Chemistry

Q18

Qstem gives you two molar masses of HCl (0.04M and 0.1M) and it says for the 0.04M the osmotic pressure is 1.9atm so what is the osm pressure for 0.1M so its just a ratio so 0.10/0.040 = 2.5 so 2.5(1.9atm)=4.8atm Educational objective: The diffusion of solvent across a semipermeable membrane during osmosis produces an osmotic pressure against the membrane, which is directly proportional to the concentration of the osmotically active solute in solution.

Q13

Qstem gave us Ksp for NaC9H7O4 = 34.9 and is asking for the conc of C9H7O4 in NaC9H7O4. Ksp=[dissolved species 1]^a x [dissolved species 2]^b **raised to their balanced equation coefficients and ONLY aqueous, no solids** look at pic for math Educational objective: The solubility product constant Ksp represents the limit of solubility for a compound, and it is defined as the product of the molar concentrations of the dissolved species each raised to the power of their respective balanced equation coefficients. This relationship can be used to find molar concentrations of ions in saturated solutions.

Q48

they asking about bonding in region 1 so looked at region 1 and it had O and N so picked hydrogen bonding Educational objective: Hydrogen bonding is one of the stronger intramolecular and intermolecular forces. It is a special kind of noncovalent dipole-dipole interaction between dipoles formed when hydrogen is covalently bound to the highly electronegative elements nitrogen, oxygen, or fluorine.

Electronegativity differences

Nonpolar: 0-0.5 polar covalent: 0.5-1.75 ionic: >1.75

Q17

(pH = −log[H+]). Based on this relationship, the molar concentration of hydrogen ions can be calculated as [H+] = 10^−pH. The molar concentration can be expressed as either moles/liter or scaled to millimoles/milliliter (mmol/mL). Multiplying a volume of the solution by its molar concentration yields the number of moles of acid solute contained in that volume of the solution. The passage states gastric acid has a pH of 1.0. Therefore, the concentration of hydrogen ions can be determined by the expression: [H+] = 10−1 = 0.1 M = 0.1 mmol/mL The number of mmol of H+ in 10.0 mL of gastric acid can be determined as follows: (0.1 mmol/mL)(10.0 mL) = 1.0 mmol of H+ Educational objective: The pH of a solution is defined as pH = −log[H+], where [H+] represents the hydrogen ion concentration in units of molarity (moles per liter). Knowing the pH of a given volume of solution allows the determination of the molar concentration and the number of moles of hydrogen ions present.

Q55

1. Look at the atom. 2. Count the number of atoms connected to it (atoms - not bonds!) 3. Count the number of lone pairs attached to it. Lone pair are treated as 1 electron not 2 electrons. 4. Add these two numbers together. If it's 4, your atom is sp3. If it's 3, your atom is sp2. If it's 2, your atom is sp. (If it's 1, it's probably hydrogen!) Educational objective: Hybrid orbitals are formed by combining the atomic orbitals of an atom. The hybridization of an atom is determined by counting the electron domains (sigma bonds and lone pairs) and assigning a hybrid name. The sum of superscripts must equal the number of electron domains.

Q25

A concentration cell employs the same electrode material and ionic solution for both the cathode and the anode, but the ion solutions surrounding the anode and cathode differ in concentration. As a result, the same half-reaction occurs at both electrodes, but it occurs as an oxidation at one and as a reduction (in the reverse direction) at the other. Accordingly, adding the two half-reactions yields a result of E° = 0.00 V. The electrons will migrate from the cell with fewer cations to the cell with more cations until the cation concentration in each cell is equal. Because oxidation occurs at the anode, this is the source of electrons for the cell, which are transferred to the cathode where reduction occurs. Because the measured potential for the cell is positive (E = 0.089 V), the ion concentrations must have a ratio of [X]anode/[Y]cathode that is less than one so that the logarithmic term of the Nernst equation will be negative. A negative logarithmic value multiplied by the negative constant in the equation results in a positive value for E. Therefore, [Ni2+] is greater at the cathode, and the electrons in this cell will flow from the anode to the cathode. Educational objective: Concentration cells use the same electrode material and ionic solution at both the anode and the cathode but operate by a concentration difference between the cells, which produces a difference in potential. Electrons will migrate from the anode (fewer cations) to the cathode (more cations) until the concentrations equalize.

Q16

All you had to do for this problem is understand the relationship bw pka and Ka. smaller pka = stronger acid higher ka = stronger acid out of the ans choices citric acid has the largest Ka value **compare exponents first, as you approach 0 the values get larger (10^-2 >> 10^-7) and then values with the same exponent you have to compare the actual values (1.4x10^-4 and 8.4x10^-4). 8.4 is >> 1.4 so its the larger value Educational objective: Acid strength increases as the acid ionization constant Ka increases, or as the acid pKadecreases. As a result, the pH of a solution at a given concentration decreases as the pKa decreases.

Q3

Accordingly, the pH of a buffered solution depends on the concentration ratio of acid to conjugate base. Adding more acid will decrease the pH and adding more base will increase the pH. In Experiment 1, the citrate buffer is made of citric acid [HA] and sodium citrate [A−]. Adding more sodium citrate will increase the pH of the solution. The passage states that oxytocin is more stable at a pH of 4.5, and Figure 1 shows that at the higher pH of 6.8 (water), less oxytocin was recovered. (Choice B) The experiment measures the percent recovery of oxytocin. Decreasing the initial amount of oxytocin in solution will lower the amount present in solution but will not change the percentage recovered. (Choice C) Cooling the samples during the experiment will not decrease the percent recovery because the results in Experiments 1 and 2 show that oxytocin is more stable at cooler temperatures. (Choice D) Increasing the buffer concentration will increase buffer capacity to resist changes in pH. Based on the results in Experiment 1, this will not decrease oxytocin recovery. Educational objective: A buffered solution consists of a weak acid and a salt of its conjugate base or a weak base and its conjugate acid. Buffers resist changes in pH by neutralizing added H+ and OH− ions. The pH of a buffered solution depends on the ratio of weak acid and its conjugate base.

Q26

Again this problem is asking about longest bond (bond length) and it corresponds to atomic radii Covalent bonds between atoms are formed by sharing electrons, which is made possible through the overlap of the atomic orbitals from each atom. Covalent bonds made by the end-to-end overlap of atomic orbitals are called sigma (σ) bonds, whereas covalent bonds made by the side-to-side overlap of p orbitals are called pi (π) bonds. A single covalent bond consists only of a σ bond, a double bond consists of one σ bond and one π bond, and a triple bond consists of one σ bond and two π bonds. Because a σ bond involves end-to-end overlap of two orbitals, the length of a σ bond can be estimated as the sum of the atomic radii of the bonded atoms. Bonding between atoms with larger atomic radii positions the atomic nuclei farther apart and results in a longer σ bond (and vice versa). On the periodic table, atomic radii tend to decrease across a row (due to an increasing effective nuclear charge) and increase down a column (due to having an additional electron shell). In this question, the given compounds all have σ bonds to a sulfur atom. As a result, the longest bond between sulfur and another element will be formed with the element that has the largest atomic radius. Because Cl has a larger atomic radius than H, N, or F, the S-Cl bond in S2Cl2 will the longest bond with sulfur among the given compounds. (Choices A, B, and D) Atoms of H, N, and F each have a smaller atomic radius than Cl. Therefore, the S-H bonds in H2S, the S-F bonds in SF6, and the S-N bonds in S4N4 are all shorter than the S-Cl bonds in S2Cl2. Educational objective: Covalent sigma bonds are made by sharing electrons through the end-to-end overlap of atomic orbitals. The length of a sigma bond can be estimated as the sum of the atomic radii of the bonded atoms. Atomic radii tend to decrease across a row and increase down a column on the periodic table.

Q32

An important reaction type is the decomposition reaction, which involves the breaking down of a compound into its elemental constituents or into simpler, stable compounds. This type of reaction usually requires the addition of energy (such as heat) to break the bonds of the original compound. In this question, calcium carbonate is isolated from the inactive ingredients in an antacid tablet and then placed over a flame. The gas produced is CO2, based on the results from the glowing splint test. This means that the white solid that remains is a new compound that does not have carbon in it. Calcium preferentially forms a metal oxide in the presence of oxygen, so the other product must be CaO. (Choice A) According to the passage, HCl was not added to perform this reaction. Also, this is not a decomposition reaction because there are two reactants, CaCO3 and HCl. Reactions with acids or bases are often double replacement reactions in which water is produced. (Choices B and D) The question asks for the decomposition of calcium carbonate, which means CaCO3 must be the only reactant (on the left side) of the chemical equation. Educational objective: A decomposition reaction is a type of reaction in which energy is added to break down a compound into either its elemental constituents or into simpler, stable compounds.

Q2

An ionic bond between two atoms requires the transfer of valence electrons from one atom to the other, creating a cation and an anion. The ionic character increases as the difference in electronegativity between two atoms increases. Electronegativity tends to increase when moving across a period (row) from left to right and when ascending a group (column) on the periodic table, with some exceptions in the transition metals. Comparing NaBr and MgBr2, NaBr has a higher electronegativity difference (and higher ionic character) because Na is further to the left of Br on the periodic table than Mg. Similarly, NaCl has a higher ionic character than MgCl2. Of the two higher-ranking sodium salts, NaCl has the greatest electronegativity difference (ionic character) as Cl is above Br in the halogens column. Therefore, the expected order of these four salts if listed from the least ionic character to the greatest ionic character is MgBr2 < MgCl2 < NaBr < NaCl. Educational objective: Electronegativity trends on the periodic table are useful to assess relative differences in electronegativity between elements. The difference in electronegativity between two bonded atoms is directly proportional to (and indicative of) the degree of ionic character of the bond between the atoms.

Q15

Another stoichiometry problem. Be careful have to use 2 moles of FE(CN)6)^3- for every 2 moles of electrons. Also can change molar mass to nay ratio units as long as you do it to both num and denom. Look at pic for umol and ug cancellations Educational objective: Given the mass of a compound sample, the number of moles present in the sample can be determined using its molar mass. Relating a known number of moles of a reactant to a balanced reaction equation, mole ratios can be applied to determine the stoichiometry of one chemical species relative to another species in a reaction.

Q9

As described in the second paragraph of the passage, when heme b (Figure 1) axially binds with oxygen and a globin histidine, the Fe2+ has three ligands (1 porphyrin, 1 histidine, and 1 O2) that together form six coordinate covalent bonds: four to amines in the porphyrin, one to histidine from the globin protein, and one to an O2 molecule. Therefore, Fe2+ and its ligands form an octahedral arrangement in which the six coordinate bonds are all at approximately 90° angles from each other. Educational objective: Coordinate covalent bonds tend to maximize geometric space around a central atom. Transition metals generally form complexes with two, four, or six coordinate bonds. Therefore, the geometry of the coordinate covalent bonds will typically be linear (two bonds), tetrahedral or square planar (four bonds), or octahedral (six bonds).

Q10

As stated in the last sentence of the passage, ZPP has an excitation maximum of 425 nm and an emission maximum of 595 nm whereas PPIX is excited at 407 nm and emits a photon at 625 nm. Because ZPP can be excited by a longer wavelength than PPIX, ZPP requires less energy to excite its electrons from the ground state. Educational objective: Electromagnetic radiation at a certain wavelength has photons of a particular energy. Energy is inversely proportional to wavelength.

Q10

Asking to find conc of NaC9H7O4 given solubility of sodium acetylsalicylate 120g per 100ml given in passage: molar mass of NaC9H7O4 so start with 120g/100ml x 1000ml/1L x 1 mol NaC9H7O4/202.2g = 5.9 mol/L Educational objective: A saturated solution contains the maximum amount of compound that will dissolve in a given volume of solvent under the given conditions. Solubility (expressed as grams of solute per 100 g of solvent) can be stated in units of molarity (moles of solute per liter of solvent) using unit conversions and the molar mass of the solute.

Q16

Atoms with small atomic radii can form short, strong bonds whereas atoms with larger radii form longer, weaker bonds. Each subsequent row on the periodic table indicates an additional electron shell, and therefore a larger atomic radius, so atoms in the first row will have smaller atomic radii than atoms in the second row, and so on. Double bonds require more energy to break than single bonds. Citrate includes bonds between carbon and carbon, carbon and oxygen, carbon and hydrogen, and oxygen and hydrogen. Of the choices given, the C-C bond is the only single bond that does not involve a first row atom (hydrogen). Therefore, the C-C bond will be the longest and require the least energy to break. (Choices A and C) These options include hydrogen, which forms short, strong bonds. Note that although O-H bonds are commonly broken in acid-base reactions, these reactions also involve the formation of a new bond to hydrogen, which provides the high amount of energy necessary to break the old bond. Educational objective: The energy required to break a bond, known as the bond dissociation energy, is related to the bond length. Atoms with smaller radii (near the top of the periodic table) tend to form shorter, stronger bonds than atoms with larger radii.

Q30

D2sp3 means there are 2+1+3 = 6 bonds to the central atom so it has to be SF6 an octahedral Educational objective: In hybridization, the number of hybrid orbitals that form must be equal to the number of atomic orbitals that hybridize. The sum of the number of bonding regions and the number of nonbonding electron pairs around the central atom in a compound must be equal to the number of hybrid orbitals employed by the central atom.

Q7

Coordinate covalent bonds are a special type of bond between a central atom, such as a metal, and a ligand. Both electrons in the bond come from the ligand. Although classified as covalent bonds, coordination bonds do share some of the characteristics of ionic bonds. For example, the metal ion maintains its oxidation state, as in the case of an ionic bond, but the ligands are not charged and are usually electronegative, as in covalent bonds. The metal and its ligands together form a complex, and the number of coordinate bonds to the metal is known as the coordination number. The metal ion Fe2+ is positively charged whereas the donor atoms (nitrogen) are neutral, resulting in a net charge of +2 for the complex. Unlike ionic bonds, the donor atoms do not give electrons to the positively charged metal. Instead counterions often surround the complex in solution to balance the metal's positive charge. (Choice A) The nitrogen ligands each have a lone pair of electrons that provide the bonding electrons in the coordinate covalent bond. (Choice C) The lone pairs of nitrogen's electrons interact with iron's d orbitals to form the coordinate bond. In a coordination bond, the d orbitals are no longer degenerate and have distinct shapes, some of which point directly at the negatively charged ligands. (Choice D) The strength of the coordination bond depends on which d orbitals have electrons in them. Educational objective: Coordinate covalent bonds are a special bond between a central atom and its ligands. The number of coordinate bonds indicates the coordination number and the ligands provide bonding electrons, which interact with the metal's d orbitals to form the coordinate covalent bond.

Q22

Coordination complexes consist of a central metal ion surrounded by one or more ions or molecules called ligands that are bound to the metal center by coordinate bonds. The ligands act as a Lewis base, and the metal center acts as a Lewis acid. Stronger Lewis bases can displace weaker Lewis bases as ligands within a complex. Stronger Lewis bases tend to be those that have lone pair electrons on atoms with a net charge and/or a lower electronegativity. Charged species tend to be less stable (more reactive) than uncharged species because forming a bond produces a more stable, lower energy state. Likewise, atoms with a lower electronegativity tend to be better Lewis bases because the electrons are held more loosely and any net charge is less stabilized. Both C2O42− and H2O have available lone pair electrons on an oxygen atom (same electronegativity aspect), but C2O42− is a stronger Lewis base because it also has a negative charge on each of two of the oxygen atoms. This charge comes from the presence of an extra lone pair of electrons on each. Therefore, C2O42− does not have fewer lone pair electrons per oxygen atom than H2O, and this choice does NOT explain why C2O42− displaces H2O. Educational objective: Stronger Lewis bases (electron pair donors) can displace weaker Lewis bases as ligands within a coordination complex. Lone pair electrons on atoms with a lower electronegativity tend be stronger Lewis bases than those with a higher electronegativity, and charged atoms with lone pair electrons tend to be stronger Lewis bases than comparable uncharged atoms.

Q12

Covalent bonds are formed between two atoms by sharing electrons through the end-to-end overlap of atomic orbitals. In contrast, ionic bonds are formed between two atoms when the valence electrons from one atom are transferred to another atom, resulting in charged ions that are held together by strong electrostatic attractions. The type of bond that forms between two atoms depends on the relative difference in electronegativity between the atoms. A large difference in electronegativity promotes ionic bond formation, but a small difference promotes covalent bond formation. As a result, ionic bonds generally form between a metal and a nonmetal, whereas covalent bonds generally form between two nonmetals; however, there are some exceptions because the ionic character of a bond exists on a continuum. Therefore, in chemical bond formation, covalent bonds are generally formed by electron sharing between two atoms with a small or moderate electronegativity difference. Educational objective: The type of bond formed between two atoms depends on the relative difference in electronegativity between them. Atoms with a large difference in electronegativity (usually a metal and a nonmetal) form ionic bonds. Atoms with a small difference in electronegativity (usually two nonmetals) form covalent bonds.

Q49

Covalent bonds are strong chemical bonds formed when two atoms share electrons. In contrast, noncovalent interactions are weak attractions between atoms with opposite partial charges within dipoles that do not share electrons. Noncovalent interactions include attractions between two permanent dipoles, attractions between a permanent dipole and an induced dipole, and attractions between two induced dipoles. These attractions are collectively called van der Waals forces. The weakest of the van der Waals forces are London dispersion forces, which are also known as induced dipole-induced dipole interactions. These interactions occur between two molecules whose proximity to each other induces a mutual attraction via the creation of a weak, temporary dipole moment. Within a nonpolar molecule, the average electron distribution does not concentrate around any particular nucleus sufficiently enough to produce a permanent dipole moment. However, when two nonpolar molecules come near each other, the electron cloud of one molecule can momentarily shift and form a localized, slightly positive region that then induces a slightly negative region in a neighboring molecule. Consequently, instantaneous dipoles with weak mutual attraction are formed momentarily. In region IV of the protein in Figure 2, there are only nonpolar alkyl side chains (no permanent dipoles). Therefore, the primary interactions in region IV are just London dispersion forces (the weakest type). Educational objective: London dispersion forces, also called induced dipole-induced dipole interactions, are the weakest of the noncovalent van der Waals forces. These interactions occur between two atoms or molecules that are near enough to each other for distortions in the electron cloud to induce weak instantaneous dipoles.

Q3

Covalent bonds between atoms are formed by sharing electrons, which is made possible through overlap of the atomic orbitals from each atom. Covalent bonds made by the end-to-end overlap of orbitals are called sigma (σ) bonds, whereas covalent bonds made by the side-to-side overlap of p orbitals are called pi (π) bonds. Although some atoms participate only in single covalent bonding (by a σ bond), other atoms can participate in multiple bonds (double or triple bonds) by the addition of one or more π bonds. In this way, a double bond consists of one σ bond and one π bond, and a triple bond consists of one σ bond and two π bonds. Comparisons can be made between single, double, and triple covalent bonds provided that the bonds being compared involve the same two types of atoms. The overall dissociation energy of a bond (the energy required to break the bond) results from the sum of all σ and π bonding contributors. Therefore, the overall bond dissociation energy and bond strength tend to decrease (relative to a triple bond) as the number of π bonds decreases (Number I). Because of the shapes of the orbitals involved, bond length tends to increase (relative to a triple bond) as the number of π bonds decreases (Number II). Because σ bonds are made by an end-to-end orbital overlap, free rotation around the bond between the two σ-bonded atoms can be achieved while maintaining the orbital overlap. Therefore, decreasing the number of π bonds also decreases the rigidity of the bond between the atoms (Number III). Educational objective: Multiple covalent bonds formed by s and p orbitals consist of one σ bond (end-to-end orbital overlap) and one or more π bonds (side-to-side p orbital overlap). Decreasing the number of π bonds between two atoms increases bond length, decreases bond rigidity, and decreases the overall bond dissociation energy and strength.

Q38

Dumb just didn't read Q properly, it said "NOT" sp3 hybridized and I also forgot that sp3 meant 4 bonds not 3. Educational objective: During hybridization, atomic orbitals combine to form new hybrid orbitals that assume spatial orientations to minimize the electronic repulsion between orbitals. The number of hybrid orbitals that form and the number of orientational vectors adopted by the hybrid orbitals will equal the number of atomic orbitals that combine.

Q2

E3 is an electrochemical cell (from passage) and it harnesses energy from an oxidation-reduction reaction by separating the reaction into half-cells. Anode one reactant is oxidized (loses electrons); Cathode, another reactant is reduced (gains electrons). Electrons flow from anode to cathode through wire that connects the two half-cells. This produces a current that can be used to do work. To determine whether a species is oxidized or reduced, examine the oxidation state of the element of interest. The oxidation state is the hypothetical charge of an atom within a molecule (ie, the charge the atom would have if it were not neutralized by other atoms in the molecule). The more electronegative atom in a covalent bond owns both electrons in the bond, and therefore gains a hypothetical charge of −1; this atom is reduced. The less electronegative atom "loses" its electron and gains a hypothetical charge of +1; this atom is oxidized. In a neutral molecule, the sum of the oxidation states of atoms must balance to zero. At E3, gaseous O2 (oxidation state of 0) reacts with aqueous protons (H+) to produce water. To balance the positive charges added by the protons, the reaction also requires the addition of two electrons. O2 is more electronegative than hydrogen, and therefore it owns these electrons and is reduced to an oxidation state of −2 whereas the protons each remain at an oxidation state of +1. By definition, this reduction occurs at the cathode. Educational objective: Electrochemical cells harness energy released by oxidation-reduction reactions. Oxidation (loss of electrons) occurs at the anode whereas reduction (gain of electrons) occurs at the cathode.

Q9

Easy just had to compare the solubilities from table 1 for the other answer choices and find the one between the values the Q wanted. A was too big, C and D were too small Educational objective: The solubility of an ionic compound is measured by the amount, in moles, that can dissolve in a given volume of solvent. The more that can dissolve, the more soluble the ionic compound.

Q54

Educational objective: Resonance structures are Lewis dot structures of a given molecule in which electrons have been delocalized. Acceptable resonance structures move only electrons and generally only to adjacent atoms, keep the number of valence electrons constant, obey the octet rule, do not break sigma bonds, and keep the overall charge of the molecule constant.

Q5

Electron orbitals are defined by a set of quantum numbers. The principal quantum number n represents the distance of an orbital shell from the nucleus. Each orbital shell can be divided into subshells, defined by the angular quantum number l and the magnetic quantum number m. The angular quantum number represents the shape of a subshell, the most common being s, p, d, and f. The magnetic number describes the orientation of a given subshell in which s orbitals can exist in one orientation, p orbitals can exist in three orientations, d orbitals can exist in five orientations, and f orbitals can exist in seven orientations. Each subshell can hold 2 electrons, so s, p, d, and f orbitals can hold 2, 6, 10, and 14 electrons, respectively. Each orbital is associated with a different energy level, and the Aufbau principle states that electrons will fill the lowest energy levels first. The periodic table is arranged to reflect the relative energies of each orbital type and, therefore, the order of electron orbital filling. The orbitals are arranged into different blocks on the periodic table: s orbitals are in the first two columns (except helium), the p orbitals are on the right, the d orbitals are in the middle, and the f orbitals are at the bottom. Therefore, the order of orbital filling can be determined simply by finding an atom's position on the table. Iron is within the 3d orbital section, which comes after the 4s section, so 4s is filled first. (Choice A) The order of orbital filling is determined by energy level, which is reflected in the periodic table. (Choice B) Although the 4s and 4p orbitals have a higher principal quantum number n, indicating they are further away from the nucleus than 3d, the periodic table shows that 4s is slightly lower in energy and will be filled first. (Choice D) An atom's position on the periodic table, not its principal quantum number, determines how its orbitals are filled. Educational objective: The Aufbau principle states that low energy orbitals are filled first. The periodic table can be used as a guide to orbital filling. Each block on the table represents different orbital types, with s orbitals in the two left columns, p orbitals in the six right columns, and d orbitals in the middle, and f orbitals at the bottom.

Q3

Equation 1 is a double replacement reaction because the bicarbonate anion (HCO3−) exchanges the sodium counter ion for an acidic proton from hydrochloric acid (forming carbonic acid, H2CO3), and the hydrochloric acid exchanges the acidic proton for the sodium ion (forming sodium chloride, NaCl). Educational objective: Chemical reactions can be categorized broadly into five main types: combination, decomposition, single replacement, double replacement, and combustion. Double replacement reactions involve the exchange of the bonding partners of two reactant compounds, forming two new product compounds.

Q44

FC = Should - Has Educational objective: A formal charge is the charge assigned to an atom based on an accounting method which assumes that the bonding electrons between two atoms are shared equally. Formal charge equals the group valence of an atom minus the number of nonbonding electrons and minus half of the number of bonding electrons.

Q58

Figure 3 shows that some molecules exhibit a greater overall response to tastant than others, as well as a greater difference in response between tastant and tastant + IMP. The increase in the ratio of the response with IMP to the response without IMP is greatest for Asp. The degree of potentiation by IMP for a tastant molecule is defined by the relative increase in response upon addition of IMP. Therefore, IMP most effectively potentiates Asp. (Choices A and B) Figure 3 shows that the ratio of the response of tastant + IMP to tastant alone for each molecule is different. Therefore, IMP potentiates different tastant molecules to different degrees. (Choice C) Although the difference in responding cells for Thr with and without IMP (~120 cell difference) is much greater than for Asp (~60 cell difference), the ratio of the response for tastant + IMP relative to the initial condition is less for Thr (~4:1) than for Asp (~60:<1). Educational objective: Drawing conclusions from experimental results requires careful analysis of the experimental data. The degree of potentiation, or a change in a system's response to a stimulus, is best measured as a ratio. This ratio shows the percent change in a signal's intensity in the presence of a potentiator.

Q19

First had to balance the equation. Thought it was a limiting reactant problem but didn't know how to go about it, got halfway there. LOOK AT PIC FOR MATH Educational objective: Theoretical yield is the maximum amount of product that can be formed in a reaction. It is governed by the limiting reactant, which is determined by the balanced equation and the molar amounts of each reactant available.

Q14

For anything to precipitate, it needs to have a higher product of Ksp than the compound itself If the concentrations of [X] and [Y] are such that [X]^c[Y]^d < Ksp, the solution has not reached the solubility limit and will permit more ions into solution (dissolve more of Z). Conversely, if the concentrations of [X] and [Y] are such that [X]^c[Y]^d >Ksp, the solubility limit of the solution has been exceeded and the solution will not accept any more ions (no more Z will dissolve). This will cause any ions in excess of the Ksp to come out of solution as a precipitate. Ksp=[dissolved species 1]^a x [dissolved species 2]^b **raised to their balanced equation coefficients and ONLY aqueous, no solids** Educational objective: In a solution formed from the dissolution of a compound into ions, the compound will dissolve until the solution reaches the value of the solubility product constant Ksp for the compound. Ion concentrations that exceed the Ksp result in precipitation of the excess.

Q19 (had trouble)

Gave us: osmolarity eqn osm pressure = 6.9atm temp = 37 C --> 310K R = 0.08 L*atm/mol*K i = 2 so had to do 6.9/(0.08)(310) osmolarity = 0.28osmol/L LOOK IN PIC for math Educational objective: Solvent diffusion across the semipermeable membrane during osmosis produces an osmotic pressure against the membrane proportional to the osmolarity (concentration of osmotically active species that cannot cross the membrane) rather than to the solute molarity due to the dissociation of some solutes. Solute concentration and osmolarity are related by the van 't Hoff factor.

Q14

Had to do stoichiometry and dont forget to always convert to scientific notation even for last answer choice!!! Look at pic for math Educational objective: Solution concentrations can be expressed as the mass of a given substance per unit of solution volume. The mass of a substance contained in an aliquot can be calculated by multiplying the aliquot volume by the solution concentration when both values are expressed with respect to the same units of volume.

Q46

Had to find the overall reaction and make sure that the electrons cancelled out so add reactions together and subtract what you need to Educational objective: Redox reactions are fully balanced when the same number of each type of atom and the same net charge are present on both sides of the reaction equation. For redox processes expressed as two half-reactions, the net reaction is found by adding the two half-reactions (or multiples of them) together so that the electron terms cancel.

Q15

Had to know that Ka is acid dissociation constant so if a weak acid (clue 1, bc they don't fully dissociate) has a low solubility (clue 2, doesn't easily dissociate or fully) that means it will have a Ka < 1 Educational objective: Strong acids ionize essentially 100% in aqueous solutions, resulting in a large acid dissociation constant (Ka > 1), whereas weak acids ionize only to a small extent, resulting in a small acid dissociation constant (Ka< 1).

Q7

Had to know that when concentration is expressed as a percent and they are asking you for ppm then you use the conversion factor 1% = 10,000ppm Educational objective: Parts per million (ppm) is a dimensionless quantity used to express concentrations as a fraction out of a basis of 1,000,000. In aqueous solutions, 1 ppm = 1 mg/L, and for concentrations expressed as a percentage, 1% = 10,000 ppm.

Q5

Had to know to look at percent of oxytocin recovery in figure 2 graph at Zn2+ at 10mM (given in Q stem) LOOK at pic for math Educational objective: Solution concentration can be expressed in terms of mass per unit volume, which can be used as a conversion factor to find the volume of solution that contains a given amount of mass. Conversions between the moles and mass of a substance can be made using the molar mass.

Q9

Had to write out the chemical equation to find ration of HCl and NH3. In this titration, 100 mL of 0.1 M HCl were required to reach the equivalence point. Accordingly, the number of millimoles of HCl required to reach this point is 100 mL × (0.1 mmol HCl/mL) =10 mmol HCl Because each mole of ammonia gives one equivalent of base and each mole of HCl gives one equivalent of acid, the neutralization reaction in this titration proceeds with a 1:1 molar ratio. Therefore, 10 mmol of HCl were required to neutralize 10 mmol of NH3 present in the unknown ammonia solution. Educational objective: The equivalence point in an acid-base titration occurs when the number of equivalents of acid equals the number of equivalents of base; it is seen as the midpoint of a nearly vertical segment of the titration curve. The volume of titrant at the equivalence point multiplied by its concentration gives the moles of titrant species added.

Q24

How to balance half equations: Electrons present as reactants in one half-reaction should appear as products in the other half-reaction. This may require multiplication of one or both half-reactions by a constant to ensure that all electrons cancel. Identical species appearing as reactants in one half-reaction and as products in the other cancel and are not part of the net reaction expression. Based on these principles, an unknown half-reaction can be found by adding the balanced net reaction to the inverse of the known half-reaction. Educational objective: The balanced net reaction for an electrochemical cell may be found by taking the sum of the half-reactions, written to be consistent with the direction of electron flow within the cell. Half-reactions may be scaled stoichiometrically to achieve an electron balance. Species matching in type and number on both sides of the summation are not included in the net reaction.

Q13

Hydrogen bonding only occurs between FON Educational objective: Hydrogen atoms covalently bonded with nitrogen, oxygen, or fluorine atoms form polar bonds that yield dipoles, which exhibit significant noncovalent dipole-dipole attractions known as hydrogen bonding.

Q11

Hydrophobic character increases as the number of nonpolar molecule increases and the number of methyl groups or carbons increases. As the number of polar groups increase, the molecule gets more hydrophillic. Thats why the answer is D bc it has 4 methyl groups. LOOK AT PIC Educational objective: Molecules with many polar bonds that promote dipolar interactions with water are hydrophilic, whereas those with mostly nonpolar bonds that lack attractive dipolar interactions with water are hydrophobic. The overall character of a compound depends on the relative number of each type of bond present.

Q13

Hydroxide (OH-) conc is 1x10^-6 so hydronium (H+) conc is 14-6 = 8 so pH =8 Educational objective: The [H3O+] and [OH−] concentrations are related by the self-ionization constant Kw of water according to the expression Kw = [H3O+][OH−] = 1.0 × 10−14 M2. Using a logarithmic approach, pH is related to [OH−] by the relationship pH + pOH = 14, where pOH = −log[OH−].

Q29

I drew out the shape of hydrogen sulfide which is H2S and S has 6 valence electrons which means it has two single bonds to H and two lone pairs meaning it will look like water. Recognize that H2S looks like H2O bc S and O are in the same group. Educational objective: Atomic orbitals can combine into hybrid orbitals, which adopt an electron geometry that maximizes the separation and minimizes the repulsion between the electron-dense orbital regions. The molecular geometry (shape) of a molecule depends on the electron geometry and number of nonbonding electron pairs of the central atom.

Q12 ask landon

I looked at chemical equation given in passage and determined the ratio that way. So 2 moles of Na2SO4 produced for every 1 mole of Cu2(C9H7O4)4 produced within the same solution volume. As a result, the Cu2(C9H7O4)4 concentration (prior to precipitation) is equal to half the Na2SO4 concentration. Educational objective: The coefficients in a balanced chemical reaction indicate the stoichiometric proportions by which each chemical combines during the reaction. Mole ratios derived from these coefficients can be used as conversion factors in chemical reaction calculations to relate the molar amounts of compounds in a reaction.

Q41

If product didn't show the two bands that distinguish cis isomer from trans then you know the product is all trans isomer. Look at pic Educational objective: Infrared spectroscopy can be used to detect structural changes within molecules during chemical conversions by monitoring successive sample spectra for the emergence or disappearance of peaks that are unique to a given structural feature of a molecule.

Q23

In a titration, a measured amount of a solution with a known concentration (titrant) is added to another solution containing an unknown concentration of the compound to be measured (analyte). In acid-base titrations, an acid is titrated with a base (or vice versa). The resulting acid-base neutralization reaction produces a change in pH, which is monitored by a pH indicator that signals the equivalence point of the neutralization. The analyte must be fully dissolved before it can be measured. Sebacic acid has low solubility in water due to a high nonpolar hydrocarbon character. A base will convert the carboxylic acid groups into highly polar ionic salts with much higher aqueous solubility. Once dissolved, the carboxylate ions can then be titrated with an acid. Subtracting the number of moles of base (such as KOH) in the initial solution from the number of moles of acid (such as HCl) required to reach the equivalence point during the titration will give the number of moles of carboxylate groups from sebacic acid in the sample. Educational objective: Acid-base neutralization reactions can be used in titrations with an indicator to determine the concentration of an acidic (or basic) analyte. By assessing the equivalence point of the neutralization, the volume of titrant required can be correlated to the amount of an analyte present in solution. Acids or bases may also aid the solubility of nonpolar analytes by forming more soluble ionic salts.

Q31

In the reaction discussed in this question, elemental aluminum Al(s) replaces hydrogen in HCl to form AlCl3. In turn, the hydrogen atoms combine to form elemental hydrogen gas (as indicated in Table 2), which exists as H2 because elemental hydrogen is most stable as a neutral diatomic molecule. Educational objective: Single replacement reactions involve the replacement of one type of atom with another in a reaction between a neutral element and a compound such that a new compound and a new neutral element are produced.

Q35

Just had to balance the equation Educational objective: The conservation of mass dictates that the number of any type of atom be the same on both sides of a chemical equation (the equation must be balanced).

Q40

Knew how to draw out (C2H3O2) but forgot the equation for formal charge FC = Should - Has FC = valence electrons - lone pair electrons - sticks Educational objective: A formal charge is assigned to an atom based on an electron accounting method that assumes the bonding electrons between two atoms are shared equally. Formal charge equals the group valence of an atom minus both the number of nonbonding electrons and half of the number of bonding electrons.

Q7

Ksp is solubility product constant Any equilibrium constant of any chem rxn describes the molar ratio of products to reactants once equilibrium is achieved. It does not include a denominator because the reactant (magnesium metal, in this case) is a solid. **the amt of solid present does not affect the Ksp equilibrium** When the product of the concentrations of ions in solution is equal to Ksp, the dissolution and precipitation rates are equal. Under these conditions, the solution is said to be saturated, and any increase in ion concentration will result in precipitation. The equilibrium expression, in this case, is given by Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH−(aq) Ksp = 8.9 × 10^−12 = [Mg2+][OH−]^2 When the product of [Mg2+][OH−]2 is greater than 8.9 × 10−12, magnesium hydroxide will begin to precipitate. Educational objective: The solubility product constant Ksp describes the concentrations of slightly soluble compounds in solution at equilibrium. At equilibrium, the rates of dissolution and precipitation are equal, and the solution is saturated with the ions of the reaction. At concentrations above Ksp, the ions combine and precipitate.

Q8

LOOK AT PIC Educational objective: A complex ion consists of a central atom, typically a metal ion, and its surrounding ligands, which form coordinate covalent bonds to the metal. The number of coordinate bonds to the central atom is the coordination number.

Q33

LOOK AT PIC FOR MATH Educational objective: The limiting reactant is the reactant that is completely consumed first in a chemical reaction. It limits and determines the amount of product that can be produced based on the stoichiometric ratios of the chemical reaction.

Q4

Le chatlier principle is used here. For hyperventilation - you will exhale more CO2 so HCO3- will decrease bc equilibirum shifts left to make up the CO2. Also less CO2 in body = less acidic (more basic, increase pH) For hypoventilation - you will exhale less CO2 than normal meaning you retain more in body so equilibrium shifts right and there will be an increase in HCO3- to buffer. More CO2 in body = more acidic (decrease pH) **KNOW bicarb buffer system equation** Educational objective: Chemical reactions are in a state of dynamic equilibrium to maintain equilibrium. Changes in conditions will shift the equilibrium to counteract the changes. If a reacting species is removed, the reaction shifts toward the removed species. If a species is added, the reaction shifts away from the added species.

Q39

Lewis Base = lone pairs of electrons also notice that cis-isomer is a coordination complex and Cu is lewis acid and ligands (glycine) are lewis bases Educational objective: Within coordination complexes, a ligand acts as a Lewis base, and the metal center acts as a Lewis acid. Ligand structures that can form more than one coordinate bond by donating more than one pair of electrons have more than one site that can function as a Lewis base.

Q4

Lewis acid = electron acceptor (so look for something with positive charge) Bronsted Lowry Acid = proton donor (Q asks NOT a bronsted acid so look for something w/o proton) Educational objective: A Lewis acid is a molecule that acts as an electron pair acceptor, and a Lewis base is a molecule that acts as an electron pair donor. The Lewis definitions more broadly define acids and bases to include some compounds that lack acidic protons.

Q25

Lewis structures symbolically depict chemical compounds by using chemical symbols to represent atoms and dots placed around the symbols to represent the valence electrons and their bonding configuration. Dots placed between two symbols represent shared electrons involved in bonding whereas dots placed elsewhere around a symbol represent unshared, nonbonding electrons. Each shared pair of electrons forms one bond (often shown as a line). In Lewis structures, dots are placed around the elemental symbols in a way that gives each atom a full valence shell of electrons. Atoms that follow the octet rule use only s and p orbitals and achieve a full valence shell by having eight valence electrons (an octet). Exceptions to the octet rule occur for small atoms that have only an s orbital (H and He) and for larger atoms in Period 3 and beyond when vacant d orbitals are utilized. To draw the Lewis structure of SO2, the two oxygen atoms can be initially joined by a single bond with S as the central (least electronegative) atom, but to achieve an octet around each atom, an additional pair of nonbonding electrons must be moved to form a double bond (two shared pairs of electrons). (Choice A) Compound 1 has 10 valence electrons around one oxygen atom. Bonding that involves only s and p orbitals and follows the octet rule will have no more than 8 valence electrons around any given atom. (Choice B) Although Compound 2 has 8 valence electrons around each atom and follows the octet rule, the overall structure has a total of 20 valence electrons and has a net charge. Atoms from Group 16 each contribute only 6 valence electrons. Therefore, SO2 should have a total of only 18 valence electrons and no net charge. (Choice C) Compound 1 does not follow the octet rule, and Compound 2 has too many valence electrons. Educational objective: Lewis structures symbolically depict chemical compounds by using chemical symbols to represent atoms and dots placed around the symbols to represent the valence electrons and their bonding configuration. Dots are arranged to give each atom a full valence shell of electrons following either the octet rule or a valid exception.

Strong bases

Like Nathaniel Knows California Surely is Bad

Q15

Looked at Ca and P position in PT and can see that Ca is metal and P is nm so ionic for sure and then look at electronegativity difference (based off of position on PT) and realized phosphate and Oxygen is more EN so would have a partial negative while Ca has a partial positive. Nonpolar covalent bonds form between atoms with a difference in electronegativity no greater than 0.5. These bonds involve equal sharing of electrons, and they frequently form between two atoms of the same type (eg, carbon-carbon bond). Carbon-hydrogen bonds are also considered nonpolar because both atoms have similar electronegativities. No such bonds exist in calcium phosphate. Educational objective: Bonds between atoms are classified generally as ionic, polar covalent, or nonpolar covalent. Ionic bonds involve complete transfer of electrons from an atom with low electronegativity to an atom with high electronegativity. Covalent bonds involve electron sharing, and are considered polar if the sharing is unequal but nonpolar if the sharing is equal.

Q18

Looked at figure 2 in passage and found the structure that has the most net dipole without being cancelled out by another dipole and that one is in between 1.25<pH<4.14 Educational objective: Molecules have a net dipole moment (separation of charge) when the individual dipoles within it do not cancel each other. Symmetrical molecules typically do not have net dipole moments. Protonation and deprotonation can change whether a molecule is symmetrical and can induce or remove a dipole moment.

Q17

Looked at reaction 1 in passage and paragraph above to find charges for the Q Educational objective: Ionic bonds form when positive charges neutralize negative charges. One ion may carry multiple charges, and must be neutralized by the same number of opposite charges.

Q8

Metals such as magnesium, in the presence of anions with which they form soluble salts, will corrode more quickly than they will in pure water. Based on the anions in the answer choices, the possible magnesium salts are MgF2, Mg3(PO4)2, Mg(OH)2, and MgCl2. Looking at table 1 which is a solubility table shows that Mg2+ will react most readily with Cl- due to the high solubility value meaning it will corrode at the fastest rate than in the presence of any other anion. In fact, the NaCl present in blood plasma is a major factor contributing to the high corrosion rate of magnesium implants in the body. Educational objective: The rate of metal corrosion increases in the presence of anions with which the metal forms soluble salts. Solubility tables provide information on the relative solubilities of potential salts to be formed. Greater solubility corresponds to a faster corrosion rate.

Q21

Moving from the parietal cell (pH 7.0) to the stomach lumen (pH 1.0) involves a change of 6 pH units. The increase in [H3O+] going from the parietal cell to the stomach lumen can be determined as follows: Factor Δ[H3O+] = 10^−(1 − 7) = 10^6 Therefore, the [H3O+] in the stomach lumen is one million (10^6) times stronger than the [H3O+] in the cell cytosol. Educational objective: The pH unit is based on a logarithmic scale defined as pH = −log[H3O+]. The magnitude of [H3O+] changes by a power of 10 for each pH unit and is given by the expression Factor Δ[H3O+] = 10^−ΔpH.

Q23

Not Lewis acids bc they are metals and metals donate electrons to become cations while Lewis acids accept electrons. So didn't understand the answer choices C + D. Metal atoms want to give up electrons to become cations so its an oxidation (loss of electrons) not a reduction (so thats why C is wrong) and its formed from two neutral metal atoms. Educational objective: Metals tend to lose electrons to form cations whereas nonmetals tend to gain electrons to form anions. The charge of an atom increases by one unit for each electron lost but decreases by one unit for each electron gained. Metal cations with a smaller ionic radius and a higher positive charge are stronger Lewis acids than those with a larger ionic radius and lower charge.

Q28

Out of all the choices, (K2SO4) contains a metal. Therefore, potassium must form a metal-nonmetal (ionic) bond in the structure in addition to the nonmetal-nonmetal (covalent) bonds present between the nonmetal atoms in the structure. Educational objective: Identical chemical species must have the same elemental composition, the same number of electrons, and the same orbital bonding configuration. The net charge of a structure is equal to the sum of all charges present.

Q27

Out of all the choices, (K2SO4) contains a metal. Therefore, potassium must form a metal-nonmetal (ionic) bond in the structure in addition to the nonmetal-nonmetal (covalent) bonds present between the nonmetal atoms in the structure. Educational objective: The type of bond formed between two atoms depends on the relative difference in electronegativity between the atoms. Atoms with a large difference in electronegativity (usually a metal and a nonmetal) form ionic bonds. Atoms with a small difference in electronegativity (usually two nonmetals) form covalent bonds

Q42

Passage stated that "Symmetric and asymmetric bond stretching give four possible vibrational modes for each isomer, as shown in Figure 2....If the stretching of the bonds during vibration causes a transient, unequal distribution of electron density between the atoms within the bond, an oscillating dipole will form. Bonds with dipoles that are not cancelled out by another dipole in the molecule will absorb infrared light at particular frequencies (often expressed in wave numbers) and are designated as IR-active." So looking at figure 2 in passage you had to find the dipoles that didn't cross each other out bc those would show up in the IR spectrum. Educational objective: Vibrational modes that do not produce a net change in a dipole are not IR-active and will not produce an absorbance peak in an IR spectrum.

Q12

Passage states that Mg(OH)2 is an endothermic process. For endothermic reactions, heat can be treated as one of the reactants because it is required for the reaction. Heat can also be treated as a product for exothermic reactions. So if you increase the temp (which is stated in the Qstem) then you drive rxn towards products (Mg2+ and OH-). OH- increase means increased pH and in this case, Ksp increases as well bc more ions are dissolved. Educational objective: Adding heat to an endothermic reaction shifts the equilibrium toward the products according to Le Châtelier's principle; the opposite is true of exothermic reactions. For dissolving salts, changes in temperature change the solubility product constant Ksp.

Q47

Q asks to find which regions have noncovalent H-bonding and ionic bonding LOOK AT PIC Educational objective: Ionic bonds are ion-ion interactions between two charged species. When two oppositely charged species are within proximity of each other, they will form an ionic bond via electrostatic attraction. Hydrogen bonding is a special type of noncovalent dipole-dipole interaction involving species with partial charges.

Q52

Q asks which change will provide the smallest change in water solubility. So isoleucine to valine is nonpolar to nonpolar Educational objective: Water solubility requires that a compound has enough polar intermolecular interactions between the structure of a compound and the water molecules that the compound can be solvated and carried into solution. Hydrophobic molecules tend to be predominantly nonpolar and are sparingly soluble in water. Structural changes that increase polar sites on a molecule will enhance water solubility.

Q6

Polyprotic acids are acids with more than one H so they can give up more than one proton. At the first stoichiometric equivalence point, the initial acid species has been fully consumed and converted into its conjugate base. For a polyprotic acid, the first conjugate base species formed is amphoteric and can act as either an acid (by donating an additional proton) or a base (by accepting a proton and regenerating the original acid). Consequently, the pH at the first equivalence point results from pKa1 and pKa2 and must be between them. In this question, the pH of the triprotic citric acid solution at the first equivalence point must be greater than 3.13 (pKa1) but less than 4.76 (pKa2) Although unnecessary for this question, the exact pH can be found by averaging the pKa values: (3.13 + 4.76)/2 = 3.9 Educational objective: In a titration of a weak polyprotic acid and a strong base, the pH at a stoichiometric equivalence point is determined by the pKa values of the acid and the conjugate base species present in the equilibrium reaction.

Q17

Q asking which one of the following would produce the lowest osmotic pressure. So the easiest way to do this is by finding out "i" by writing out chemical equation for each and then multiplying by the molar mass to find the changeable part of the equation to find the lowest value on one side of the equation bc it will equate to the lowest on the other side of the equation as well. LOOK AT PIC IT DOES A GOOD JOB AT EXPLAINING Educational objective: Osmosis is the diffusion of solvent across semi-permeable membrane from a solution of lower solute concentration into a solution of higher solute concentration until the solution concentrations equalize. During osmosis, solvent diffusion imparts an osmotic pressure that is calculated by Π = iMRT, which is the product of the van 't Hoff factor, the solute molar concentration, the ideal gas constant, and the absolute temperature.

Q1

Q stem gives you 0.075g of oxytocin monoacetate and they want you to find what volume of the solutions tested would contain this 0.075g. in passage they give you 0.3mg/ml oxytocin monoacetate look at pic for math Educational objective: Solution concentrations are given in terms of the amount of solute per unit volume of solution. Solute amounts are typically given in terms of moles or mass, whereas solvent amounts are given in terms of volume. The volume of a solution containing a certain amount of solute is found by converting the solute amount to matching units of mass (or moles) and then dividing by the solution concentration.

Q6

Q stem says that HCO3- is a weak acid and is the analyte and NaOH is the titrant and the graph will show the analyte. Have to recognize that H2CO3 is a diprotic acid and will have two equivalence points and starts at pH of 6 bc its a weak acid also cuz the first pka occurs below 6.35. - The flat part of the curve corresponds to the buffering zone. At the center of this zone, the pH is equal to pKa, and the concentrations of the acid and the conjugate base are equal. - Beyond the buffering zone, the pH rises sharply. In the center of this region is the equivalence point, at which an equal molar quantity of base has been added and the acid is entirely converted to its conjugate base. For a polyprotic acid (a compound that can lose multiple protons), there are multiple pKa values and equivalence points. In this example, H2CO3 is diprotic, and therefore has two pKa values. pKa1 is 6.35 and corresponds to equilibrium between H2CO3 and HCO3−. pKa2 corresponds to equilibrium between HCO3− and CO32−. As visualized on the curve, the first buffering (flat) region is seen around pH 6.35, consistent with the pKa1 of 6.35 given in Reaction 1. pKa2 is seen at a higher pH. The first equivalence point occurs when all 5 mmol (50 mL at 0.1 M) of H2CO3 is converted to HCO3−. This requires 5 mmol of NaOH, corresponding to 25 mL of a 0.2 M solution. A second equivalence point occurs after another 25 mL of NaOH has been added, converting all of the HCO3− to CO32−. (Choice D incorrect bc) This curve shows the titration of a monoprotic acid with a pKa of 6.35 instead of a diprotic acid. Educational objective: On a titration curve for a weak acid, pKa corresponds to pH at the center of the buffering zone. Beyond this zone, the sharp rise in pH corresponds to the equivalence point. For polyprotic acids, multiple pKa values and equivalence points can be visualized.

Q20

Qstem mentions that its a weak base meaning its a proton acceptor based off of Bronsted Base definition so then you know its going to decrease the conc of H+ in body so A+B you can cancel out. And its gonna reduce the hydronium (H3O+) conc in body, also can figure it out from the Q stem saying it helps reduce the risk of uric acid stones meaning its gonna increase basicity and decrease acidity. Alsoooo The passage states that citrate alkalinizes (increases the pH of) the urine, and Table 1 indicates that uric acid becomes more soluble with increasing pH. The more soluble a compound is, the less likely it is to precipitate and form kidney stones. Citrate is a weak base and therefore a proton acceptor. It can alkalinize a solution by removing protons from hydronium ions, converting those ions to water and reducing the total hydronium concentration. Therefore, citrate may increase the solubility of uric acid by reducing the hydronium concentration and may decrease the risk of uric acid kidney stones. Educational objective: The pH of a solution is equal to the negative logarithm of the hydronium ion concentration. It is related to hydroxide concentration by pH = 14 − pOH and can be increased by adding hydroxide ions or removing hydronium ions.

Q21

Qstem says "two new coordinate bonds are formed..." Complex ions (coordination complexes) consist of a central metal ion surrounded by one or more ions or molecules called ligands that are bound to the metal center by coordinate bonds. The coordinate bonds form because the ligands surrounding the metal center act as Lewis bases and donate a lone pair of electrons to the metal center, which acts as a Lewis acid. These Lewis acid-base coordinations hold the complex together, but stronger Lewis bases can displace weaker Lewis bases as ligands within the complex. In the [Fe(H2O)6]3+ complex ion, an uncharged water molecule serving as a ligand is a weak Lewis base and can be displaced by the oxalate (C2O42−) anion, a stronger Lewis base. Each C2O42− anion displaces two water molecules and decreases the overall charge of the resulting complex ion by −2. (LOOK AT PIC FOR CHARGE) Displacement of water within the complex occurs in a stepwise sequence. In the first step, two water molecules are displaced from [Fe(H2O)6]3+ by a single C2O42− anion to form an intermediate complex ion: [Fe(H2O)6]3+(aq) + C2O42−(aq) → [Fe(H2O)4(C2O4)]+(aq) + 2 H2O(l) In the second and third steps, additional C2O42− anions displace the remaining water ligands to form the final product. (Choice B) This is the intermediate ion formed during the second step of the reaction, not the first step. (Look at hyperlink in problem) Educational objective: When a reaction occurs in a stepwise sequence, the species formed as products in earlier steps and then subsequently consumed as reactants in later steps are intermediate species. Intermediates do not appear in the overall net reaction.

Q36

Qstem says some solid precipitate is lost meaning that actual yield will be lower and its over the theoretical x 100 so it will be even less than 100% Educational objective: The percent yield is a description of the efficiency of a reaction. It is the ratio of the mass obtained from an experiment (the actual yield) to the calculated mass (theoretical yield) expressed as a percentage.

Q11

Raising the pH from 7.4 to 12 would increase OH- in solution so Le chatliers principle would shift the rxn to the left so less Mg would corrode, so it will corrode more slowly. This phenomenon is known as the common ion effect because it involves an ion that the salt and the solvent have in common. Educational objective: In solubility equilibrium, the common ion effect results in a decrease in solubility of the solid salt due to Le Châtelier's principle. Although the Ksp of the substance remains the same, the presence of one of the reaction products (the common ion) will push the equilibrium toward the reactants and decrease overall solubility.

Q34

Showed in the passage. Because the glowing splint burned more brightly in the gas that was produced when the MnO2 was added, one of the products of the decomposition of H2O2 is O2. From this, it can be inferred that the other product is H2O. Educational objective: Reaction types follow a general pattern and allow for the prediction of the products of a reaction that can then be verified in the laboratory.

Q19

Super easy, just read the Q wrong and thought they asked for hydronium conc but they asked for hydroxide (OH-) so pH = 6 pOH = 8 [OH-] = 10^-pOH ---> 10^-8 Educational objective: The [H+] and [OH−] concentrations are related by the self-ionization constant Kw of water according to the expression Kw = [H+][OH−] = 1.0 × 10−14 M2. Using a logarithmic approach, pH is related to [OH−] by the relationship pH + pOH = 14, where pOH = −log[OH−].

Q53

Sigma bonds, which are generally the first bond formed between atoms, are created when the electron bonding pair is between the two atoms such that their atomic orbitals overlap end-to-end along the internuclear axis. Because of the efficient overlap of the atomic orbitals, sigma bonds are low in energy and are very stable. Every additional bond formed between two atoms after a sigma bond is a pi bond. Pi bonds are created by the sideways overlap of p orbitals along a plane perpendicular to (ie, above and below) the internuclear axis. Because the overlap in electron density is not as efficient as for sigma bonds, pi bonds exist in a higher energy state and are not as stable as sigma bonds. As a result, they require less energy to be broken than sigma bonds (ie, they have a smaller dissociation energy) (Number I). Although individual pi bonds are weaker than sigma bonds, a double bond is composed of both a sigma and a pi bond, and therefore is stronger than a single bond (Number II). (Number III) Because the pi bond in C=O is higher in energy than the sigma bond, it is less stable. Educational objective: Sigma bonds are lower in energy and more stable than pi bonds, and therefore have a higher dissociation energy. Sigma bonds form first between two atoms, and every bond formed thereafter is a pi bond. Although individual pi bonds are weaker than sigma bonds, a double bond is composed of both a sigma and a pi bond, and therefore is stronger than a single bond.

Q14

Stronger bonds require more energy to break (dissociate) than weaker bonds. The overall strength of a bond results from the sum of all σ and π bonding contributors. As a result, when comparing bonds involving the same two types of atoms, a triple bond is stronger than a double bond, and a double bond is stronger than a single bond because double and triple bonds are composed of both σ and π bonds. However, if the strengths of the σ bond and π bond contributors are considered separately, π bonds are weaker than σ bonds. The end-to-end orbital overlap in σ bonds is more efficient than the side-to-side orbital overlap in π bonds. This causes σ bonds to exist in a more stable, lower energy state. As a result, breaking a σ bond requires more added energy than breaking a π bond (ie, a σ bond has a greater dissociation energy). Educational objective: Sigma bonds are lower in energy, more stable, and have a greater dissociation energy than π bonds. Although individual π bonds are weaker than σ bonds, double and triple bonds are composed of both σ and π bonds and are therefore stronger overall than a single bond.

Q56

Tetrahedral - sp3, 4 bonds, no lone pairs Trigonal pyramidal - sp3, 3 bonds, one lone pair Trigonal planar - sp2, 3 bonds, no lone pairs Bent - looks like water Educational objective: The geometry of a molecule can be determined according to valence shell electron pair repulsion (VSEPR) theory, which states that the distribution of bonds and lone pairs will minimize repulsion between electrons. To find the geometry of a molecule, count the number of electron domains and determine the electron group geometry, keeping in mind that lone pairs have greater repulsive forces than bonding pairs.

Q4

The best buffer to choose for maintaining any pH should have a pka that is within one unit of the desired pH. The passage says that the buffered solutions were adjusted to pH 4.5 so buffer acetate has a pka of 4.7 so best answer. The acid dissociation constant Ka of a weak acid indicates the relative strength of the acid. For convenience, this value is often expressed as the pKa, which is defined as the negative logarithm of the Ka. When there are equal concentrations of the weak acid (or weak base) and its conjugate in a buffer solution, pH = pKa Educational objective: A buffered solution consists of a mixture of a weak acid and a salt of its conjugate base or a mixture of a weak base and its conjugate acid. When the concentrations of the weak acid (or weak base) and its conjugate are equal, pH = pKa. The best buffer for maintaining a given pH should have a pKa that is within one unit of the desired pH.

Q8

The best indicator to use for titration is one that has a pH range closest to the equivalence point. equivalence pt = 6 or slightly less Educational objective: Indicators are added to a titrated solution during an acid-base titration to visually detect when a titration is complete. The best indicator for a titration is one that has an endpoint (color change) pH range nearest to the pH of the equivalence point.

Q20

The bicarbonate ion (HCO3−) is a base (proton acceptor) capable of neutralizing gastric acid through the formation of carbon dioxide and water: H+ + HCO3− → CO2 + H2O The [H+] decreases as bicarbonate consumes the available H+ from the gastric acid, and the relative [OH−] increases. As a result, the duodenum (pH 6.0) is less acidic than the upper stomach (pH = 1.0). Therefore, going from the highly acidic stomach to the less acidic duodenum gives a higher pH and a lower pOH, meaning [H+] decreases and [OH−] increases. Educational objective: Analogous to the pH scale, the pOH scale is defined as pOH = −log[OH−]. The pOH scale assesses acid-base measurements with respect to [OH−] whereas the pH scale operates with respect to [H+]. The pH and pOH scales are inversely correlated (as one decreases, the other increases), such that pH + pOH = 14.

Q22

The buffering range is generally 1 pH unit away from the buffer's pKa (the pH at which the species in the buffer are half protonated and half deprotonated moving in either direction along the pH scale). Outside this range, the constituent buffer species are either fully protonated or deprotonated and the buffer loses the ability to resist pH changes. The pKa of a buffer is related to its acid dissociation constant (Ka) by the equation pKa = −log(Ka). The question states that the medial Golgi maintains a pH of approximately 6.3. To stably maintain this pH during an experiment, researchers must use a buffer with a pKa within 1 pH unit of 6.3 in either direction (pH 5.3-7.3). A buffer with a pKa of 6 has a Ka of 10^-6, and a buffer with a pKa of 7 has a Ka of 10-7, so a buffer with a Ka value between 10-6 and 10-7 would have a pKa near 6.3. MES meets this requirement with a Ka of 7.1 x 10-7 and would, therefore, be the best of the available buffers for mimicking the pH of the medial Golgi. Educational objective: Buffers are mixtures of a weak acid or base and its corresponding conjugate salt that can resist changes in pH by neutralizing added hydronium or hydroxide ions. They are effective within 1 pH unit of their pKa. The pKa of a buffer is related to its acid dissociation constant (Ka) by the equation pKa = −log(Ka).

Q4

The compounds H2O, H2S, H2Se, and H2Te each contain bonds with hydrogen. The difference among these compounds is which of the Group 16 elements (oxygen group, or chalcogens) is bonded to hydrogen. Because electronegativity tends to decrease moving down a group (column) on the periodic table, tellurium is the least electronegative of the four chalcogens. As a result, the electronegativity difference between hydrogen and tellurium in the H−Te bond is the smallest of these four hydrogen-chalcogen bonds, and therefore it has the smallest dipole moment. Educational objective: The difference in electronegativity between two covalently bonded atoms is proportional to (and indicative of) the magnitude of the dipole moment along the bond between the atoms. Larger differences in electronegativity invoke larger dipole moments between atoms.

Q18

The free energy change equation was given in the passage. Convert the pHs in each cell into conc for eqn so [H+] and math is in the pic Educational objective: The equation ΔG = −kRT log(Q) describes the free energy (ΔG) of a system, where R is the gas constant, T is the absolute temperature, and Q is the reaction quotient (concentration ratio) related to the system. For systems involving hydrogen ion concentrations, the pH correlates directly to [H+] by the relationship pH = −log[H+] (or [H+] = 10^−pH), and ΔG for H+ transport is related to differences in pH.

Q2

The only freely rotating bonds are single bonds. Double and triple are stuck in place Educational objective: Covalent bonds are formed by sharing electrons between atoms through the overlap of atomic orbitals in an end-to-end or side-to-side configuration. Multiple bonds formed by s and p orbitals consist of one σ bond and one or more π bonds. Free rotation around a bond is possible with σ bonds (single bonds) but is prevented by the structure of π bonds (double and triple bonds).

Q2

The picture explains it really well Educational objective: Molarity, the number of moles of solute per liter of solution, describes the concentration of a solution. Stoichiometric mole ratios and molarity can be used to assess the moles in an aliquot of solution.

Q3

The question asks for the concentration of HCO3− given a pH of 7.35 and a PaCO2 of 40 mm Hg. The passage describes, the H2CO3 concentration in mEq/L can also be calculated from PaCO2 in mm Hg. [H2CO3] = 0.03×PaCO2 = 0.03×40 = 1.2 mEq/L and then use Henderson-Hasselbalch equation to get 12 (passage also said to find HCO3- conc to apply this eqn after finding the [H2CO3] Look at pic Educational objective: Consisting of a weak acid and its conjugate base, buffers help maintain the pH of a solution near the pKa. The Henderson-Hasselbalch equation is used to relate pH, pKa, and the concentration ratio of the acid and conjugate base.

Q57

The umami sensation is triggered by binding of tastant molecules to the GPCR T1R1+T1R3 heterodimer, ultimately leading to the release of Ca2+ ions. The number of cells that respond to a given tastant was determined using a calcium dye binding assay with fluorescence microscopy. As shown in Figure 3, the y-axis indicates cell response, and it is evident that D-amino acids elicit only a small response, if any, from cells; of the D-amino acids, only D-Ala caused a noticeable cell response. Because responses are caused by tastant binding, it can be concluded that a majority of D-amino acids do not bind effectively to T1R1+T1R3 GPCR cells. (Choice A) According to the experiment, each cell was exposed to an equal amount of both L-amino acids and D-amino acids. (Choice B) The FURA-2AM dye causes fluorescence when in the presence of calcium, not amino acids. (Choice C) Although D-amino acids do not trigger DAG formation, cell response is measured according to the amount of calcium release, which is triggered by the presence of IP3, not DAG. Educational objective: Interpretation of graphic data requires an understanding of the underlying mechanisms. Cells may have altered responses to stimuli for a number of reasons. In this case, amino acids induce a response by binding to a receptor, so failure to induce a response indicates a lack of binding.

Q50

Threonine has OH for side chain and alanine has valine has a branched chain so since threonine is polar it will have a strong dipole moment Educational objective: Dipole-dipole interactions are a type of noncovalent interaction that occurs between neighboring polar molecules with polar bonds and a net dipole moment. The partial charges within nearby dipoles experience a mutual attraction and align in such a way that partial negative charges orient toward partial positive charges.

Q51

Valine is branched (has more hydrocarbons), and alanine does not. Broadly speaking, van der Waals interactions are any noncovalent interaction between two neutral molecules in which the molecules themselves do not change. Van der Waals forces include attractions between two permanent dipoles (dipole-dipole interactions), attractions between a permanent dipole and an induced dipole (dipole-induced dipole interactions), and attractions between two induced dipoles (induced dipole-induced dipole interactions). Induced dipole-induced dipole interactions (also called London dispersion forces) are a type of van der Waals interaction in which molecules exhibit a weak mutual attraction. This attraction is due to the formation of instantaneous dipoles induced by momentary distortions in the electron distribution within the molecules. As a result, London dispersion forces tend to be more pronounced in larger molecules because larger molecules have a larger electron cloud and are more polarizable. Because the side chains of valine and alanine are both nonpolar, lacking polar bonds with permanent dipoles, London dispersion forces are the only van der Waals interactions present in region IV of the protein in Figure 2. The side chain of valine is larger (more hydrocarbon units) than the side chain of alanine. Consequently, replacing the larger valine with the smaller alanine in region IV will cause the extent of London dispersion forces (van der Waals interactions) to decrease. (Choice B) Alanine is a nonpolar, hydrophobic amino acid, and therefore does not have a dipole moment. (Choice C) Neither valine nor alanine have side chains with an -OH group or an -NH group that can engage in hydrogen bonding. The number of hydrogen bonds would not change with this amino acid replacement. (Choice D) Although both valine and alanine are nonpolar, the extent of van der Waals interactions will change upon replacing valine with alanine because of the difference in the number of hydrocarbon units in the side chains. Educational objective: London dispersion forces are a type of van der Waals interaction by which molecules exhibit mutual attraction because of instantaneous dipoles induced by momentary distortions in the electron distribution within the molecules. London dispersion forces tend to be more pronounced in larger molecules with a larger, more polarizable electron cloud.

Q1

Went answer choice by answer choice and added them to step 1 in Q and checked to see if they had the right answer and crossed out any product that appeared on both sides. Educational objective: According to the law of conservation of mass, atoms are neither created nor destroyed in a chemical reaction. The same number of each type of atom must be present on both the left and right sides of a reaction arrow. Multistep reactions that go to completion must consume all intermediates produced before or during the final step.

Q5

When a membrane permeable to an uncharged species is used to separate two solutions, the species diffuses down its concentration gradient until its concentration is equal on both sides. For instance, if 0.2 mL of solution with a PaCO2 of 40 mm Hg diffuses into 0.3 mL of solution, diffusion continues until the CO2 is dissolved at the same concentration in both volumes on each side of the electrode membrane. Essentially, this diffusion scenario can be viewed as a dilution of the CO2. Initially, all of the CO2 is confined to the 0.2 mL volume of the blood sample. After diffusion, the CO2 is distributed within a total volume of 0.5 mL (0.2 mL sample + 0.3 mL electrode solution). As such, the final concentration after diffusion can be calculated as follows: PaCO2(initial)×Vinitial = PaCO2(final)×Vfinal (40 mm Hg)(0.2 mL) = PaCO2(final)×(0.5 mL) PaCO2(final) = 16 mm Hg In this example, CO2 diffuses from blood into a reaction solution, where some of the CO2 is removed as it is converted to HCO3−. As CO2 is removed from the reaction solution, it will continue to diffuse from the blood into the reaction solution. Therefore, the final CO2 in the blood sample will be less than 16 mm Hg. Educational objective: An uncharged species will diffuse across a semipermeable membrane until its concentration is equal on both sides. If the species is removed from one side of the membrane, diffusion will continue in that direction until equilibrium is regained.

Q6

When placed within a coordination sphere, the atomic d orbitals of a metal are no longer degenerate, so they will have different energies. This occurs because some orbitals are pointing toward the ligands, which are electron donors, and other d orbitals are pointing away from or between the ligands. The orbitals that point toward the ligands are higher in energy than the orbitals pointing away from or between them because electrons repel each other. The energy of the orbitals determines which wavelengths of light can be absorbed. Some ligands cause greater differences in the energy of the d orbitals than other ligands, resulting in the absorption of different wavelengths of light. These wavelengths are usually in the visible spectrum. When heme binds O2, the nature of the O2 ligand changes the energy of iron's d orbitals. This energy change causes heme to absorb blue-green light and reflect red light. (Choice B) Oxygen's polarity does not affect color. Color is due to electron transitions. (Choice C) Oxygen's lone pairs are a factor in the difference in energy in iron's d orbitals, but the red color is due to electron transitions between the d orbitals, not the lone pair of electrons on O2. (Choice D) Although hemoglobin's protein conformation does change upon binding O2, this change in conformation is not the cause of the red color in blood. Educational objective: The nature of the ligands in a coordination sphere causes the metal's d orbitals to have different energies. This energy difference determines the wavelength of light absorbed by the bonds. The wavelength of light that is reflected is often in the visible region of the electromagnetic spectrum.

Q37

coordination number = number of coordinate bonds When all of these nearest neighboring atoms are from separate molecules or ions, the number of ligands will equal the coordination number. However, if two or more of these nearest neighboring atoms are joined to the same coordinating ligand unit, then the number ligands will not equal the coordination number. In both cases, the number of nearest neighboring atoms and coordinate bonds is unchanged, but the number of ligand units is different. LOOK AT PIC TO UNDERSTAND Educational objective: Coordinate covalent bonds are formed between two atoms when one atom donates both shared electrons. Such bonds are often formed between electron-poor metal ions and electron-rich atoms in ligands in a complex, and the coordination number of the complex refers to the number of coordinate bonds formed with the metal ion. The coordinately bonded metal and its ligands are called a complex.

Q1

did not know what a reference electrode was. A reference electrode is an electrode that has a known electric potential and standard conc of reacting species. The passage states in P2 that the mechanism used at E2 to measure CO2 is a modification of the mechanism used at E1 to measure pH. - E1 is a concentration cell, meaning both electrodes contain the same chemical species. In our case, H+ which diffuses from blood sample into E1. pH is calc from voltage diff bw E1 and its reference elctrode, which is maintained at a standard pH (stated all this is passage). pH is directly related to the conc of H+ by the eqn pH = -log[H+] - CO2 diffuses from blood sample into a rxn solution at E2. The rxn produces H+, and the change in pH is determined by the voltage diff bw E2 and its reference electrode. Therefore, the reference electrode for E22 should also be maintained at a standard conc of H+ (Choices B and C are incorrect) As stated in the passage, CO2 reacts to produce H+, and it is this change, not the change in CO2 or HCO3−, that is measured. Therefore, the reference electrode should be maintained at a standard concentration of H+. Educational objective: A reference electrode contains a known electric potential and concentration of ions. When both electrodes in a cell contain the same chemical species (a concentration cell), the electric potential difference (voltage) between the electrodes depends only on the relative concentrations of these species.

Q43

dsp3 means 5 bonds Not square planar bc it has 2 lone pairs Educational objective: Hybrid orbitals form when two or more atomic orbitals (s, p, d, f) combine into new orbitals. The number of hybrid orbitals (electron-dense areas) around an atom dictates its hybridization and electron geometry, whereas only the orientation of bonds around the central atom determines the molecular geometry (shape) of a molecule.

Q1

highest bond dissociation energy = strongest bond Educational objective: Covalent bonds are formed by sharing electrons between atoms through the overlap of atomic orbitals in an end-to-end or side-to-side configuration. Multiple bonds formed by s and p orbitals consist of one σ bond and one or more π bonds, and the overall bond dissociation energy of a bond tends to increase relative to a single bond with each additional π bond.

Definitions of Acids and Bases

look at pic

Q11

look at pic for math but asking for what volume of Copper II sulfate should be used to deliver 0.8g CuSO4 into the rxn mixture. In passage: Molar mass of CuSO4 Molarity of CuSO4 look at pic for math Educational objective: The molar mass expresses the mass present in 1 mol of a substance whereas the molarity of a solution expresses the number of moles of a substance dissolved in 1 L of a solution. The molar mass and the molarity relate to each other by the number of moles, and both can be used as conversion factors in chemical reaction calculations.

Q16

pH < pka = protonation pH > pka = deprotonation at physiological pH (7.4) the amino group is (+) and carboxylic acid is (-) the pkas of the side chains are 2.2 and 7.2, both are less than 7.4 so they will be deprotonated and looking at the structure there are 2OH's so they will both give off (-) charges so: (+)+(-)+(-)+(-) = 2- Educational objective: Some amino acid functional groups act as proton donors or as proton acceptors. Each proton donated by an atom decreases its formal charge by one unit, but each proton accepted by an atom increases the formal charge by one unit. The net charge is the sum of all nonzero formal charges in a molecule.v

Q24

shortest bonds = the strongest bonds Educational objective: Covalent bonds are formed by sharing electrons between atoms through the overlap of atomic orbitals in an end-to-end or side-to-side configuration. Multiple bonds formed by s and p orbitals consist of one σ bond and one or more π bonds, and bond length tends to decrease relative to a single bond with each added π bond.

Q10

stupid, didn't read the Q properly. It's asking which one IS NOT an accurate test for the rate of Mg corrosion. A. The chem rxn is given in passage and measuring H2 gas is valid bc Mg breaks down into H2 B. Measuring mass of MG disks is valid bc it should decrease in mass over time indicating corrosion C. In another rxn in the passage it shows the breakdown of products of rxn 1 in rxn 2 into Mg2+ so it is valid as well D. is correct that it is not a valid way to measure Mg corrosion bc the addition of acid will neutralize hydroxide ions in solution, which will shift the equilibrium toward Mg(OH)2 dissolution, and HCl will directly react with Mg(OH)2 as follows: Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l). As a result, the addition of HCl will increase the corrosion rate both by directly dissolving Mg metal and by dissolving the passivating Mg(OH)2 coating. Therefore, titrating the beaker contents with HCl during the experiment is not a good way to determine the rate of corrosion because the acid alters that rate. Educational objective: In a good experimental design, the method of measurement must not alter the experimental outcome.

Q45

write out the double displacement reaction: MgCO3 + HCl --> MgCl2 + H2CO3 This causes the formation of water-soluble MgCl2 (an ionic salt) and H2CO3 (an unstable acid). The unstable H2CO3 undergoes a decomposition reaction and breaks down (decomposes) into H2O and CO2 (carbon dioxide), which then evolves from the solution. Educational objective: Double replacement reactions involve the exchange of the bonding partners of two reactant compounds, forming two new product compounds. Decomposition reactions involve a single compound breaking down into two or more new compounds. Understanding reaction types and bonding patterns allows the prediction of reaction products.