CHEMISTRY COMMUNITY

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Yes. Because they are formed and used up they cancel each other out in the overall equation.

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A catalyst exits in the beginning of a set of reactions and also comes out at the end of a set of reaction, so it is used up and then produced. An intermediate is formed and then used up.

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To see if the slope is positive or negative you take the value of k from the equation.

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When you write out the half reactions and add them to make the complete equation the answer will result in the reaction that you base Q off of.

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You need to use the NERNST equation album make sure you set up Q as [anode]/[cathode].

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You need to take the known charges of the elements in the molecule and then find the different between that and the overall charge of the reaction.

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Balance it the same as you would with acidic but add OH- at the end to cancel out the H+.

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You are only supposed to flip E when you are given reduction potentials and you are looking for oxidation potentials. This is almost all the time becase our appendix gives reduction potentials.

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The rate limiting step is the slowest reaction that occurs.

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Lowering the activation energy and increasing the rate are the main functions of a catalyst. It’s important to note they go into the reaction and come out; in other words, they are not consumed.

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You need to add H2O to balance the oxygens and H+ to balance the hydrogens. When you add OH- because the solution is basic you must add it to both sides of the reaction to make H2O by combing H+ and OH-.

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Because E is intensive it does not rely on the amount products or reactants present, so you would NOT multiply the E value. The only time you alter the E value is when you flip the equation; then you would multiply it my -1.

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Graphite and gold are some other metals that also work in place of platinum.

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The cathode is the one being reduced and the anode is the one being oxidized.

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Whenever you are deciding which gas constant to use, look at your units and pick the R that will cancel them out.

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If it is a spontaneous galvanic cell, flip whichever equation will give you a positive Ecell. If they say it is non spontaneous, or the batter is dead, make the Ecell negative.

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Yes, Delta G = -nFEcell so if Ecell is positive Delta G must be negative and therefore spontaneous. Also keep in mind a negative delta G and positive Ecell means that the battery is working well.

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If it is a battery you want the Ecell to be positive so flip whichever reaction will give you a positive Ecell value.

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Use H2O to balance the Oxygen in the reaction and then use H+ to balance the Hydrogen as a result of adding H2O. Then balance the charges by adding e-. Multiply the half reactions so that the charges cancel each other out then add the reactions together. If it is a basic solution add OH- to both sid...

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You add H+ to the half reactions to balance the hydrogens on either side. You add OH- to both sides to the entire creation to get rid of H+ and turn it into H2O when you are balancing in basic solution.

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Only cancel the H+ with OH- when it says to balance the equation in basic solution.

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The Pt electrode simply acts as an inert conductor to facilitate e- transfer.

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If the reaction spontaneous E is positive because deltaG is negative.

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Potential difference = work/charge. From this you can derive Wmax=deltaG=-nFE which can tell you if the reaction is favorable or not.

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E standard is the E only under standard conditions.

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The splitting of the redox reaction into two half reactions makes it easier to balance the reaction by adding electrons and therefore adding h+ to the opposite side.

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Typically entropy will increase as heat is added and vice-versa because deltaS= delta q/T

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K reverse : 1/k
K*n : K^n
K*m: K^(1/m)

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Many times you will need to use PV=nRt to find the partial pressures of all the gases, then use an ICE table just like you would with contractions and a Kp value.

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Adding an inert gas increases the pressure of the system but will not affect the equilibrium constant K or Q because these gases do not react easily because they have a full octet

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No it shouldn't matter because the volume is no longer changing. You can think of it as doing the volume change with constant temperature then the temperature change with the constant final volume.

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The bond enthalpy method is the least accurate because all the bond enthalpy es are averages of many molecules.

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Yes. I’m pretty sure W is number of states^number of particles.

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When calculating the enthalpy change you need to add the enthalpy as a result of change in temperature as well as the enthalpy that comes from a phase change.

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q=enthalpy only when the pressure is constant because no other work is being done.

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Hess’ Law is essentially the definition of a state function for enthalpy.

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If you had the concentration of an acid or base you would need to find the [H3O+] or [OH-] then find the PH of the acid. If the PH or POH are very very low they are most likely strong acids/bases.

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ka*kb=kw=10^-14
pka+pka=pkw=14 (you get this by taking the -log of the first equation)
Also, these equations are on the equation sheet just replace [H3O+] with ka, [OH-] with kb, pH with pka, and POH with pkb.

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All acids/bases have a Ka or a Kb but strong acids and bases have such large Ka and Kb's that they are considered to fully dissociate.

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Whenever you see heat added or released from a system, you can treat it just like the concentrations of products and reactants using Le Chatelier's.

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You may also need to use an ICE table when asked for the PH. Indirectly they are asking for the [H3O+] concentration to find the PH.

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No delta S is entropy which measures change in disorder of a system. Delta H is enthalpy which measures change of heat in a system.

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Phase changes take up a large amount of heat so although heat is being supplied, the temperature does not change until the threshold for the new phase is reached.

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Hess' Law essential proves that enthalpy is a state function because you are allowed to add enthalpies without any regard for how the value was obtained.

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Percent ionization is the x, or [H3O+] concentration divided, by by the initial acid concentration.

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The energy required to turn a liquid into a gas is a lot more than the energy that is required to turn a solid into a liquid.

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When an inert gas is added, the concentrations of the products and the reactions are the same because there are the same moles present in the same volume container; therefore, K does not change.

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The ratio does not matter, but using whole number coefficients goes along better with the balanced equation.

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When 10^-3<K<10^3, neither the products nor the reactants are significantly favored. However, if K< 1, the reactants are slightly favored and the reaction lies to the left. Also, if K> 1, the products are slightly favored and the reaction lies to the right.

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When Q is less than K there are more reactants because a larger denominator in [products]/[reactants] will lead to a smaller value Q. Therefore, this reaction will shift to the right to regain equilibrium. When Q is greater than K there are more products because a larger numerator in [products]/[rea...

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Removing SO3 will lead to increase in not only NO but also SO3 because when the products are removed the reactants are favored, which means the reaction must shift to the left to once again reach equilibrium to accommodate for the higher concentration of the reactants.

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Strong acids are HCl, Her, HI, H2SO4, HClO3, HClO4, and HNo3.
The strong bases are group 1 and 2 oxides and hydroxides.

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ms tells you if the electron is spin up or spin down. If it is the first time filling the orbital it is spin up. If you need to pair an electron it will be spin down.

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You should convert nm to m so that the units cancel with the m in J.

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NH4+ cannot accept anymore electrons because its octet is full.

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HClO2 is the stronger acid because it pulls electrons away from the H-C bonds making them easier to break which results in more H+ ions in solution.

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If there is an acid in solution it will result in H+(H3O+) and if there is a base in solution it will result in OH-.

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The higher the H3)+ concentration the lower the pH and the higher the pOH. The higher the OH- concentration the higher the pH and the lower the pOH.

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A bent shape can have either 1 or 2 lone pairs.

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In this case it would be ferrate because the coordination compound has a negative charge.

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Myoglobin has the biological capacity to transport 1 single Oxygen molecule.

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To find the oxidation, take the charge of the one of the atoms you know for sure. O is always -2 and H is always +1. Essentially, take what you know and find the charges of the other atoms by adding or subtracting it to the charge of the molecule.

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The stronger the acid the more it will dissociate in water. The weaker the acid the less it will dissociate in water

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Do all the ligands in alphabetical order not paying attention to the greek prefixes then put the central metal atom. All of these must be in the coordination sphere. Following the items within the coordination sphere, put the other ions and molecules.

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A dipole cancels when a molecule has no lone pairs and the same attachments. It also will cancel when the net dipole moments cancel when they are facing the opposite directions. In addition, it is important to know that 2-3 linear geometry and square planar are non-polar when they have the same atta...

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If you look at Be, Al, Ga, Sn, Pb, and Sb they form a rough diagonal line on the periodic table

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The coordination number is the amount of atoms or molecules attached directly to the central atom.

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Because the transition metals are all good Lewis acids, they can all form coordination compounds, but I'm pretty sure he said we only need to concern ourselves with the first row.

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The coordination sphere is the sphere around the molecules and ions attached to the transition metal atom. The ligands need to be in the sphere because ligands are the ions attached to the transition metals. Anything within the "[]" is inside the coordination sphere

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H3O+ is has an electron arrangement tetrahedral because there are 4 regions of electron density. However, the shape of H3O+ is trigonal planar because there are 3 bonds and 1 lone pair in H3O+. Be sure when looking at a molecule that you differentiate whether or not you are looking for shape or elec...

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Yes. All single bonds are sigma bonds. As you add an extra bond, you add a single pi bond to the sigma bond, so a double bond is sigma and pi and a triple bond is a sigma with 2 pis.

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H20 has a bent shape because it has 2 lone pairs and 2 single bonds. When trying to determine shape, count all the bonding regions and lone pairs. I assume we will learn about this in class on Friday.

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Yes, you should know the shape names as well as the bonding angles. These are easiest to memorize when you look at them in terms of total bonding regions as opposed to individually. You do not need to know how to draw the structure in 3d space, only the lewis structures; 3d will be reserved for orga...

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In 3-d Space when all the H atoms are equally spaced from C and when all the H atoms are as far apart as possible, the angle of the H-C-H bonds are approximately 109.5 degrees.

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When electrons are close together, their negative charges repel each other which leads to a more unstable molecule. When the electrons are father apart, the electron electron repulsion is much less and in some cases 0.

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When trying to determine shape, only pay attention to bonding regions, not the amount of bonds.

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For a molecule with a +1 charge such as ammonium, NH4+, you would subtract one e- from the valence e- so the total valence e- would 8. For a molecule with a -1 share like sulfate, SO4^2-, you would add 2 extra valence e-. The result is 32. To draw the lewis structure, just be sure to satisfy the amo...

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Lone pairs are pairs of electrons that are not shared with other elements. They are represented by dots around an element. To find the amount of lone pairs, count the number of dots and divide that number by 2.

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The double bonds are shorter than the single bonds because the extra e- creates more attractions between 2 elements. However, for benzene there are not true single or double bonds. Benzene actually has delocalized bonds, so rather than drawing a hexagon with alternating double bonds, the actually st...

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Yes, for Ca the electron configuration would en with 4s^2. For Ca^2+, the electron configuration would end with 3p^6.

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Bond dissociation is the energy required to break a chemical bond. For example, to break an H--H bond, the bond dissociation energy is 436D.

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Whenever you are writing an electron configuration the orbital with the lower n should come first.

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The 2 double bonds on the sulfate atom bring the formal charge of sulfate to 0. It is more important to have a lower formal charge on the central atom because it has a lower electron affinity.

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The Px, Py, Pz should only be used when showing if the orbitals are full.

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A single bond is counted as one electron although a single bond has a value of 2 electrons because 1 electron goes to one atom and the other electron goes to the other atom.

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The electrons closer to the nucleus experience a stronger attraction and shield the outer electrons from receiving the same type of attraction.

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The valence electrons are electrons in the outer shell. If n=3, all of the electrons in that shell are the valance electrons.

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The Pauli Exclusion Principle essentially means that there cannot be more than 1 quantum number per election because, even if they share the same subshell, they will have different spin. If more than 2 electrons were to exist in an orbital they would have the same quantum numbers.

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If the ion is positive, electrons must be taken away from the electron configuration because electrons have a negative charge. When the ion is negative, electrons must be added to the next available subshells.

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You should not have to memorize anything since it can all be found on the periodic table.

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Ground state means that all the electrons are in the lowest possible energy levels. An excited state is when the electrons exist in a higher energy level.

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The 3d orbital has less energy than the 4s orbital so it must come before because electron configurations are written in order based on amounts of energy.

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The smallest detectable wavelength is a gamma ray at 10^-14 meters. Professor Lavelle said anything smaller than 10^-15 meters is irrelevant and we should consider it to have an undetectable wavelength.

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You need to use stoichiometry. (Unit A)(Unit B/Unit A). This will work for any unit so long as they are similar. For example: meters with millimeters or grams with kilograms.

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Combustion, or burning, occurs whenever a hydrocarbon interacts with Oxygen. The products are Carbon dioxide and Water. The more oxygen present the more successful the combustion will be.

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You should apply sig figs at the end of the calculation so that you do not get a rounding error for your final answer. It will be more accurate if you wait until the end.

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You need to memorize it based on where the element shows up on the periodic table. The first S orbital begins with 1. The first P orbital starts with 2. The first D orbital starts with 3. The first F orbital starts with 4. For each row you go down on the periodic table these numbers increase by 1.

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The theoretical yield is calculated using stoichiometry to find the ideal mass of a product under ideal lab conditions. The percent yield shows how accurate the actual yield is in comparison to the theoretical yield. The percent yield is calculated using the following equation: percent yield= (actua...

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Scientific notation should be used to provide a more simplified view of an extreme number. .000006 is difficult to understand quickly so 6.0 x 10^-6 is used so that scientists can understand the number and it’s ramifications more easily.

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You can have the final units in whichever unit makes the most sense. For example, you would not want a final mass to be 9.8 x 10^-2 kg because you could just say 98g unless it explicitly asks for kg. As you work the problem, you should use whichever units are easiest to use. Grams are usually easies...

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