CHEMISTRY COMMUNITY

Morgan Baxter 1E

Solids are not the only forms elements can be in to have an oxidation state of zero. For example, H2(g) has an oxidation state of zero.

Morgan Baxter 1E

It might be useful to know, but on the equation sheet he does give us the conversation between kPa and atm (1 atm = 101.325 kPa), so this is not something you would have to memorize.

Morgan Baxter 1E

If you understand what parts of the equations represent y=mx+b the graphs should make sense!

Morgan Baxter 1E

Keep in mind that the k you get using the isolation method is the pseudo rate constant (k') not the actual k since you are making assumptions that one reactant will be in such large excess.

Morgan Baxter 1E

I'm not sure, but would it be if the concentration of the reactant doubled, but the rate halved? Would this be an example of a -1 order reaction?

Morgan Baxter 1E

If a reaction is zero order it should not be included in the rate law. If it was there, it would falsely report that concentration of the reactant influences the rate of reaction. It does not, so it should not be included.

Morgan Baxter 1E

The mmol G means the amount of moles of product G. The L^-1 s^-1 is per liter second. Mol/L s is a typical unit for rate. The only difference is that this is millimol instead of mol, and it is specifying that it is millimol of G, as opposed to product F.

Morgan Baxter 1E

If the number that you get does not easily come out to a number that can easily round to whole number, it is likely that you have made an error in calculations. If you get an answer like that, make sure you go back and check your work.

Morgan Baxter 1E

The only time coefficients can be used in order to determine the order of a reaction is when it is an elementary reaction.

Morgan Baxter 1E

An electrode is a metal that is used conduct electricity.

Morgan Baxter 1E

You use platinum when the products and reactants of a half-reaction are in the same state (ex. aqueous).

Morgan Baxter 1E

Yes, this is the definition of a galvanic cell. They convert things from chemical energy to electrical energy. The cell potential needs to be positive. An example of a cell that has a negative cell potential is an electrolytic cell.

Morgan Baxter 1E

If something is being oxidized it is losing e-. It if is the oxidizing agent, it is causing something else to be oxidized. The only way to cause something to lose electrons (be oxidized) it to accept them (and thereby being reduced). So, being an oxidizing agent is the same as being reduced.

Morgan Baxter 1E

There are still 2 half reactions present.

Morgan Baxter 1E

Since Cu is on the right side of the cell it is the cathode. That means this is where reduction occurs. Therefore, there will be an accumulation of Cu(s) on the right side on the electrode, as it is reduced from Cu+2(aq). On the left side of the cell with Zn, it is the anode, so oxidation is occurri...

Morgan Baxter 1E

Sometimes the book expects you to know info that we are not always required to know for tests. I believe Professor Lavelle would give us this information, since it is needed for the problem, if a problem like this were to appear on the midterm.

Morgan Baxter 1E

Since he has not covered it in lecture, I would be very surprised if this was tested on the midterm. The only way I could see it being tested is perhaps in the background in a conceptual question.

Morgan Baxter 1E

I just remember that W= (# of possible states) raised to the power of (# of particles).

Morgan Baxter 1E

You are correct in thinking that the initial pressure (P1) is 15 and the final pressure (P2) is .5. However, you are using an incorrect equation, which resulted in the wrong answer. The equation is change in entropy= nRln(P1/P2). The equations for change in volume and temperature have ln(V2/V1) and ...

Morgan Baxter 1E

It is important to remember that q(sys)= -q(surroundings).

Morgan Baxter 1E

You are correct. When H is positive and S is negative, having a high temperature will increase G. In fact, since this equation is uses temperature in K, there can't be a negative temperature, so in these conditions, the G will always be positive. The book must have been referring to a case in which ...

Morgan Baxter 1E

I think either type of question can be asked. He can either give us the equation, or lets say there is a problem on the change in enthalpy of combustion, we would be expected to know how to write and balance the equations of combustion that correspond to the enthalpy of combustion data we are given.

Morgan Baxter 1E

The universe is isolated because neither matter or energy can be exchanged with it. This is because the universe encompasses everything, so it does not have surroundings it could exchange matter or energy with.

Morgan Baxter 1E

Also, more work is done in a reversible reaction. Looking at the graph of volume vs pressure for brother reversible and irreversible reactions, the area under the curve (aka the integral) is work. It makes sense that more work is done in a reversible reaction because the area under the curve is visi...

Morgan Baxter 1E

It is important that you understand both, because different problems will give you different units to work with. For example, if the problem gives you mass of iron in grams, use the specific heat (since that is in terms of grams). If you only have access to the molar heat capacity and are given iron...

Morgan Baxter 1E

I believe that the enthalpy of formation is a type of reaction enthalpy. Every formation reaction is a reaction, but not every reaction is a formation reaction. Therefore, enthalpy of formation is just one example of reaction enthalpy.

Morgan Baxter 1E

The first law of thermodynamics states that energy can't be created or destroyed. This means that throughout a calculation, the amount of energy at the beginning of the problem is the same at the end. Knowing this, you know that energy does not disappear, but instead can change forms (for example in...

Morgan Baxter 1E

For ice changing to steam you have to do several calculations. First you have to calculate energy needed to bring the ice to 0 degrees Celsius. Then calculate energy needed to phase change from ice to liquid. Then you have to calculate the energy needed to bring the water from 0 degrees Celsius to 1...

Morgan Baxter 1E

Also another way to think about it is that at 100 degrees Celsius, that is when boiling begins, not when all of it is 100% boiled. So more energy needs to be added to get it to be boiled 100%.

Morgan Baxter 1E

Another example is the earth. We are an open system because we get to both matter (meteors) and energy (sunlight) from outside of the earth itself.

Morgan Baxter 1E

In high school, I learned that these were the strong acids, and then you could assume that the rest are weak acids, since it is much more common for something to be a weak acid than a strong one.

Strong acids:
HClO4
HClO3
HCl
HBr
HI
HNO3
H2SO4

Morgan Baxter 1E

Yes, I think it would be better to use a double arrow to indicate that you know that weak acids are not completely dissociated in H2O, so there is an equilibrium present.

Morgan Baxter 1E

I agree, I think by omitting the double arrows and instead just having one forward arrow better shows your understanding of the reaction that is occurring. It shows that you have recognized that this is a strong acid/base.

Morgan Baxter 1E

For a strong acid, for example, HCl, that means nearly 100% of the acid dissociates into H+ and Cl- ions. This would mean there is a high K value for the dissociation of a strong acid.

Morgan Baxter 1E

First I count all of the valence electrons involved in the bonding. If you do not get this step right, your structure will be off. Then I figure out what is the central atom and place the other atoms around it. Then I put a single bond between each, because you know that there will at least be a sin...

Morgan Baxter 1E

K and Q are calculated the same way. If the reaction is aA + bB --> cC + dD, then K= [C]^c x [D}^d / [A]^a x [B]^b . K is when the equation is at equilibrium, Q is when the reaction is not at equilibrium. Based upon the values of K and Q, you can determine if a reaction is at equilibrium or how the ...

Morgan Baxter 1E

Since many of the homework problems in chapter 4 ask for the bond angle of familiar shapes, it will be necessary to know these for test #4 and the final.

Morgan Baxter 1E

Yes, Gianna's diagram is really helpful. There are two instances where the shape is linear:
1. 2 areas of electron density where none of which are lone pairs
or
2. 5 regions of electron density where three of which are lone pairs

Morgan Baxter 1E

Yes, you should use the diameter not the radius to find the uncertainty in position. So make sure if the question gives you the radius, double it to get the diameter. This is a small little detail, but it is important that you remember to do that so that you receive full credit for the problem.

Morgan Baxter 1E

You do not count the f-block square, and it is 5d6.

Morgan Baxter 1E

Yes this is correct. The 3d will be written before the 4s and 4p because principal quantum number n=3 is lower energy than principal quantum number n=4, and it is standard to write it in order of increasing energy.

Morgan Baxter 1E

Yes, it does not make sense/is not a reasonable model when the speed of the electron is faster than the speed of light.

Morgan Baxter 1E

The trend in ionization energies is that ionization energy increases from left to right and from bottom to top. You are correct in that the rule is to use the element with the lowest ionization energy as the central atom. Therefore, Cl has a higher ionization energy than C, so C should be the centra...

Morgan Baxter 1E

If you have a transition metal, the number of valence electrons you have will come from the s and d blocks depending upon how many electrons are lost. The 4s orbital has slightly lower energy than the 3d block. In iron, the 3d6 has higher energy than the 4s2. In Iron, the 4s2 electrons will be lost ...

Morgan Baxter 1E

I agree, I think that for this test we only need to know the two that he mentioned in lecture and that are mentioned in the book. My TA went over two others, but I think it is best to spend your time studying the two that we are most familiar with.

Morgan Baxter 1E

My TA told me that in terms of visible light, we should know the top and bottom of the range of wavelengths for visible light, and know which end corresponds to red and which corresponds to violet, but that we do not need to know, for example, exactly what is the wavelength for green and what is ora...

Morgan Baxter 1E

To do this problem you use the equation wavelength=h/(mv). You must look up the mass of an electron (in kg) to put in the equation for m. For v you put in the velocity, and h is a constant. Then you are able to solve for wavelength.

Morgan Baxter 1E

You say that you have memorized " long wavelengths cannot and short wavelengths can" in terms of wavelengths of light that can eject electrons from metal. From my understanding, it is not that all short wavelenghts can, it is just that having a shorter wavelength makes it so the frequency ...

Morgan Baxter 1E

Once you have the ratio in decimal form such as 1:1:1.333, in some instances it could be helpful to convert the decimals to fractions. Then the ratio is 1:1:4/3. Written like this, it is easier to see that multiplying by three would get rid of all the denominators and turn into a ratio of whole numb...

Morgan Baxter 1E

Yes I completely agree. The response above was stated very well. When thinking about net ionic equations, an important part is to remember the solubility rules. If you know your solubility rules (outlined within the fundamentals), then you will be able to determine which of the compounds in the prod...

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