CHEMISTRY COMMUNITY

Q Scarborough 1b

While it will change the total pressure, it does not affect the individual concentrations of the reactants leading to no effect on Q.

Q Scarborough 1b

We will not have to memorize Ka or Kb values. If they are needed to solve a problem we will be given them. Otherwise, we should be able to solve for them with the information given.

Q Scarborough 1b

I believe 10^-3 is the cut off, but during lecture Lavelle did say it was safer to use 10^-4, but I doubt he would give us any questions where there was question about whether or not you could ignore it.

Q Scarborough 1b

Temperature will change the K value, but pressure will not. Pressure can change the concentrations which could shift the reaction to the left or right, but it will not change the K.

Q Scarborough 1b

We would need to know if the reaction is endothermic or exothermic and then we could determine which is favors based on the temperature change.

Q Scarborough 1b

I had the same question. I kept getting -.176 as well and I don't see this one listed as an error on the solution manual error list.

Q Scarborough 1b

Here is the link https://lavelle.chem.ucla.edu/wp-conten ... hedule.pdf
Our first midterm is Jan 29 and our second midterm is Feb 19. Our final is March 14.

Q Scarborough 1b

You are correct. Small equilibrium constants will lie to the left because the reactants will be greater than the products and reactions with large equilibrium constants will lie to the right because the products will be greater than the reactants.

Q Scarborough 1b

Once you are on sapling, go to the home page. Under the resources tab there is a PDF named Atkins 7e SSM. This is the solutions manual that you can use!

Q Scarborough 1b

Hi there! I just realized that my sapling has a different value for the initial values of N2 and O2. So while mine would be k=36, it seems yours is K=4. Sorry for the confusion!

Q Scarborough 1b

Hi there,

For this problem you would set the initial value of NO as .900M. We can determine the value of K from the given concentrations, which turns out to be K=36. From there, you can use an equilibrium table to find the final NO, and use K=36, since that value does not change. Hope this helps.

Q Scarborough 1b

Hi there, You're on the right track but to get started you will want to use the 1st and 3rd equations. First I multiplied H2(g) + I2 (g) ⇌ 2HI by 3 to get 3[H2(g) + I2 (g) ⇌ 2HI]. By multiplying the entire equation by 3, the K value is put to the power of 3, so 160^3. The second equation I used was ...

Q Scarborough 1b

Hello, I used the same equations as you. First I multiplied H2(g) + I2 (g) ⇌ 2HI by 3 to get 3[H2(g) + I2 (g) ⇌ 2HI]. By multiplying the entire equation by 3, the K value is put to the power of 3, so 160^3. The second equation I used was also N2(g) + 3H2(g) ⇌ 2NH3(g). For this equation I just invert...

Q Scarborough 1b

When he says a reaction is at equilibrium, it means that both products and reactants are being formed at the same rate. This does not mean that they have the same concentrations. Instead, both the forward and backward reactions are happening at the same rate causing the reaction to stay at one equil...

Q Scarborough 1b

As stated before, Amphoteric substances can act as either acids or bases. The generally follow the same diagonal pattern as the metalloid band but do not line up exactly.

Q Scarborough 1b

That is correct. The longer the bond, the weaker and more easily the bond is broken. That makes the weaker bond the stronger acid.

Q Scarborough 1b

Hydrogen bonds are the strongest intermolecular force. This means that the molecules are more tightly held together and require more energy to break the bond between them. This results in a higher boiling point.

Q Scarborough 1b

Hello, First, I would classify the pH values, remembering that pH<7 is acidic, pH=7 is neutral, and pH>7 is basic. Next, to classify the [H+] values, remember that acidic solutions have a high [H+], specifically greater than 1*10^-7M, basic solutions have a low [H+], less than 1*10^-7M, and neutral ...

Q Scarborough 1b

Hi,

If we are given the pH of the solution, we can use [H+]=10^-pH. From there, we can also use pOH=14-pH to find the pOH. And then we can use [OH-]=10^-pOH.

Hope this helps

Q Scarborough 1b

I see what you're saying. However, I think it is meant to be Co (cobalt) and it may just be a typo. Otherwise, there would be no metal to bond to.

Q Scarborough 1b

Trigonal bipyramidal is associated with the coordination number 5. It has 5 bonds not 4.

Q Scarborough 1b

The coordination number is the number of bonds. Both square planar and tetrahedral have four bonds. This is why is it is both of these. Hope this helps.

Q Scarborough 1b

Yes you got it. The energy levels go s<p<d<f

Q Scarborough 1b

The coordination number is the number of bonds on the central atom. So the number of ligands attached to the transition metal. Based on the number of ligands attached, we can use our knowledge of molecular geometry to determine the geometry of a complex.

Q Scarborough 1b

Hi, The sum of the charges in the complex must be equal to the overall charge on the compound. The one chloride ion outside the brackets has a charge of -1, so the overall charge of the coordination sphere must be +1. The ammine ligands are neutral, and there is one negative charge on the Cl. So the...

Q Scarborough 1b

For a molecule with 5 regions of electron density, with 3 lone pairs and 2 bond pairs, the structure would be linear. This makes the smallest bond angle 180.

Q Scarborough 1b

With an overall charge of -1, the central I atom has 3 lone pairs and two bond pairs. This makes the geometry trigonal bipyramidal, but the shape would be linear.

Q Scarborough 1b

The units for molarity are mols/liters. This is only ever used to determine the concentration of a solution and is not used for solids or gases.

Q Scarborough 1b

Also, how can you calculate what grade you need on the final to get an A? You can figure out how many points you have already lost by looking on My UCLA. There are a total of 500 points so you can miss 35 points for an A, and 50 points for an A-. So you can count up your total and see the maximum n...

Q Scarborough 1b

As stated above, Ozone's central oxygen has three regions of electron density, both the other oxygens as well as a long pair. This makes the molecular geometry trigonal planar, but because there is one lone pair the shape is bent. This makes the angle <120 degrees.

Q Scarborough 1b

I feel the same way. Personally, I like going back through the old sapling homework because they give detailed solutions and how to solve them. I also think the textbook problems are a great way to practice and make sure you understand the concepts.

Q Scarborough 1b

Hi,
That is correct. X is for bonded atom and E is for lone pairs around the central atom.

Q Scarborough 1b

Hi,
You are correct in that you are supposed to find the largest difference in electronegativity. For this question, that largest electronegativity difference would be BaBr2, making it the most ionic.

Q Scarborough 1b

Hi,

Lewis acids accept electrons whereas Lewis bases donate electrons. Metal cations accept electrons in acid-base reactions. Molecules with an excess of electrons, such as anions or molecules with a lone pair electrons, can donate electrons. Such as H2O

Q Scarborough 1b

Hello,
While both butanol and diethyl ether have the correct elements for hydrogen bonding, you have to look at the structure. Butanol has a H bonded with the O while diethyl ether has the H bonded to the C, not the O. It definitely helps to draw out the structures and visualize them. Hope this helps!

Q Scarborough 1b

You are correct that the 4d state has 5 orbitals, but this problem specifies the ml number as -2, which corresponds to only one orbital.

Q Scarborough 1b

Hello,
We may not have to draw them because the test will most likely be multiple choice. However, you should be able to identify them and which ones contribute to the overall structure. This includes finding the formal charge to determine the resonance structure.

Q Scarborough 1b

Gallium is in group 13 which has 3 valence electrons. The ion most likely formed would want to give away 3 electrons to make a full valence shell. So the charge of Ga will be +3.

Q Scarborough 1b

The formal charges for each molecule are given. To find the most plausible resonance structures you want to find the ones with the least formal charges. In this case it would be C and D, both of which only have one negative charge whereas others have multiple.

Q Scarborough 1b

The answer is C because it has two negative charges and one positive charge. Both A and B only have one negative charge which lowers the total energy.

Q Scarborough 1b

Hello, for one resonance structure you will simply switch the double bond from the one Oxygen to the other, making sure you also include the formal charge. For the other there is a double bond between the Nitrogen and Carbon, and single bonds between the Carbon and Oxygen, also make sure to include ...

Q Scarborough 1b

Each atom has its own formal charge, and the summation of them give you to total formal charge

Q Scarborough 1b

Yes you need to calculate the formal charge for each atom. The overall charge of the molecule is made up of the individual charges of the atoms so you need to find those.

Q Scarborough 1b

That would be super helpful. I would love to be a part.

Q Scarborough 1b

Yes we will have to memorize it. Anything not on the formula sheet needs to be memorized.

Q Scarborough 1b

Discussions are not mandatory so it will not affect your grade. You may want to contact your TA or other classmates to ensure you didn't miss anything important.

Q Scarborough 1b

You can also store constants in your calculator which I find very helpful. That way you do not have to type out long numbers like planks constant or avogadro's number. Just type the number, push sto-> and any letter. Then use that letter the same way you would the number going forward.

Q Scarborough 1b

The value of l can be 0,1,2,3,...,(n-1). With 0 corresponding to the s-orbital, 1 with the p-orbital, 2 with the d-orbital, and so on.

Q Scarborough 1b

The first step here is recognizing that 102.557nm is part of the Lyman series, meaning that the first energy level will be n=1. I use the following equation: v=-R((\frac{1}{n^2})-(\frac{1}{n^2})) You then need to convert the wavelength 102.557nm to frequency, giving you 2.93x...

Q Scarborough 1b

Im pretty sure Prof Lavelle said that if the energy of the photon is less than the work function that it just passes through and is not absorbed.

Q Scarborough 1b

While Professor Lavelle prefers the En one, I think we can use either. The test is also multiple choice so he wouldn't be able to tell which one you used.

Q Scarborough 1b

If you know the radius of a Hydrogen atom, you can just find 1% of that to find your uncertainty. In this case it would be .05nm * .01= 5*10^-13, which would be your uncertainty position.

Q Scarborough 1b

I am confused about this as well. I hope it is not on the midterm because he hasn't gone over anything other than the very basic descriptions of the levels, but I am not sure.

Q Scarborough 1b

If the energy of the photon is less than the threshold energy, the photons will just pass through the metal.

Q Scarborough 1b

The correct answer is D. The uncertainties in position, speed, and momentum of a moving object are not noticeable or measurable. Heisenberg’s uncertainty principle deals with moving objects. The formula is λ = h/p where p is mv. There has to be some velocity or else the denominator would be zero, so...

Q Scarborough 1b

We will have a constants/equation sheet that we will have to print out before hand. You can find it on his website. https://lavelle.chem.ucla.edu/wp-conten ... ations.pdf

Q Scarborough 1b

There is a solutions manual on sapling under resources.

Q Scarborough 1b

Yes, I believe you are correct. The midterm will be multiple choice.

Q Scarborough 1b

Check to make sure you have all your units correct. 1 MHz is equal to 10^6 Hz, and the answer for this questions is supposed to be 1m.

Q Scarborough 1b

The second half of question 4 is asking for the maximum number of electrons that could be ejected from the metal by a burst of photons with a total energy of 3.84*10^-7. We know that the work function of this metal is 5.28*10^-19 from part one of this question. To find the maximum number of electron...

Q Scarborough 1b

I have the same problem as well. Sometimes im not sure if I'm meant to round up or find another value to multiply it by. In general, if the values are .95 or above I will round up. Otherwise I will try to find another number to multiply by. Most of our textbook/homework questions will have answers t...

Q Scarborough 1b

You can either click on your name or you can click on quick links in the top left and then your posts.

Q Scarborough 1b

MHz stands for Megahertz, which is 10^6 hertz.

Q Scarborough 1b

We use 100g simply because it is easier to visualize the mass percentage composition. You could use other values but it would make it more difficult. It just depends on the method that works best for you.

Q Scarborough 1b

Hello,
The post-assessments are just for your benefit. They are not graded and do not count as homework, but it is good practice and Dr Lavelle recommends everyone completes them. Sapling is the only homework that we have.

Q Scarborough 1b

Like what was said above, you should wait to round until the final answer. If you round before then, especially if the question has many steps, you can cause your answer to be skewed.

Q Scarborough 1b

950000 only has 2 sig figs because the last four zeros are trailing zeros. The only sig figs here are the 9 and 5. 80060 has 4 sig figs because the two zeros between the 8 and the 6 are significant. The last zero is not because it is a trailing zero.

Q Scarborough 1b

In this question the M from M(OH)2 is a placeholder for a metal, it is not an actual element. The question asks to find the molar mass of the metal, which would just be 74.1g/mol-34.02g/mol = 40.08g/mol. This is the molar mass of Calcium.

Q Scarborough 1b

Hello,

He uses 100g to better represent the mass percentage composition. He used the values of each element of Vitamin C and expressed it as a percentage of the whole compound.

Q Scarborough 1b

Hello,

Yes Mole and mol are the same thing. Mol can be shorthand for Mole.

Hope this helped!

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