mol If we simply double this reaction, we find: Br2(l) → 2Br(g) ; ∆Hf = (2)(+193) = +386 kJ/mol [C] is the answer.

10 g of an unknown compound are added to water to form a 7.89 molar solution. If 2 liters of solution are present, what is the molar mass of the unknown compound?

Calculate the molarity of the solution: M = mol of solute / L of solution Plugging in the given values: 7.89 = mol of solute / 2 mol of solute = (7.89)(2) Dividing the grams given in the question by the moles calculated will give the unknown compound's molar mass, in g/mol: molar mass = 10 / (7.89)(2) [E] is the answer.

An unknown solution is found to have a pOH of 4.12. Which of the following would be equal to the [H+] of the solution? A. 10-pOH B. 10-(14-pOH) C. 10(14-pH) D. 10-(14-pH) E. 10(14-pOH)

Calculate the pH: pH + pOH = 14.00 pH + 4.12 = 14.00 pH = 9.88 Calculate [H+]: 10-pH = [H+] 10-9.88 = [H+] This expression is equal to 10-(14-pOH) = [H+] [B] is the answer.

The decrease in atomic radius from left to right across a period in the periodic table can be best explained by A. an increase in the number of protons. B. an increase in atomic mass. C. an increase in the number of inner shell electrons. D. an increase in the number of neutrons. E. a decrease in atomic mass.

Changes in atomic size are due to effective nuclear charge. Effective nuclear charge is the net attraction that a valence electron experiences due to two opposing forces: nuclear attraction from positive protons drawing valence electron inward, and electron repulsion from inner shell electrons that repel valence electrons. Moving from left to right across a period, the number of inner shell electrons, which are responsible for the repulsive force, does not change. However, the number of protons within the nucleus increases. There is an increase in the attractive nuclear proton force, drawing the valence electrons in, and decreasing the atomic radius. [A] is the answer.

All of these are colligative properties EXCEPT for one. Which one is the EXCEPTION? A. Change in vapor pressure B. Freezing point depression C. Osmotic pressure D. Boiling point elevation E. Density

Colligative properties are based on the number of solute and solvent particles in a solution, not the identity of the particles themselves. Molality, measured in mol of solute/kg of solvent, is used for vapor pressure, boiling point elevation, and freezing point depression. Molarity, measured in mol of solute/L of solution, is used for osmotic pressure. Thus, dissolved species can reduce vapor pressure, decrease freezing point and increase boiling point, and change the osmotic pressure. These are colligative properties. Eliminate [A], [B], [C], and [D]. Density is an intensive property. It does not depend on the number of particles you have, but rather the identity of the substance itself. If you have a stick of copper and cut it into two pieces, each of the two new pieces still has the same density as copper and each other. [E] is the answer.

A researcher is running an experiment within a bomb calorimeter. The researcher noted that during the reaction process: pressure increased, entropy decreased, enthalpy increased, temperature increased, and heat energy remained constant. What type of reaction process was the researcher observing? A. IsenthalpicB. IsobaricC. IsentropicD. AdiabaticE. Isothermal

D. Adiabatic An adiabatic process is one in which heat energy remains constant (∆q = 0). An adiabatic process is a process where no heat is lost or gained. This matches the conditions described in the question. [D] is the answer. A. Isenthalpic An isenthalpic process is one in which enthalpy remains constant (∆H = 0). Eliminate [A]. B. Isobaric An isobaric process is one in which pressure remains constant (∆P = 0). Eliminate [B]. C. Isentropic An isentropic process is one in which entropy remains constant (∆S = 0). Eliminate [C]. E. Isothermal An isothermal process is one in which temperature remains constant (∆T = 0). Eliminate [E]. Topic:Thermodynamics and Thermochemistry

What is the OH-(aq) concentration in 9.0 M CH3NH2? The Kb for CH3NH2 is 4 × 10-4. A. 3 × 10-4 B. 3 × 10-2 C. 6 × 10-2 D. 6 × 10-3 E. 6 × 10-4

DAT Pro-tip: The formula for weak bases with Kb values and [OH-] is very similar to the formula used for weak acids. Formula for weak acids:

Combustion analysis of a sample compound containing only C and H determines there was 18 g of C and 1 g of H in the sample. What is the empirical formula of this compound? A. C2H2B. C3H2C. CH3D. C1.5HE. C2H3

Determine how many moles of carbon are present: Determine how many moles of hydrogen are present: There is 1.5 mol C and 1 mol H. Chemical formulas cannot contain decimals, so multiply each value by 2. The empirical formula is C3H2. [B] is the answer.

Water molecules in an aqueous solution will have the strongest interactions with ions with which of the following characteristics? A. Small charge and small sizeB. Small charge and large sizeC. Large charge and small sizeD. Large charge and large size

First, examine charge. A polar water molecule will be more likely to interact with an ion with a higher charge. This expands on "like dissolves like". The high polarity molecule will be attracted to a highly charged ion, and this will result in a greater interaction between the two. Eliminate [A], and [B]. Second, examine size. Think about the charged ion and water molecule as two magnets. When these magnets are spaced far apart, they barely have an effect on each other. As you bring the magnets closer together, you feel their forces increase. The same principle applies to a water molecule and a charged ion in an aqueous solution. Minimize the distance between the water molecule and charged ion by choosing an ion with a smaller size. Eliminate [D].

Which of the following compounds has the same sulfur oxidation number as SO2? A. SO3 B. S2O3 C. SO3-2 D. SO4-2 E. HSO4-

For SO2, assign oxygen an oxidation number of -2. If the oxygens each have an oxidation number of -2, and the overall compound is neutral with no charge, the sulfur must have an oxidation number of +4. A similar process can be used for each answer option, assigning an oxidation number of +1 to hydrogen and an oxidation number of -2 to oxygen. A. SO3 S + (3)(-2) = 0 S = +6 B. S2O3 2(S) + 3(-2) = 0 S = +3 C. SO3-2 S + 3(-2) = -2 S = +4 D. SO4-2 S + 4(-2) = -2 S = +6 E. HSO4- 1 + S + 4(-2) = -1 S = +6 Answer choice [C] has a sulfur with the same oxidation number as SO2. [C] is the answer.

An object releases heat into the surroundings but is NOT doing work. What best describes a quality of this object? A. Negative ΔE B. ∆U = 0 C. Positive ΔV D. Positive q E. ΔT = 0

For any system undergoing a chemical or physical change, the change in internal energy (∆E) can be calculated using the following equation: ∆E = q + w where q is the system's change in heat, and w is the amount of work done by or to the system. The question tells us that an object releases heat into the surroundings. When heat is transferred from the system to the surroundings, this means that: The reaction is exothermic q is negative Furthermore, the question tells us that the object is not doing work. If w = 0 and q is negative, then we can use the equation ∆E = q + w to conclude that ∆E is negative. Choice [A] is the answer. Further Details - useful if you had trouble with the question Q: An object releases heat into the surroundings but is NOT doing work. What best describes a quality of this object? A. Negative ΔE Correct. The question tells us that an object releases heat into the surroundings (q is negative) and does not do work (w is zero). If w = 0 and q is negative, then we can use the equation ∆E = q + w to conclude that ∆E is negative. B. ΔU = 0 Incorrect. In equation form, the first law of thermodynamics is ∆U = q + w, where ∆U is the system's change in internal energy. If w = 0 and q is negative, then we can use the equation ∆U = q + w to conclude that ∆U is negative. C. Positive ΔV Incorrect. The following equation relates work to pressure and volume (V): w = -P∆V, where P is the system's internal pressure, and ∆V is the system's internal change in volume. However, since the object is not doing work, we cannot conclude that the system is experiencing an internal change in volume. D. Positive q Incorrect. Because the object releases heat into the surroundings, q is negative. If heat were transferred from the surroundings to the system, then q would be positive. E. ΔT = 0 Incorrect. Heat and temperature are directly related. Because the object releases heat into the surroundings, q is negative and the system's temperature correspondingly decreases (∆T <

Ca2+(aq) + SO4-2(aq) → CaSO4(s) Which of the following must be true for the above reaction? A. ΔS > 0 B. ΔS < 0 C. ΔG < 0 D. ΔH > 0 E. ΔS = 0

In this reaction two aqueous solutions are coming together to form a solid precipitate. This product is more organized than the two aqueous reactants, therefore the reaction has a negative change in entropy (-ΔS). There is not enough information to determine whether this reaction is exothermic or endothermic, or the sign of ΔG. [B] is the answer.

Calculate the standard enthalpy change for the combustion of methane from the following data. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) C(graphite) + 2H2(g) → CH4(g) ΔHo = 75 kJ/mol C(graphite) + O2(g) → CO2(g) ΔHo = -570 kJ/mol H2(g) + 1⁄2 O2(g) → H2O(l) ΔHo = -390 kJ/mol A. -75 + 570 + 780 B. 75 + 570 + 780 C. -75 + (-570) + 780 D. 75 + (-570) + (-780) E. -75 + (-570) + (-780)

Manipulate the given thermochemical equations so that when they are added, the target reaction is found: C(Graphite) + 2H2(g) → CH4(g) ΔH° = 75 kJ/mol C(graphite) + O2(g) → CO2(g) ΔH° = -570 kJ/mol H2(g) + ½ O2(g) → H2O(l) ΔH° = -390 kJ/mol If an equation is reversed, the enthalpy should be multiplied by -1. If the equation's coefficients are multiplied by a value, the enthalpy should be multiplied by the same value. Add the three reactions and their enthalpy change according to Hess' Law:

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)? A. 0.10 molB. 0.20 molC. 0.30 molD. 0.40 molE. 0.50 mol

NaCl has one chlorine atom, so when 0.20 mol NaCl dissociates it will produce 0.20 mol of Cl-. MgCl2 has two chlorine atoms, so when 0.10 mol MgCl2 dissociates it will produce 0.20 mol of Cl-. In total there is 0.40 mol of Cl- in the solution. There is only one silver atom in AgNO3, so 0.40 mol of Ag will be needed to react with all 0.40 mol of Cl-, producing the precipitate AgCl (s). [D] is the answer.

Consider the following reaction: MgCO3(s) → MgO(s) + CO2(g), which of the following would be the correct equilibrium constant expression Keq? A. Keq = [CO2] B. Keq = [MgO] C. Keq = [MgO][CO2] D. Keq = ([MgO][CO2]) / [MgCO3] E. Keq = [MgCO3] / ([MgO][CO2])

Pure solids and liquids are not included in equilibrium expressions. Therefore, both the MgCO3 (s) and MgO (s) are not included in the equilibrium constant expression. The expression is: Keq = [CO2] [A] is the answer.

Aqueous ammonium sulfide is added to a solution of iron (II) chloride. What are the spectator ions in this reaction?I. NH4+II. S2-III. Cl-IV. Fe2+ A. I and III B. I and II C. II and IV D. III and IV E. II and III

Remember three solubility rules: 1. Most Group 1 metal cations, NO3-, ClO4-, C2H3O2-, and NH4+ salts are soluble.2. Most Ag+, Pb2+, S2-, OH-, Hg2+, CO32-, and PO43- salts are insoluble.3. The solubles generally trump the insolubles. Thus the reaction will be: NH4+ + S2- + Fe2+ + Cl- → FeS + NH4+ + Cl- NH4+ and Cl- appear as both a reactant and product without participating in the chemical reaction. They are spectator ions. The actual reaction that occurs is: Fe2+ + S2- → FeS [A] is the answer. Topic:Chemical Solutions

A reaction has a rate law of: R = k[A]3[B] Which of the following would happen to the rate, if the concentration of A is doubled and the concentration of B is held constant? A. It would stay the same B. It would increase by a factor of 2 C. It would increase by a factor of 4 D. It would increase by a factor of 6 E. It would increase by a factor of 8

Substance A is to the third power in the rate law. Doubling the concentration would result in a rate increase by a factor of 23 or 8. If the concentration of substance B is held constant, it will have no effect on the rate. [E] is the answer.

In the Lewis dot diagram for SF4, how many total electrons, shared and unshared, are surrounding the central element? A. 4 B. 6 C. 8 D. 10 E. 12

Sulfur is the central element and has the ability to have an expanded octet. Unlike other elements, that can only have a maximum of 8 electrons in the valence shell, sulfur can have as many as 12 electrons in its valence shell. Sulfur has 6 valence electrons, and each F has 7. According to the Lewis dot diagram, there a total of 4 S-F bonds, and one lone pair on sulfur, for a total of 10 electrons. [D] is the answer.

Which molecular shape is characterized by having 3 bonding domains and 1 non-bonding domain? A. Tetrahedral B. Bent C. Seesaw D. Trigonal planar E. Trigonal pyramidal

The central element is bound to 3 other elements. There is 1 non-bonding domain, so there is one lone pair. The total non-bonding and bonding domains is 4, so the electron-pair geometry is a tetrahedral. However, the molecular geometry of this molecule is trigonal pyramidal. [E] is the answer.

The first, second, and third ionization energy of an atom of a period 3 element is 738 kJ/mol, 1451 kJ/mol, and 7733 kJ/mol, respectively. The element is most likely A. sodium. B. magnesium. C. aluminum. D. sulfur. E. phosphorus.

The ionization energy of an element increases dramatically upon removal of its inner-shell electrons. There is a dramatic increase from second to third ionization energy, indicating that the element has two valence electrons. The only answer choice with two valence electrons is magnesium. [B] is the answer.

A Lewis structure for O3 is: O = O - O Including this structure, the total number of ground state resonance structures for this molecule is:

The lone pair from the O- will create a new double bond with the central O. A pair of electrons from the original double bond will become a lone pair, converting an O to an O-. There are two resonance structures:

Which particles are commonly referred to as nucleons? A. Electrons and protons B. Protons and neutrons C. Neutrons and electrons D. Beta particles and positrons E. Alpha particles and beta particles

The nucleons are the particles that make up the nucleus. The nucleus is comprised of neutrons and protons. Alpha particles are comprised of nucleons, but are not referred to as nucleons themselves. [B] is the answer.

During a titration experiment, the titrant is placed in which of the following? A. Flask B. Buret C. Beaker D. Graduated cylinder E. Pipette

The titrant is the stock or known solution. The concentration of the titrant is known, however the concentration of the titrand (solution being titrated) is unknown. The titrant is poured into a buret and the volume is recorded before titration begins. The titrand is then measured and added to an Erlenmeyer flask along with a pH indicator before the beginning of the titration experiment. [B] is the 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 change for the formation of one mole of the compound from reactants in their most stable forms at standard state conditions (298K and 1 atm). Sodium is a solid and chlorine is a gas at standard state conditions. To form one mole of the compound use 1 mole of Na (s) and 1/2 mole of Cl2 (g) to form one mole of NaCl (s).

A student heated a hydrate, MgCl2 • H2O, and recorded the following results. What was the percent change in water for the 4th trial?

To calculate the percent change the following equation can be used: Plug in the values for the fourth trial:

The boiling point of a 1.0m aqueous solution is 101.5°C. The molal boiling point constant for water is 0.512°C•m-1. Which of the following substances is most likely dissolved in the water? A. NaClB. Na2SO4C. C6H6D. CaCO3E. K3PO4

Use the boiling point elevation formula: Where ΔTb represents the change in temperature, "i" represents the van't Hoff factor, m represents the molality, and Kb represents the molal boiling point constant for water. Plug in the given values: The van't Hoff factor is equal to 3. Look to the answer choices to determine which species will dissolve into 3 parts. Na2SO4 will break down into two Na+ ions and one SO4-2 ion, for a total of 3 ions. [B] is the answer.

Liquid mercury in a glass buret is more strongly attracted to itself than the walls of its container. All of the following are correct steps for measuring the volume of liquid mercury in a buret EXCEPT for one. Which one is the EXCEPTION? A. Place the buret as straight as possible B. Position oneself eye-level to the meniscus C. Focus on the center of the meniscus D. Take a reading from the bottom of the meniscus E. Report the volume based on the buret's smallest line markings

When placed in a glass buret, liquid mercury exhibits a strongly convex meniscus. This is due to mercury's high cohesion and low adhesion to the glass. Recall that cohesion is the attractive force between "like" molecules, and adhesion is the attractive force between dissimilar molecules. https://en.wikipedia.org/wiki/File:Stativ.svg In order to read a meniscus accurately, one should follow specific steps: Clamp the buret as straight as possible. Position oneself eye-level with the meniscus to prevent parallax error. Measure the center of the meniscus.Take a reading from the bottom of a concave meniscus.Take a reading from the top of a convex meniscus. Report the volume based on the buret's smallest line markings (50-mL burets are divided into 0.1 mL increments). https://en.wikipedia.org/wiki/File:Reading_the_meniscus.svg Because mercury exhibits a convex meniscus when placed in a glass buret, one must take a reading from the top of the meniscus. Choice [D] is the answer.

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. A. 0.025 MB. 0.050 MC. 0.075 MD. 0.125 ME. 0.250 M

Write the balanced reaction that produces the precipitate: Ca2+ + SO4-2 → CaSO4 Calculate the number of moles of both ions in the solution: (0.100 L)(0.25 M) = 0.025 mol SO42-(0.100 L)(0.50 M) = 0.050 mol Ca2+ 0.025 moles of Ca2+ will react with 0.025 moles of SO42- according to the balanced reaction. 0.025 moles of Ca2+ will remain in the solution. Convert from moles to molarity. The final volume is 100 mL + 100 mL = 200 mL. 0.025 mol [Ca2+] / 0.200 L = 0.125 M [Ca2+]. [D] is the answer.