CHEMISTRY COMMUNITY

Alex Nechaev 1I

Yes, and the electrodes should go on either side as well.

Alex Nechaev 1I

The anode is usually on the left for our purposes, and they probably will not switch it up since we haven't seen any examples in class or on homework proving otherwise.

Alex Nechaev 1I

K can be used calculate the final concentrations at equilibrium for a reaction using an ICE table and the natural progression of the reaction, from left to right or from right to left. Q expresses the relative ratio of products to reactants at a given instant.

Alex Nechaev 1I

Since there is already a negative sign in the formula for determining the reaction rate of a reactant, the subtraction of concentrations that would come out negative becomes positive. I think that reaction rates are just always expressed as positive values.

Alex Nechaev 1I

I thought they were related at first as well, but the coefficients actually play no role in determining reaction order. For this, you look at concentrations and reaction rate.

Alex Nechaev 1I

Coefficients do affect unique rates, but not the general rate law or determining formation or consumption of products or reactants.

Alex Nechaev 1I

It could have something to do with the fact that for unique rate you have to divide by the stoichiometric coefficient, but I do think that in this case you are supposed to multiply by 3.

Alex Nechaev 1I

The question will usually be asking for the formation of products if asking for reaction rate, so generally whichever species is the product will be the one to which the rate refers.

Alex Nechaev 1I

Adam Enomoto 1L wrote:Reaction A: k = 2.3 M-1s-1
Reaction B: k = 1.8 Ms-1
Reaction C: k = 0.75 s-1

How do we know which one is a zero order reaction just given the units?

I would just try to memorize the units for at least the first few orders.

Alex Nechaev 1I

Here's the general layout of a cell diagram

Alex Nechaev 1I

The electrodes are also included on either side of the cell diagram as the first and last elements.

Alex Nechaev 1I

The problem will usually tell you whether the solution is acidic or basic.

Alex Nechaev 1I

Just remember opposites. The element being oxidized is the reducing agent and the element being reduced is the oxidizing agent.

Alex Nechaev 1I

Just remember opposites. The oxidizing agent is being reduced, and the reducing agent is being oxidized.

Alex Nechaev 1I

I always like to use OIL RIG (Oxidation Is Losing, Reduction is Gaining) to remember which elements are being oxidized and reduced. Once you know this, remember that the element being oxidized is the reducing agent and the element being reduced is the oxidizing agent.

Alex Nechaev 1I

Knowing whether a system is closed or isolated only tells you whether it can exchange energy with its surroundings or not. Don't assume that pressures and volumes are constant unless the problem specifically states that they are.

Alex Nechaev 1I

204918982 wrote:So should we always put a + sign on positive answers for change calculations on tests/exams?

I would say yes, since the examples we have done in class and the solutions in the manual all include the + sign for positive answers.

Alex Nechaev 1I

Are we going to need to know the difference between translational, rotational, and vibrational energy?

Alex Nechaev 1I

Negative entropy simply means that there is a decrease in disorder, or rather an increase in order of the system.

Alex Nechaev 1I

Delta S is the entropy of the system, and can be positive or negative to describe that change. Delta U is the change in the internal energy of the system.

Alex Nechaev 1I

This is basically just a mathematical explanation of the second law of thermodynamics. It gives us a way to determine whether a cyclical process is reversible or irreversible.

Alex Nechaev 1I

If a system's temperature increases, that means that heat is being released from the reaction to increase the temperature of its surroundings. Don't think of the system itself as the reaction, but rather as the surroundings of that reaction whose change in temperature determines whether the reaction...

Alex Nechaev 1I

Also note that the equation given above is used for irreversible processes.

Alex Nechaev 1I

The equations used to calculate work also vary depending on whether the process is reversible or irreversible. Reversible processes are isothermal, with the equation w=-nRT(ln(vf/vd)). Work for irreversible processes can be calculated using w=-Pext+Δv.

Alex Nechaev 1I

I would say that potential energy is highest before combustion, as it will all be converted to kinetic energy during the reaction.

Alex Nechaev 1I

If you add energy to a solid, it can become a liquid. Liquids have more energy and entropy than solids. Any chemical reaction that increases the number of gas molecules also increases entropy. A chemical reaction that increases the number of gas molecules would be a reaction that pours energy into a...

Alex Nechaev 1I

Steam causes more severe burns than boiling water does because the temperature required to convert water to steam is much higher than the temperature required to bring water to a boil. Steam quickly transfers the heat of vaporization to whatever it comes in contact with, which will result in severe ...

Alex Nechaev 1I

I don't think it really matters whether you leave your final answer in kJ or Joules, but it's probably more likely that you'll see values in kJ, especially larger values of heat energy.

Alex Nechaev 1I

It's always best to find Q to check the direction of the reaction. This way you just eliminate any chances of making mistakes at the beginning and know for sure that you're calculating correctly.

Alex Nechaev 1I

It wouldn't be wrong to round up, since I'm pretty sure the rounding rules still apply to sig figs. Both would probably be acceptable.

Alex Nechaev 1I

If there's a resonance structure, the pi bonds will be delocalized. Sigma bonds don't become delocalized, so resonance has no effect on them.

Alex Nechaev 1I

At the TA review session on Monday, they mentioned that the best way to determine whether a ligand is a polydentate is to try to draw the Lewis structure. This way, you can actually see the bonding sites in order to know.

Alex Nechaev 1I

^What process do we need to use to determine this though? Is there a way to see without drawing the Lewis structures?

Alex Nechaev 1I

Haocheng Zhang 1C wrote:Also, how to identify a polydentate ligand?

A polydentate ligand can be recognized by having more than 2 lewis base sites, such as multiple lone pair donating sites used to bond to a central atom or ion.

Alex Nechaev 1I

The way that we determine hybridization is generally by drawing the Lewis Structure of the given compound and counting the number of regions of electron density that the compound has. If we have COF2, which has three regions of electron density, the number of orbitals filled when they merge together...

Alex Nechaev 1I

This chart is also really helpful when it comes to VSEPR and hybridization: http://web.gccaz.edu/~chriy68124/VSEPR%20handout.pdf It depicts the lewis structures of many examples, which allows you to see the regions of electron density, and then includes a column with the hybridization of each exampl...

Alex Nechaev 1I

Yes. A single bond such as (C-H) has one sigma bond whereas a double (C=C) and triple (C≡C) bond has one sigma bond with remaining being pi bonds.

Alex Nechaev 1I

Lone pairs and bonding pairs both count as regions of electron density. Lone pairs help determine the shape when you take electron repulsion into account and will also enable you to calculate the bond angles.

Alex Nechaev 1I

This website has a really good description of the differences and includes diagrams: http://www.majordifferences.com/2014/10/difference-between-sigma-and-pi-bond.html#.WgQO1ROPLfY Basically, a sigma bond is formed by end to end overlapping while a pi bond is formed laterally, with the two p orbitals...

Alex Nechaev 1I

Even though the periodic trends indicate that elements closest to the noble gases have higher ionization energy and electron affinity, the noble gases themselves are exceptions to this trend. They have full octets, therefore they are stable, which means they are less reactive and do not need to have...

Alex Nechaev 1I

I read in another discussion topic that the trends don't really apply to the noble gases because they are stable, therefore they are much less reactive than the atoms in other groups and don't need to follow the same trends.

Alex Nechaev 1I

#67 in Chapter 2 asks which element has the higher electron affinity. Part b asks you to compare nitrogen and carbon. I thought the answer would be nitrogen because nitrogen is more towards the right of the periodic table (I believe that electron affinity is higher towards the top right of the peri...

Alex Nechaev 1I

Another notation style includes listing each sub-orbital (i.e. px, py, pz) with the number of electrons in that orbital (i.e. 2px1). That's also just a convention, and it's fine to just write p instead of separating it into 3 sub-orbitals. It's alright if we don't write the notations with the as px...

Alex Nechaev 1I

We went over this question in one of the Peer Learning Sessions, and I just need some clarification. The question reads: Which ion of the following pairs has the larger radius: a) Ca^2+ , Ba^2+ b) As^3- , Se^2- c) Sn^2+ , Sn^4+ ? What effect do the charges of these ions have on their atomic radius? ...

Alex Nechaev 1I

When electrons jump from one energy level to the next (n=1 to n=2, for example), how do we know when the energy exerted or used will be a negative value? If it goes down an energy level, will that finding of the energy from that jump be negative? Or is it the opposite?

Alex Nechaev 1I

To add on to the first response, we are generally unable to observe wave-like behaviors in large objects like baseballs. Because electrons are so small, we can measure wavelength and frequency, but not enough wave-like properties are visible in a larger object to actually make a difference. They are...

Alex Nechaev 1I

The first thing I did was use v=c/λ to find the frequency of the wave. The value you get here can then be used to evaluate for n(final) in Rydberg's formula, since you know that n(initial) is 1 (because hydrogen starts at n=1). After plugging in all of your calculated values, all you have to do is s...

Alex Nechaev 1I

I would definitely use 6.626*10^-34 when doing calculations since it is more precise. If you're concerned about sig figs, just remember that they don't matter so much during the process as they do in your final answer. Use the least number of sig figs in the given values to determine how many you in...

Alex Nechaev 1I

The reason that you can't simply add another O to the reaction is because matter can neither be created nor destroyed. Therefore, you cannot end up with more than what you started with if it was not involved in the reaction in the first place. Adding more atoms to the product than there were in the ...

Alex Nechaev 1I

When working with significant figures in a problem, say, calculating the mass % of an element in a sample, always stick with the LEAST number of sig figs in the problem. For example, if an 8.00 g sample of Vitamin C contains 3.27 g of Carbon, 0.3661 g of Hydrogen, and 4.36 g of Oxygen, the least num...

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