CHEMISTRY COMMUNITY

gwynlu1L

also, if you need to you can take the concentrations at various times during the rxn and see if they have the same rxn rate. Graphing your findings also helps, since for a graph comparing time to [A] is a linear graph with negative slope.

gwynlu1L

I believe it is because you are trying to observe the instantaneous rate at the beginning of the rxn, before the product starts to go back to reactants (reverse rxn). If we tried to include products in the rate law, as in looking at the rates at a later time in the reaction, it would overcomplicate ...

gwynlu1L

another way to tell the order of the rxn is if the question gives you k. Based on the units of k, you can tell the order of the rxn.

gwynlu1L

pay attention to the y axis of a graph, since it can tell you what order rxn it is. but for the most part, associate ln[A] with first order rxns.

gwynlu1L

as a general idea, you use [A] for zero order, ln[A] for first order, and 1/[A] for second order

gwynlu1L

Also, if you're given a graph and it's comparing time vs concentration (not ln [a] or 1/[a]), then it is a zero order rxn

gwynlu1L

I think the question would have to specify if the cells use a salt bridge or a porous disk. I don't think you can just make that assumption on your own.

gwynlu1L

Yes, If the question doesn't specify if the solution is in an acidic or basic solution, than it is pretty safe to assume there will be no H+ or OH- in your equation. Only if it says that the solution is in an acidic or basic solution will you need to add H+ or OH-. Usually, you can deduce this by se...

gwynlu1L

also, remember that neutral elements, like O2 or Cl2 have an oxidation number of 0. Any single ion such as Na+ has an oxidation number equal to the charge of the ion.

gwynlu1L

Also, while balancing redox rxns, it helps to know the typical oxidation states of for example, group 1, group 2, halogens, etc. You need to know these to help you balance charges

gwynlu1L

also, if you're given the values for delta H and delta S, you can make a guess as to what T values will make the rxn spontaneous or not spontaneous. For example, if delta H is negative and delta S is positive, the rxn is spontaneous at all temps bc delta G will always be negative. However, if delta ...

gwynlu1L

All you really need to know that in order to calculate gibbs free energy you use the equation
delta G= delta H -TdeltaS
All of the other values in the equation can be measured, but i believe the only way to calculate delta G is through this equation

gwynlu1L

For Cp and Cv, you also have to pay attention to if the gases you're working with are monoatomic, linear, etc. That also changes that value of Cp and Cv.

gwynlu1L

I think you can make the assumption that isothermal reactions have deltaU = 0. That also means that q = -w

gwynlu1L

I believe that for monatomic ideal gases, Cv is 3/2 R and Cp is 5/2R

gwynlu1L

the 2nd law of thermodynamics is that entropy always increases in an isolated system. The 3rd law of thermodynamics is when the system reaches absolute zero, the entropy is zero; basically, the entropy of a pure crystalline substance (in perfect order) at absolute zero is zero.

gwynlu1L

I think it's better to think of these questions as 2 separate questions, not just 1 question. Reversible and irreversible reactions are not same, so their methods of solving are different. solving that kind of question is knowing that 2 separate experiments would be taking place.

gwynlu1L

also, Cp,m will be larger than Cv,m because at constant pressure the system is allowed to expand so the some of the heat is used to do work. on the other hand, at constant volume the system cannot expand.

gwynlu1L

typically, compression will have a positive work value since work is being done on the system to compress it. Expansion will typically have a negative work value bc the system has to put in energy in order to work against the surroundings and push the piston up.

gwynlu1L

For example, for questions involving the compression/expansion of air, there is work done on the system, but usually no heat involved. So, based on the equation, the internal energy change of the system would be the same as the work.

gwynlu1L

I would think you would usually just stick to the units given already in the problem, unless specified otherwise. although, it may be better to use J because it is the SI unit and in calculations.

gwynlu1L

Basically, you have to understand that a phase change takes heat to reach the vaporization point and more heat to actually change the molecules into the desired phase. Once enough heat is added to water that it reaches the vaporization temp, any xtra heat added will cause some of the molecules to ch...

gwynlu1L

I think the easiest way to explain it is that a phase change requires heat to raise the energy of a substance to the melting point/vaporization point, and then once this temperature is reach, the phase change requires more heat to actually change the state of the molecules into the desired phase. Ba...

gwynlu1L

ICE tables are almost always used to find the equilibrium concentrations, hence the E standing for Equilibrium in ICE (Initial, Change, Equilibrium). Questions involving Q would be giving you initial concentrations and asking you to determine which direction the rxn will proceed, and then possibly u...

gwynlu1L

I believe that these kind of questions would have to give you K, or at least have you solve for K by giving you the equilibrium concentrations to solve like as a first part. You need K to compare Q to, since depending on if Q is greater than or less than K, the rxn will proceed left or right.

gwynlu1L

To understand why by changing the volume, if the number of moles is different between reactants and products thus shifting the reaction left or right, you have to understand that volume and pressure are inversely related. You increase the volume, the pressure will decrease. So, in this circumstance,...

gwynlu1L

It also helps to remember that concentration = n (moles)/L (volume). Since moles do not change, a change in volume would change the concentrations. If it helps to understand the short cuts, you can try plugging in the new concentration values you'd get from changing the volume into the Q equation.

gwynlu1L

The reason why you solve for Ka by doing 10^-(pKa) is bc of the fact that pKa = -log(Ka), so the anti log would be Ka = 10^(-pKa). Remember that log has a base of 10.

gwynlu1L

hello, I know today in lecture Lavelle talked about how in weak acid/base equilibrium questions we ca make the approximation that x is not significant enough so we can simplify the quadratic and make solving for x easier, as long as x is less than 5% of the initial. can we assume this pretty much fo...

gwynlu1L

You will either be finding Kc or Kp in a given question. Usually, the question will list concentrations if it wants you to find Kc, but you do not want to mix up using concentrations and partial pressures. When you think about it, the two values are in different units, so putting them together in th...

gwynlu1L

I believe that if the question gives you concentrations of gases and asks for Kc and doesn't specify to find Kp instead, you can still find Kc instead of converting to partial pressures to find Kp. Only if the question specifies to find the Kp if given the concentrations would you probably have to c...

gwynlu1L

I believe you meant solvent. In the equilibrium constant, pure substances such as solids, solvents, or pure liquids are omitted bc their concentrations do not change significantly. Since pure water is usually the solvent used in many reactions, you omit putting pure liquid H2O into the equilibrium c...

gwynlu1L

You also have to consider the arrangement/structure of certain compounds to determine how many spots can bond to the metal center. For example, although water has 2 lone pairs that can potentially bind to the metal center, once one lone pair binds, the other one is not in a position to bind to the m...

gwynlu1L

There is also a list of common ligand names, both with neutral and negative charge that are useful when trying to identify a common ligand. It's not much but I used it a bit on the homework and it did help. I would say the only real trick is to just do a bunch of examples and get used to the differe...

gwynlu1L

I believe so. Consider what the circumstances may be in order for a compound to be considered amphoteric; It would probably need a lone pair available for a proton (H+) to bind to. But it would also need to have a proton (H) available to leave and bond to another compound. Try visualizing it with wa...

gwynlu1L

The question asks to identify the hybrid orbitals used by the atom in boldface red type. Im having trouble understanding how to figure out the hybrid orbitals, since in 2F.5 it was fairly easy, since all I did was count how many atoms are around the central atom to figure out hybridization (which di...

gwynlu1L

Sigma bonds also have the ability to rotate, while pi bonds are fixed.

gwynlu1L

I would think that the only thing you would have to consider with organic compounds is that there may be more than 1 central atom, and thus you would have to account for 2 or more shapes/structures within 1 molecule. For example, C2H4 I believe would include 2 carbons as central atoms, with both hav...

gwynlu1L

For the most part, hybridization is used to ensure that the correct number of electrons are free for bonding. Hybridization also helps with stability. Typically, All you need to understand is that if 2 bonds need to be formed, the hybridization is sp; 3 bonds is sp2; and 4 bonds is sp3.

gwynlu1L

You can't make the assumption that there are no lone pairs on a linear molecule without either knowing the lewis structure, or the VSEPR formula. When drawing a lewis structure, it becomes more clear that some molecules will have different amounts of electron density yet when taking into account lon...

gwynlu1L

For the most part, you won't have to compare bond angles with lone pairs to other shapes with lone pairs - mainly you are one comparing the shapes with lone pairs to shapes with only bonding pairs (same arrangement but different shape due to lone pairs). The questions from the homework at least will...

gwynlu1L

For shapes like see saw or trigonal pyramidal, you know there has to be a lone pair in order to make those shapes. Without the lone pair, the shape wouldn't make sense since there would basically be a gap in the shape. However, for a shape like bent, it is possible that the VSEPR formula is AX2, or ...

gwynlu1L

It’s good to remember that polarizing paper is the ability of an ion to pull electrons and distort the electron cloud. So, the more a cation can pul lat the electrons, the higher the power; so that must mean it needs either a large positive charge to pull on electrons or be small enough to get close...

gwynlu1L

Also, the distortion is describing the electron cloud of a bond. So, if more loosely held electrons (probably big ions they have lots of shells for shielding) are present, a cation will pull those electrons into the space between the two ions more, distorting the electron cloud.

gwynlu1L

Just remember that high polarizing power means either small size or big charge; smaller size means the cation can get closer to the electrons to pull stronger and bigger charge is fairly self explanatory I believe.

gwynlu1L

I don't remember Lavelle ever discussing that equation in class, so I would say no. Even if it may be in the text book, unless Lavelle doesn't really go over it in lecture I think it would be safe to assume you don't have to worry about it. You can also go to some of the review sessions the rest of ...

gwynlu1L

I wouldn't really focus on how many bonds an atom can potentially make, all you should really focus on is what number of bonds/how many electrons in the expanded octet give the atom the most stable formal charge.

gwynlu1L

you can think of it less as the electron being in 2 bonds at once, and more as the electrons having equal probability to be in a certain bond at a certain time. Don't focus much on the lewis structures for a visual sense, since that is the limitation of the drawings.

gwynlu1L

From what I remember from lecture, Lavelle said atoms in the 3rd row have the potential to have an expanded octet because the electrons in the d orbitals can become available for bonding. In addition, molecules that have an odd number of electrons such as NO or NO2 may be exceptions to the rule as w...

gwynlu1L

In general, px py and pz are just there to help distinguish which orbitals are filled, wether that be completely or partially. It is not crucial which ml number you associate with which p orbital.

gwynlu1L

yes, typically during your homework problems you can assume that for the p-orbitals, if ml=-1 it is px, if ml=0 it is py and if it is ml=1 it is pz. This just helps to distinguish which specific p orbitals are filled.

gwynlu1L

I agree with you, since it says the velocity was 5+/- 5 m/s, the range of velocities is from 0-10, making change in velocity = 10m/s. I was confused by the solutions manual too.

gwynlu1L

Lavelle only mentioned up to the f orbitals in lecture, and even then we will mostly be focusing on the s p and d orbitals bc the elements with those orbitals are the ones we will most use in the future. You shouldn't worry about the g orbitals.

gwynlu1L

Lavelle did say in class that the importance of how to derive the subscripts is not that important, since those kinds of concepts are not important to the class we are in, and are quite advanced. He mainly just told us that so that we have an idea of what the different d orbitals are and what certai...

gwynlu1L

Just be aware that most of your SI units are in kilograms, because depending on the question, they may ask for an answer in grams. This won't always be the case, but it's good to be aware of the units in different measurements like joules and watts, it makes conversions much more clear.

gwynlu1L

yeah, the mass of an electron acts like a constant, you always use 9.1095 x 10^-31 kg if the question calls for it.

gwynlu1L

You know what the work function for the given problem is, since they give you the lowest possible frequency. So, hv= work function. First, use 1.09 x 10^15 Hz to calculate the energy of the work function using E = hv. Then, use the wavelength they give you to calculate the frequency of the given lig...

gwynlu1L

yes, typically UV light is used. I believe in the original experiments, they used visible light but with no results. So, using UV light, with shorter wavelength and higher frequency, there was then enough energy to remove electrons. You can also think about how UV light is dangerous to our skin, sin...

gwynlu1L

You will also need Ek= 1/2 m(e)v(e)^2, since the question gives you velocity. using m= 9.109 x 10^-31kg, you can calculate the kinetic energy of the electron and then use that value to calculate the Energy of the photon. Then use E=hv, and c=wavelength x v to calculate what the wavelength is.

gwynlu1L

Yes, you need a balanced chemical formula in order to complete limiting reactant questions, but the equation only gives molar ratios, not the specific amounts for the given question. So the 2 from the chemical equation is not the amount you have in the question, it is just part of the ratio.

gwynlu1L

Be careful to use the right molar mass for the calculations. Pay attention to the wording of the question. Although chlorine usually exists as chlorine gas, Cl2, the question doesn't say that it is chlorine gas. Thus, you would use the molar mass of just chlorine, or 35.45 g/mol.

gwynlu1L

Yes, the audio-visual focus-topics are not due. It doesn't take a whole lot of time to do, and it does make the homework much easier. Plus, the topics covered in his lectures are on the modules as well, so after lectures it is helpful to do the modules to review what was taught in class.

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