CHEMISTRY COMMUNITY

aashmi_agrawal_3d

Adding an inert gas won't shift the reaction because an inert gas does not affect the concentrations of reactant and products.

aashmi_agrawal_3d

It is favorable because endothermic reactions require heat to take place and when you add heat they are able to reach the activation energy to start the reaction more easily.

aashmi_agrawal_3d

I think we usually refer to the long term change because the question is already stating the short term change.

aashmi_agrawal_3d

It changes depending on the temperature. However, the standard temperature is 25 degrees Celsius and at that temperature it is 10^-14.

aashmi_agrawal_3d

I think the question would specify the units. Lavelle also mentioned that we can use them interchangeably in this class because they are approximately equal.

aashmi_agrawal_3d

Aqueous solutions are included, but solvents indicated by (l) aren't.

aashmi_agrawal_3d

It is rare for the concentrations of reactants and products to be equal.

aashmi_agrawal_3d

K does not change because it is a constant. If Q is greater than K, then you know more reactants will form to reach equilibrium and if Q is less than K, then you know that more products must be formed.

aashmi_agrawal_3d

I think they would usually be given but if not they will probably give you the moles and the volume and then you would just have to divide in order to get the concentration.

aashmi_agrawal_3d

Homogeneous reactions just have all of the reactants and products in the same phase whereas heterogeneous reactions do not. When calculating the equilibrium constant, you will just have to make sure that you do not include liquids and solids in the calculation.

aashmi_agrawal_3d

Either way they would cancel out but it would be the simplest in terms of calculations to use net ionic equations.

aashmi_agrawal_3d

Yes, the lower the pH, the stronger the acid.

aashmi_agrawal_3d

I think they are both basically the same thing.

aashmi_agrawal_3d

I think it's a strong acid.

aashmi_agrawal_3d

I don't think so. I think that's what we start with in 14B.

aashmi_agrawal_3d

Water can act as either an acid or a base since it is amphoteric.

aashmi_agrawal_3d

It has a carboxyl group which makes it an acid.

aashmi_agrawal_3d

I think we should have a general idea of the oxidation numbers of common ligands but I think that will just come automatically from doing homework/practice problems.

aashmi_agrawal_3d

There are four Br atoms bound to the Fe atom and the coordination number is the number of bonds to the central transition metal so it is 4.

aashmi_agrawal_3d

We can figure that out by looking at the periodic table and seeing which column it is in. For example, Na is in the first column so it has a +1 charge.

aashmi_agrawal_3d

I think it only refers to the central atom.

aashmi_agrawal_3d

The coordination number is basically the number of bonds attached to the central metal.

aashmi_agrawal_3d

I think that since there are pi bonds and those bonds can't rotate, it has to stay on the same plane.

aashmi_agrawal_3d

I don't think it really matters since both are correct. I think sp3d is more of the convention though.

aashmi_agrawal_3d

Yes. Every bond and lone pair is a region of electron density. However, it doesn't matter if the bond is a double or triple bond. It is still one region of electron density either way.

aashmi_agrawal_3d

I'm pretty sure there are but I don't think we learn about them in this class.

aashmi_agrawal_3d

There are 3 bonds and one lone pair which basically means that there are 4 groups. This means the answer would be sp3 because there is one s orbital and 3 p orbitals which makes up 4 groups.

aashmi_agrawal_3d

Hybridization is basically mixing atomic orbitals to form new hybrid orbitals.

aashmi_agrawal_3d

I think the order you listed is correct.

aashmi_agrawal_3d

Yes, all molecules have LDFs as temporary dipoles are often formed.

aashmi_agrawal_3d

I think it is because iodine is much larger so there are stronger london dispersion forces since neither of them have dipole dipole or hydrogen bonding.

aashmi_agrawal_3d

Ionic bonds are stronger than hydrogen bonds. This means that it would take more energy to break them and therefore they would have a higher boiling point.

aashmi_agrawal_3d

A dipole is basically where there is a partial positive or negative because electrons are shared unequally.

aashmi_agrawal_3d

It technically can form two hydrogen bonds but I don't think it is very likely.

aashmi_agrawal_3d

Yes, but I think all you really need to know are the hydrogen bonds are much stronger than dispersion forces.

aashmi_agrawal_3d

The hydrogen just needs to be bonded to N O or F. The lone pairs on N O or F are not relevant.

aashmi_agrawal_3d

I think you have to examine it more carefully and look at other characteristics if the bond is in between these two values but I don't think it is likely that he will give us questions like that at this point.

aashmi_agrawal_3d

It increases to the right and decreases going down. You can also remember that fluorine is the most electronegative element to help you remember this trend.

aashmi_agrawal_3d

Expanded octets can occur from the period 3 or after because the d orbital starts.

aashmi_agrawal_3d

You would place the double bond where it would result in the lowest formal charges.

aashmi_agrawal_3d

No, we don't need to worry about shape for now.

aashmi_agrawal_3d

Radicals just have one unpaired electron so if there are an odd number of electrons, you can see that it is a radical.

aashmi_agrawal_3d

I don't think the side matters. You just need the correct number of electrons and bonds.

aashmi_agrawal_3d

This is because single bonds have less electrons so they are not as strong and cannot pull the atoms as close to one another. However, a double bond is stronger and can pull the atoms closer together, making the bond shorter.

aashmi_agrawal_3d

I think they are different because the electrons are still technically in different positions.

aashmi_agrawal_3d

Brackets are used so that the charge of the entire molecule can be displayed right outside of the bracket. Since they are used to show the charge, they are used for both anions and cations.

aashmi_agrawal_3d

I think we should use lines to show bonds and dots for lone pairs as that is the proper notation.

aashmi_agrawal_3d

The charge represents the charge of the entire molecule. Since the charge of the molecule was -1, he put a minus. This allows for one to understand the number of electrons being represented in the lewis structure.

aashmi_agrawal_3d

You should covert to meters as that is the SI unit and will be consistent with the other measurements.

aashmi_agrawal_3d

You multiply by 2 when there is a +/- because the uncertainty can be double the amount due to this. You use the given uncertainty if it gives you a percentage without the +/- or doesn’t explicitly tell you to double the value.

aashmi_agrawal_3d

I don't think you change anything when plugging numbers into problems for maximum and minimum uncertainty. The solution to these problems is always >= when plugging numbers into the equation so if it asks for the minimum, you simply put in the answer you get when you solve the equation.

aashmi_agrawal_3d

You do multiply by 2 because the uncertainty is double the amount when it could either be plus or minus. There was actually a problem in the solution in the textbook for that problem.

aashmi_agrawal_3d

There is an inverse relationship between the certainty of position for electrons and momentum for electrons because knowing the speed makes it harder to know the position. In order to know the speed, photons must hit the particle which moves the particle’s position, making it harder to know the posi...

aashmi_agrawal_3d

For determining the number of spectral lines that could appear, you just have to think about which energy levels it could drop down to. There are only 4 energy levels below it so only 4 spectral lines could appear. The wavelength range would also be the difference in the wavelength when it is a drop...

aashmi_agrawal_3d

This is because the wavelengths for larger objects are so small that they are basically undetectable.

aashmi_agrawal_3d

I think it’s multiple choice, closed note and we have around 50 minutes to take the test. I don’t think there is any partial credit and I am pretty sure that some questions are weighted differently than others. We will also be using Respondus on our main device with the test on it and Zoom on anothe...

aashmi_agrawal_3d

I think you mean E=hc/λ and that is basically a combination of E=hv and C=λv. This is because v can also be represented as c/λ.

aashmi_agrawal_3d

I think that we just have to know that the visible region is between 400 and 700 nm and red is 700 while violet is 400. I think we should also know that smaller wavelengths usually fall under ultraviolet and longer wavelengths usually fall under infrared.

aashmi_agrawal_3d

This is true. I think this was mentioned in a module but I think you mainly should just know that as the intensity increases, the amplitude increases as well for this class.

aashmi_agrawal_3d

Yes, you should be able to solve for any of the variables in the equation which can be done by plugging in the other values that you know.

aashmi_agrawal_3d

It is 10^-3 of a mole.

aashmi_agrawal_3d

They are basically the same but molar mass is measured in grams while molecular mass is measured in amu.

aashmi_agrawal_3d

The ratio is actually 1:1 between moles of Cu ions and CuBR2 so you don’t have to divide the value by 2.

aashmi_agrawal_3d

Quantized is basically where there are discrete values so it is not continuous.

aashmi_agrawal_3d

A transferable unit is basically the smallest unit of water that can be transferred to another object like the bucket in this example. That would be one water molecule.

aashmi_agrawal_3d

It is used when you are asked to convert to the number of molecules, atoms or particles.

aashmi_agrawal_3d

There are 2 moles of H20 but 4 moles of H.

aashmi_agrawal_3d

I don't think heat would be factored in because it does not fall under any of the four states of matter (solid, liquid, gas, aqueous).

aashmi_agrawal_3d

I think you would always balance the equations so that the coefficients are in the lowest whole number ratio possible. If there was a coefficient of 8.5, I think you would just multiply every coefficient in the reaction by 2 so that this coefficient becomes the whole number 17.

aashmi_agrawal_3d

In order to figure out with reactant is an excess, you will have to use the coefficients next to the reactants to see what the ratio between the reactants should be. In this reaction the ratio is 6 moles of CIO2 to 2 moles of BrF3. If you have 12 moles of Cl02, you should have 4 moles of BrF3. Since...

aashmi_agrawal_3d

Yes because O2 is always a reactant in a combustion reaction.

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