CHEMISTRY COMMUNITY

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Yes we would have to assume that in order to be able to calculate the rate law

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can someone elaborate on what the solutions manual is talking about when we say that the second elementary reaction is fast and it does not affect the reaction order?? the rate law is determined by the slowest elementary step, which is step 1 in this case, so step 2 doesn't have any effect on it be...

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This is section 15.12 of the textbook which is one of the sections that we are responsible for knowing according to the syllabus, but I don't think we have covered it in class

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yes according to the syllabus we are responsible for knowing all of 15.1 to 15.15, except 15.9, for the final

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Yes I'm pretty sure this will always work, I've never thought about it like this so thanks for the tip!

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If you don't remember the units for k for certain orders you can always just divide the rate by the concentration. For 3rd order reactions, the units are (M/s)(1/M^3) which simplifies to (1/M^2)s

204918982 · Re: Quiz 3 yes the only problems we will be tested on are from 15.1-15.6

yes the only problems we will be tested on are from 15.1-15.6

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No but we don't have to worry about these until the final

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The concentration of reactants is negative so there needs to be a negative sign to cancel it out and make the rate positive

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you use a single line because they are different states

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standard potential has to be positive for the reaction to be spontaneous

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There will always be a negative sign in front of the reactants in order to make the reaction rate positive

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I just checked that link and the class that is bruincasted is not Lavelle's

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Yes, they are the same

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I think we will be given a chart with half reactions and their cell potentials and then we will have to come up with the reactions for the specific problem and find them on the chart

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So should we always put a + sign on positive answers for change calculations on tests/exams?

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Delta S=0 because the amount of heat lost by the system is equal to the amount of heat gained by the surroundings.

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Yes I think that's what we are supposed to use when it says to assume ideal behavior

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In a reversible reaction, the change in entropy of the system is the negative of the change in entropy of the surroundings, so the total change is zero

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It's not on our review sheet and he hasn't mentioned it in class, so we probably don't need to memorize it.

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Yes I think that's what we are supposed to do.

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It's usually liquid H2O because the temperature of the reactions in our homework problems is stated or assumed to be 25 degrees Celsius

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I got that same answer as well

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Yeah I think the standard units are just J/K and J/K/mol is used when the question asks for the entropy per mole.

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The cylinder itself is losing energy so the heat is negative

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atm and liters would probably be the best units since the conversion of L.atm to joules will be given to us on the constants and formulas sheet.

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This is the conversion of kPa to atm that we have to use in order to get our answer in joules.

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Yes this is only true when the pressure and volume are constant

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So thinking in terms of which one has more bonds is incorrect?

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Heat depends on the path that is taken not just its initial and final state, so it is not a state property

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yes, Bronsted acids donate a proton and Bronsted bases accept a proton while Lewis acids accept an electron pair and Lewis bases donate an electron pair.

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lone pairs are included and double/triple bonds are only counted as one area of electron density

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There is no difference between the formulas, but they calculate different things. K is calculated by using the equilibrium concentrations and Q is calculated by using the concentrations at a different time during the reaction.

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What does the R stand for again?

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Yes, they mean the same thing.

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We use cobalt in our answer for b and d because there is a positive charge on the complex and we use cobaltate in our answer for c because the complex has a negative charge.

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why can't pi bonds rotate?

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Any bond (single, double, or triple) only counts as one region of electron density.

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sofiakrylova3j wrote:So technically we don't know if a certain electron has positive or negative spin?

I don't think so. Unless you are given ms= 1/2 or ms= -1/2 then you know it's spin up or spin down.

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The two resonance structures can exist because the actual structure is a blend of all three, even though most of the time the sulfur dioxide molecule is in the state of the one with the formal charge of zero.

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I learned this method in high school too and I think it will work in this class, but it might be better to calculate the electronegativity difference.

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Yes they still can form compounds, but under extreme circumstances.

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I think he might prefer if we write it in the Px, Py, Pz form but I don't think we'll be penalized if we write it the other way

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the energy of (n)s is greater than (n-1)d so you write the d block before the s block. For example: 3d is written before 4s because the energy of 4s is greater than the energy of 3d

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n1 is always the ending energy level: E(final)-E(initial)

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I think the only ones we are expected to remember for now are red(700 nm) and violet(400 nm) and maybe blue(470nm)

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Josh Moy 1I wrote:If we have kg or km should we convert to g and m?

If the problem specifically asks you to convert to g or m then you should, but if it doesn't ask you to then I would suggest just keeping the units that you are given in the problem.

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Yes the question is asking for all the conditions that apply so C and D are both right since the energy of the photon can be equal to or greater than the energy needed to remove the electron.

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A high percent yield is better because it suggests that there were fewer impurities that occurred during the experiment and can help with the accuracy of calculations.

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Yes I got those answers too and I checked them in the solutions manual :)

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