CHEMISTRY COMMUNITY

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Dr. Lavelle said in lecture that we do not need to know what that species is- it will be given to us. That is way outside the scope of our knowledge at the moment.

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Yes the bolded times are the review sessions while the others are the TA's office hours

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for zero-order rxns (n=0), [A] vs time in a decreasing linear graph

for first-order rxns (n=1), ln[A] vs time in a decreasing linear graph

for second-order rxns (n=2), 1/[A] vs time in an increasing linear graph

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Your answer is correct, but you do not need the parentheses around the product.

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The equation for half life of something in the first order is t1/2 = ln(2) / k, which = 0.693/k

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We use H+ to balance redox reactions occurring in acidic conditions and OH- for redox reactions in basic conditions. Dr. Lavelle has two handouts about this that are super helpful on his website towards the bottom. The section is called Redox Reactions.

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The problem deals with basic solution, therefore you should not have H+ ions in your equation.

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The oxidizing agent is reduced and the reducing agent is oxidized.

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As we learned in the first unit of 14B, solids are not accounted for in Q nor K. Therefore, it would just be 0.768/0.0150 for Q

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You should try deleting the parentheses. I had the same answer except without the parentheses.

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If there are no solids, there needs to be an inert conductor, and the most common one used is Pt(s)

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You will typically use the gas constant according to what is given in the problem. You want to cancel out units that are not in the final answer.

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A state property means that the path taken does not matter. You can directly calculate the value with the initial and final values without regarding what is going on in between.

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Uppercase K means the equilibrium constant, or it could mean Kelvins. lowercase k would be the boltzmann constant.

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It depends on what the problem is asking. You typically should look at the units to determine which gas constant R to use.

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It depends on which constant you're using. Typically the constant is R = 8.314 J/ Kxmol, so you would have to use Kelvins in that case.

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Yes, m is mass and n is moles. If you're given m, you should use the specific heat capacity and if given n you should use molar heat capacity. You could also convert the mass to moles and moles to mass to use either of the heat capacities.

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They're the same equation, just with different units. So it would depend on what units you get in the question, and then you will have to either use the specific heat capacity or the molar heat capacity.

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You should use Hess's law since this is a state property. So you would manipulate equation 2 by reversing the reaction (x-1) and multiplying by 2 (x2) to both the delta S and delta H values. Then you would calculate delta G for each value, and then get the delta H of the rxn by adding them together....

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The most accurate is 273.15K = 0 C which is given on the equation sheet. The textbook and sapling problems do use 298 K as 25 C though.

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It's much more straightforward to calculate delta S using the riemann sum equation to get delta S reaction.

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delta U would = n Cv delta T, Cv is 4R - R = 3R so the equation would be delta U = (n mol) (3R) (delta T)

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w = -P ext x delta V.

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You decide which value to use by looking at the units. You choose which is the most desirable to get the units of the value you want.

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You would use the initial pressure to calculate the number of moles.

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When the volume is constant, there is no work being done. U = q + w, and when U = q + 0, therefore U = q

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It depends on the units the problem gives you. You should know what units the answer should be in and then use the appropriate R to cancel out the units for the given information.

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When bonds are formed, energy will be released which would make it exothermic. When bonds are being broken, energy is required to break them, which means it is endothermic.

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We make X negligible when X is 10^-4 or smaller. There is a guideline that says 10^-3, but if it were something like 9 x 10^-3, that would be closer to -2, which is why using 10^-4 is safer.

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If you're given the initial concentrations for both reactants and products, you would calculate Q in order to see which way the reaction will proceed towards. If Q > K, reactants will be favored. if Q< K then products will be favored. If you are only given the initial molar concentration for the eit...

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Inert gases are not reactive, therefore they would not "shift" the equilibrium.

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pKa would be -log ([H30+])
Ka would be 10^(-pKa)

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HClO3 is considered to be one of the 7 strong acids.

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That equation requires the usage of Kelvins.

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Sapling included HCLO3 as a strong acid last quarter.

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Solids and Liquids are not included in the equilibrium tables, only aqueous solutions and gases.

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We are focusing on Bronsted-Lowry acids and bases, because we're dealing with the proton transfer.

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Depends on the reaction. If concentrations of both products and reactants are given, you could calculate the reaction quotient(Q) and see whether it is small, equal to, or bigger than K. This way you could determine which direction the reaction is proceeding. If only one of the two (R&P) concent...

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The x's are determined by the stoichiometric coefficients.

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The concentrations should stay the same because the rate of the forward and reverse rxn's should be equal.

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Dr. Lavelle tells us to write the chemical reaction with the (l) or (aq), as an easier way to identify if something is a pure liquid. Typically problems will include the chemical reaction, specifying which of the reactants and products are liquids or solids, and which are aqueous and gases.

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Kc is the equilibrium constant, so we use it to describe the forward reaction. When we are describing the reverse reaction(when products are making the reactants), we will use 1/Kc to describe the equilibrium constant. Another two to remember are when we multiply the moles of the entire reaction by ...

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We do not use solids or liquids because they do not have concentration. We will include all gases, and aqueous solutions as well because they have concentration (we can calculate their molarity)

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We do not exclude any gases. We will only exclude H2O when it is in liquid form. All liquids and solids are excluded because they do not have concentration. All gases and aqueous solutions will be included in the equilibrium constant/expression.

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From all the examples we have done so far, when there are two positive values, one of the x values will be larger than the initial concentration we need to subtract x from. Since it's not possible to have a negative concentration, the one that is smaller than the initial concentration will be the an...

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Since we will probably be dealing with weak acids and bases this quarter, every acid and base that is not part of the strong acids and bases is a weak acid/base. So the strong acids are HCl, HClO4, HClO3, HNO3, HBr, HI, H2SO4. and the strong bases are the groups one and two metal oxides.

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The conjugate seesaw essentially says
- the stronger the acid, the weaker the conjugate base.
- the stronger the base, the weaker the conjugate acid.
- the weaker the acid, the stronger the conjugate base
- the weaker the base, the stronger the conjugate acid.

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The answer is C, because even during equilibrium, there are chemical reactions happening at a "molecular level." Equilibrium is achieved when the rate of the forward reaction equals the reverse reaction, therefore C is false.

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Depends on what the problem gives you. If it gives you units in bar/atm, it will most likely be Kp. If it gives you units in M or x moles in x L, it will probably be Kc.

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It's easier to use whole numbers because you will usually end up with a quadratic equation already, and it's easier to calculate when you have as many whole numbers as possible. Theoretically though, you could, as long as you don't mind the decimals.

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Yes, if the number is between 10^-3(or 10^-4 in some cases) and 10^3, they are not strongly favoring neither reactants or products.

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We should be using Kelvins

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K is the equilibrium constant and can be Kp or Kc. Kc is used for concentrations.

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F is very electronegative, so the bond between H and F are stronger than with the other H-halogen acids, which means its a weaker acid!

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They are basically the same thing. When we say we donate a proton, typically means donating an H+ which will form H3O+, so theyre essentially saying the same thing.

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ABCDE were just points in the data collection where S is the pH at 7 where the solution is at the stoichiometric point.

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The more oxygen atoms there are, the stronger the acid.

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The bond between HF is very strong because F is very electronegative. A strong acid dissociates/ionizes almost completely. It would be hard for the aqueous solution to dissociate HF, therefore it is not a strong acid.

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I also had this problem with today's lecture. I took a break and refreshed it 40 minutes later and it was fine again. It might be due to the amount of people on CCLE since finals are coming up.

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In terms of looking at salts, you can tell which salts will behave as acids or bases. For example, NH4+ Cl- + H2O -> NH3 + H3O+ Cl-, chlorine does not affect the pH which means it has no effect as an acid or base. However, H3O+ concentration increases, which lowers the pH. You will need to see whic...

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CO2 and H20 mix to make carbonic acid which is H2CO3. H2CO3 can give off protons and become HCO3-, another common acid (bicarbonate).

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25 degrees is considered the standard, much like we consider room temperature.

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Na2SO42- would be the neutral salt that occurs when the strong acid and strong base are combined into one solution. The result of the combination of a strong acid and strong base would be the salt and H20 at the stoichiometric point where the pH is neutral

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In the textbook problems that Dr. Lavelle has assigned, specifically 9C.9, it asks us to determine the coordination number of the metal ion in the following complexes. For c and d, it gives [PtCl2(en)2]^2+ and [Cr(edta)]^-. Without knowing the structure and formula of ethylenediamine and EDTA, how w...

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You see how many bonds are formed with the Transition metal, by counting the number of atoms/molecules inside the square brackets.

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The longer the bond length, the weaker the bond, which means it is more easily dissociated/ionized. In lecture, Dr. Lavelle said that strong acids are almost completely ionized/dissociate in solution, therefore the longer the bond, the stronger the acid. (In most cases)

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The coordination number is the number of ligands attached to the central TM atom, which is essentially the stuff inside the square brackets.

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A strong acid(or even base) is almost completely ionized/dissociated in solution. The longer the bond, the more easily the acid dissociates, which is why HBr would be a stronger acid than HCl

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coordination number is the number of ligands attached to the transition metal, essentially the atoms inside the square brackets.

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Hemoglobin is in blood, and myoglobin is in tissue ie. muscles etc. essentially myoglobin gets oxygen from hemoglobin, and hemoglobin gets oxygen when it passes through our lungs when we breathe in the oxygen, the local partial pressure of oxygen is the highest in the lungs which is why hemoglobin b...

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Yes, bent and angular are both terms used for that shape.

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The hw is due on Sunday at 23.59!

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Tetrahedral should be correct, maybe check your spelling!

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It is sp^2 because there are three regions of electron density. The double bond with the other N, the lone pair, and the single bond with C. The double bond still counts as 1 region of electron density, so there are 3 regions which means it is sp^2.

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The dipole moments do not cancel out in NO2 similarly to SO2, but unlike CO2 because CO2 has a linear shape. NO2 and SO2 both have lone pairs, which due to electron repulsion cause the shape to be bent/angular, therefore the dipole vectors do not completely cancel out and you will have a polar molec...

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We should know the approximate angle, ie. greater than, less than 109.5, etc.

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The most stable lewis structure is the structure with the lowest formal charge. The more electronegative atoms would be more likely to hold the negative formal charge in the molecule.

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You would look at the electronegativity of the atoms. The most electronegative would be the more negative pole of the dipole moment.

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This kind of shape can be called angular or bent.

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Bond angles will depend on what molecule you are examining.

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You would draw the dipole moments on the lewis structure(by determining en) and then using vector addition, you could see which of the dipole moments cancel.

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Hydrocarbons are typically nonpolar, and molecules that have cancelling dipole moments are also nonpolar. It would be much more efficient to know how to draw the lewis structures and the dipole moments to know which cancel out.

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Lone pairs count as e- density regions as well.

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calculate the number of electrons, then find the atom with lowest ionization energy to put in the center, then find the number of lone pairs on the central atom and how many bonds the central atom forms, and then you draw the structure out with as much symmetry as possible. Then you can calculate th...

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The electronegativity of H is 2.2, F is 4.0(the most electronegative element), and C is 2.6. The EN difference between C and H is only 0.4 while the EN difference between C and F is 1.4. So CH4 could even be classified as a nonpolar covalent bond, while CF4 would display more ionic character.

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These are both intermolecular forces, but LDF and dipole-dipole are different. LDF could be used interchangably with induced dipole-induced dipole/van der waals forces/dispersion.

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Radicals have an unpaired electron, you can usually tell when the total number of valence electrons you calculate out before you draw the lewis structure are odd.

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It is of course best for the whole structure to have a formal charge of 0, but if that doesn't happen, i think the second most stable form is when the central atom has a formal charge of 0.

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The equation is the most straightforward way to find the formal charge and find the most stable lewis structure.

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n gives the energy level, and every orbital can hold up to two electrons. l=0 would correspond to s sublevel, l=1 would be p, l=2 would be d, and l=3 would be f, etc. for instance, n=5 l=2 would be 5d based on the definition we gave for n and l.

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We will need to know the trends, but it is not expected of us to memorize the specific electronegativity of each element.

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Yes, the first three quantum numbers can be the same, but the fourth must be different as each electron on that orbital would have a different spin.

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In the example Dr. Lavelle gave during lecture, the lone pair donor(lewis base) was NH3 and it was on the N atom, and the lone pair acceptor(lewis acid) is B from BF3.

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Lucy's analogy of people standing around the fire pit is a good example! Since s-orbitals are closer to the nucleus, they are more effective at shielding.

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Sunday at 11.59 pm or 23.59.

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We don't have to draw them to be longer vs shorter, but we should know double bonds are shorter and single bonds are longer.

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Honestly, going to UA sessions and office hours is super helpful. Going through each of Dr. Lavelle's lectures and processing the information, and then APPLYING the actual stuff we learned is also important. Doing all the homework problems in the outlines. Going to as many UA sessions as possible re...

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Yes, during lecture 12(10/30), Dr. Lavelle used the example of Potassium Chloride to show the lewis structure of an ionic compound.

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It's probably better to draw the lone pairs as pairs of dots, and to leave the lines to denote bonds.

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Dr. Lavelle has said that he would prefer for us to use the longer notation because it will tell us more about the configuration ie. which orbital the electron is occupying rather than writing 2p3.

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