CHEMISTRY COMMUNITY

Chem_Mod

Delta U represents the change in internal energy, which consists of work W and heat Q. The units will be in joules.

Chem_Mod

U(you) are solving for U. If U=q+w, then that means we need to solve for q and w to get U. By reading the question carefully, you will find the calorimeter is constant volume. If w=P*deltaV, then deltaV is 0 (constant volume) and then w=0. Thus U=q+0, which is U=q.

Chem_Mod

Not all calorimeters are completely sealed. For instance, one of the best known examples of a constant-pressure calorimeter is a coffee cup calorimeter, which is simply two coffee cups stacked on top of each other and a lid on top with two holes for a stirrer and a thermometer. However, bomb calorim...

Chem_Mod

This is a more mathematical way of saying any change at all in the system can reverse the direction of the process/reaction

Chem_Mod

Yes, the bond enthalpy for C=C is the energy it takes to completely break the double bond. In many reactions that you will study in organic chemistry, it will actually look like just the pi bond that breaks and reforms two new bonds. However, it is difficult to partially break just the pi bond of a ...

Chem_Mod

Simply restating the question is not a valid Chemistry Community post; please state more specifically what you are having trouble with concerning this problem, or let us know what you have tried so far or what step is confusing you

Chem_Mod

Chem_Mod

Come see me in office hours.

Chem_Mod

Most likely no, just because that would be very difficult and time consuming to do on a scientific calculator.

Chem_Mod

Significant figure rules apply to everything, including pH and pOH. The number of significant figures in your answer depends on the numbers given in the original problem. Remember that pH significant figures are slightly different from what you are normally used to. For example, pH 9.74 has two sign...

Chem_Mod

See https://lavelle.chem.ucla.edu/wp-conten ... _Bases.pdf

In class I discussed how to make a buffer and calculate the pH of a weak acid and its salt, or a weak base and its salt.
This is level at which we covered this topic and expected to know.

Chem_Mod

See my answer here:
viewtopic.php?f=49&t=56786&p=212111

Chem_Mod

In Friday’s class I did equilibrium examples applying them to titrations (salt/pH at the stoichiometric point) and buffers (I did weak acid and its salt example). Sometimes I discuss osmotic pressure and ATP hydrolysis. This quarter I will cover these in a review session. The osmotic pressure of a d...

Chem_Mod

So the I in ICE table stands for initial, so in this row you want to put in the initial concentrations of each species that will be involved in K. The C stands for change, so the amount of concentration that each reactant will lose and each product will gain will depend on the different stoichiometr...

Chem_Mod

Do acid and base homework problems Week 3.
Great preparation for Test 1.

Chem_Mod

rgk2300 wrote:Thank you for sharing! As someone who is on the fence about medicine, this was very inspiring!

I'm happy this helped. If you have any other specific questions or concerns/worries regarding the medical school process or Chem 14A and 14B, I'm happy to answer them.

Chem_Mod

Adding a catalyst lowers the activation energy of the forward and reverse reaction; however, this is a kinetic property. Equilibrium is a thermodynamic property so it is not affected by the addition of a catalyst. The system may reach equilibrium faster, but the equilibrium composition will be the s...

Chem_Mod

An ICE table is useful in solving this problem; the difference compared to most other problems is that you're given an equilibrium concentration, so you kind of have to work backwards. If there is no mention of a species being present initially, we assume that its concentration is equal to zero.

Chem_Mod

Most of the time you can disregard the deprotonations other than the first for polyprotic acids because the Kas for subsequent deprotonations are extremely small compared to the first deprotonation and can therefore be ignored. However, in the case of H2SO4, the Ka for the second deprotonation is re...

Chem_Mod

Here is the worksheet for Monday. We will go through it at Covel.

Chem_Mod

Hello Community!

This is UA Jonathan Pai and I will be filling in for Lyndon on his Monday workshop. Once the worksheet for Test 1 is finalized, I will post it (hopefully tonight). Please print it out and bring it with you to the workshop. There will not be an answer key. Study well.

Chem_Mod

There is a Ka/pKa list in Table 6D.1 located on page 478 of the 7th edition textbook.

Chem_Mod

The Sapling Learning you have from last quarter is valid for 24 months (2 years), and no further purchases are needed.
I will contact Sapling Learning.

Chem_Mod

See my previous post/answer: Monday campus is closed which means all the on campus peer learning sessions are cancelled Monday. However, I have organized the following 5 hours of Covel Peer Learning Sessions: Covel Commons (ALL Room 210) ALL STUDENTS WELCOME! Sunday: • 7-9pm (Step-Up) Monday: • 6-8p...

Chem_Mod

Monday campus is closed which means all the on campus peer learning sessions are cancelled Monday. However, I have organized the following 5 hours of Covel Peer Learning Sessions: Covel Commons (ALL Room 210) ALL STUDENTS WELCOME! Sunday: • 7-9pm (Step-Up) Monday: • 6-8pm (Workshop) • 8-9pm (Drop-In...

Chem_Mod

This means Cl- is inert in aqueous solution. Think back to Chem 14A reactions with salts. Group I and II metal cations are unreactive in water, as are conjugate bases of strong acids, like Cl-, which is the conjugate base of the strong acid HCl. It is important to be able to recognize which species ...

Chem_Mod

If the species gains a proton, then it is the base. If the species loses a proton, then it is the acid. Salts will dissolve in water; identify what ions will result and which are reactive/non-reactive (ie Group 1 metal ions are nonreactive in water).

Chem_Mod

A buffer is a solution containing both acid and conjugate base that can thus resist changes in pH upon the addition of another acid or base. This is because the acidic species in the buffer neutralizes added base and the conjugate base neutralizes any added acid. You may be asked to find the pH of s...

Chem_Mod

Please include the entire question with your post; this will better assist us in helping you

Chem_Mod

You need to take into account autoprotolysis of water when you either have an extremely weak acid/base or a very dilute acid/base added (or both). Generally, if the concentration of H+ that you get from doing ICE table with the acid is 10^-6 or less, then you should consider the autoprotolysis of wa...

Chem_Mod

Please post the entire textbook question with your post. This will aid us in assisting you

Chem_Mod

Strong base will neutralize the weak acid. Depending on the concentration of the weak acid and the base, you will end up with only one of them left Then calculate pH as you normally would

Chem_Mod

Interested to know how was this Friday's class/lecture?

Chem_Mod

You first need to multiple .025M by the volume 200.0mL (or .200L), not divide. You want to multiply so you get the number of moles, which you can then divide by the new volume to find mol/L.

Chem_Mod

For part a, you would have 200.0mL of a .025M solution, so you don't need to calculate a new molarity since the solution is at the desired concentration.

Chem_Mod

Are you given any K values?

Chem_Mod

Strong bases and acids will completely dissociate. For example, .0146M HNO3 will dissociate to produce a solution of .0146M H+ and .0146M NO3-. You can solve for pH by substituting [H+] into -log [H+].

Chem_Mod

If you can approximate using the 5% rule, then that would be your simplest route, but if K is large enough such that initial concentrations significantly differ from equilibrium concentrations, you would have to use quadratic.

Chem_Mod

Heat can act as a "reactant" in endothermic reactions or "product" in exothermic reactions. For example, if you add heat to an exothermic reaction, the reverse reaction will be favored.

Chem_Mod

You can compare the ratio of [P] and [R] to 1 to approximate if products or reactants are more favored, but you can use 10^-3 to determine whether there is a negligible or significant change between initial concentrations and concentrations at equilibrium.

Chem_Mod

You would need some sort of volume to determine the number of moles in your solution, and then convert that to mass using molar mass.

Chem_Mod

In the book problems, they explicitly say to ignore the second deprotonations bc the Ka values are extremely small. Likewise, if it is not explicitly said on a problem, look at the Ka value and determine if there is any significant protonation occurring.

Chem_Mod

pH is a measure of acidity/basicity. Although you substitute molarity into the equation, it is unitless because it does not refer to a quantity, ie 1 liter of a solution with pH 8 is not more basic than 5 liters of a solution with pH 8. Moreover, it involves logarithms, which do not have units.

Chem_Mod

They just multiplied both sides of the Kc expression by (.020-2x)^2, resulting in x = 1.6 x 10^2 (.020-2x)^2. When you multiply those two together, you get x = .064 - 13x + 6.4 x 10^2 x^2.

Chem_Mod

Use these Learning Outcomes to prepare for Test 1: Outline 1: Chemical Equilibrium https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Chem14B_Outline1_Chemical_Equilibrium.pdf Outline 2: Acids and Bases https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Chem14B_Outline2...

Chem_Mod

See email from your TA.

Your TA will announce a make up session.

Chem_Mod

Hi All,

See Announcements on my class website.

Correct I always announce this in class. :-)

Test 1 covers Equilibrium and Acid & Base Equilibria.

Chem_Mod

I sent class email Monday stating Peer Learning Sessions in Young Hall started today (Tuesday).
Are you not getting class email?

Make sure to check your spam/junk folders. Maybe your email is blocking me … I hope not. :-)

Chem_Mod

PV= nRT
the moles (n) are multiplied by the R and T. This means if you divide volume from both sides, you would be dividing the entire product of n, R and T by the volume V: P= (nRT)/V

Chem_Mod

We are technically using chemical activities to calculate K. Activities are unitless, so that makes K unitless. To be be more exact, you should be writing the equilibrium constant by using activities. K equals the activities of products to respective stoichiometric coefficients over activities of re...

Chem_Mod

Solids and liquids are pure substances, and their concentrations do not change over the course of the reaction (i.e. do not affect amount of reactant present). Yes, you are correct that solids will not have a concentration because concentration is mol/L, which doesn't make sense for an undissolved s...

Chem_Mod

According to Le Chatelier's principle, an increase in pressure will lead to the reaction shifting to the side with less moles of reactants or products. In this case, there is one mole of a reactant and two moles of product, so the reaction will shift to the left and produce more reactants.

Chem_Mod

You need to talk to Dr. Lavelle in person in his office hours to resolve this.

Chem_Mod

When you do your calculations with the ICE table it is OK to use moles or molarity since the proportions are the same (molarity is just moles divided by volume). However, when you calculate the value of Kc it is necessary to use the concentrations in molarity. So if you did your ICE table calculatio...

Chem_Mod

The Test/Exam schedule can be found on the class website: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Chem14BTestSchedule.pdf
Test 1 starts in Week 3 Discussion Section.

Chem_Mod

I remember in high school we learned that when doing ICE tables and having a the final concentration (in the ice table) being something like 3.00 - x, we would just approximate it to 3.00 or something like that. Is there any situation we would apply that approximation to in class or are we only usi...

Chem_Mod

Althea Zhao 1B wrote:I saw dates for the midterm and final on CCLE, but does anyone know the dates for Test 1 and Test 2?

The complete Test/Exam schedule can be found on the class website: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Chem14BTestSchedule.pdf

Chem_Mod

Kassidy Ford 1I wrote:How do you tell which gases are inert gases? is it any gas that doesnt react in the chemical equation?

Inert gases are also known as the noble gases, ie Ne, He, Ar, Kr. Thinking back to Chem 14A, these gases are not very reactive because they have a complete octet of valence electrons.

Chem_Mod

Helium is an inert gas, but we (as a global society) are actually running out of it, which is a problem. Liquid helium is actually vital for superconducting magnets which are very important in synthetic organic chemistry, and the medical field (MRI scanners). For the purpose of increasing the pressu...

Chem_Mod

In lecture Dr. Lavelle also talked about how you can have pressure that does affect whether the reactants or products are favored in the reaction and pressure that does. You can increase pressure by changing the volume, which will shift whether or not the reactants or products are favored. If you in...

Chem_Mod

The equilibrium constant is what remains the same in both reactions. It is a number describes the point at which the forward and reverse reactions are equal, using a different ratio that is not the equilibrium constant or a manipulation of the equilibrium constant like (K^-1), would not result in a ...

Chem_Mod

K is an equilibrium constant which is constant for a given temperature. The above answer is mostly correct, just one technicality: it should be noted Q can be measured at any point in the reaction, including at equilibrium, which is when Q=K.

Chem_Mod

When dealing with a homogeneous gas equilibrium solving for K means solving in terms of partial pressure, if you were dealing with a homogeneous gas equilibrium and you used concentrations to find the equilibrium constant you would be obtaining Kc. Kc and Kp are not always equal, you can read more a...

Chem_Mod

The above reply is true.
Solving for the equilibrium constant K should tell you what the ratio of partial pressures should be for both experiments, this will also help you in solving part d and e.

Chem_Mod

Sat Jan 11, 2020 1:28 pm Your equation for Kc should be [H2][CO]/[H2O]. Find the starting concentration of H2O using given moles and volume. Use final concentration of H2 and stoichiometric coefficients of the reaction to determine how much H2 was made and how much H2O was used up. Then use this inf...

Chem_Mod

So K<10^-5 tells you it may be worth your time to use the approximation, but after using the approximation to solve for x, you will have to check for the 5% rule itself, which states that x may be approximated as 0 (when added or subtracted) only if x < 0.05 of the initial concentration. So use the ...

Chem_Mod

Everything you need to know will be covered in class. That being said, I think knowing the Ideal Gas Law is a good way to remember all the other laws, but may get confusing to use. The key is to remember to keep all other variables constant while you’re focused on the relationship between 2 particul...

Chem_Mod

Here is Week 1 lecture note.

Chem_Mod

Yes, the delta H indicates whether heat is released or absorbed. Since it's negative, heat is released and can be seen as a "product." Therefore, increasing heat would push the reaction in the reverse direction.

Chem_Mod

It may be helpful to post your steps to see where you might have missed something! But first, write out the balanced chemical equation and set up your equilibrium expression. Make sure the exponents match the stoichiometric coefficients. Keep in mind that you should be using concentrations, so you w...

Chem_Mod

Thanks so much for the advice! So nice to hear that these courses helped you with your career and path in medicine - and that there are so many helpful resources! Take advantage of all the resources Dr. Lavelle offers. You will notice in future courses that there are limited resources compared to t...

Chem_Mod

Coming into the chemistry 14 series was kinda scary for me, since I’ve never taken chemistry before. But after discovering chemistry community and posts like this, I’m very reassured that even if I struggle with the course it will be worth it in the long run(like when I eventually take the mcat). j...

Chem_Mod

I will discuss this in class in both the equilibrium and kinetics sections. Changing the volume in a gas phase reaction changes concentrations, and changing concentrations does not change K. I will discuss this in detail this Friday and Monday in class. The activation barrier is different in the for...

Chem_Mod

It does not make a difference where it was bought. Enter the Sapling Plus code that came with your textbook package.

Most students are already setup as Sapling Plus was used in Chem 14A and is valid for 2 years and includes the E-text.

Chem_Mod

Great to read the positive feedback. Much appreciated.

Enjoy your break and I'll see you in Chem 14B.

Chem_Mod

Several lectures on acids and bases. An entire lecture on salts and how they affect pH. Then a 2nd lecture on using salts to determine pH at the stoichiometric point. There were no other types of titration questions, no calculations. Only salts at the stoichiometric point. That is, [acid] = [base] =...

Chem_Mod

Never be numb, always reach out and share, even if its something small, its a start.

Enjoy your break.

Chem_Mod

See lecture notes. Same example.

Chem_Mod

An entire lecture on salts and how they affect pH.
Then a 2nd lecture on using salts to determine pH at the stoichiometric point.

There were no other types of titration questions, no calculations. Only salts at the stoichiometric point.

Chem_Mod

Final grades will be posted on MyUCLA in about 10 days.

Chem_Mod

Great post and thank you!
See you in class next quarter. :-)

Chem_Mod

Thank you for the positive feedback which encourages me to keep assisting students.
It was a great quarter!

Chem_Mod

Grading on final exam is done very carefully and takes a lot of time.
In about 10 days after the final exam.

Chem_Mod

A chelating ligand is a ligand that forms more than one bond to the same transition metal.
Results in a ring of atoms that includes the transition metal atom.

See lecture notes and textbook for multiple examples.

Chem_Mod

Yes. I covered them in class.

Chem_Mod

This question has been omitted as it is an organic addition reaction.

Chem_Mod

Usually we ask for the most stable structures which are the optimal (low) formal charge.

Chem_Mod

See pages 490 and 491 for acid rain.

See pages 506 and 507 for blood.

Chem_Mod

Because there is no electron-electron shielding/interaction.

Chem_Mod

Not part of the syllabus.

Chem_Mod

Great!

Chem_Mod

Use the measured wavelengths to identify the element(s) in a sample.
Maybe arsenic was used to poison someone …

Should I use this example in class?

:-)

Chem_Mod

Don’t worry about this equation. Just know the atomic radius trends as you move from left to right in a row and down a column. When moving left to right in a row it’s the Z or number of protons that we care about. As you increase number of protons moving left to right in a row then atomic radius dec...

Chem_Mod

Yes, make sure to keep the equations for light separate from the equations for particles. Light has the following two equations: c= wavelength* frequency and Energy=Hf. These equations can only be used for light. Similarly, for a particle you have the De Broglie wave equation wavelength=h/mv. This m...

Chem_Mod

Correct.

Chem_Mod

The Heme complex is an example of a biological coordination compound, which makes up hemoglobin and myoglobin. Both of these molecules use their heme complexes to bind and release O2 depending on environmental conditions, which is what oxygenates your organs and muscles.

Chem_Mod

No, only one of them have to have a H bound to it. However, the other one must be one of these electronegative atoms.

Chem_Mod

Write as you would any other proton donating or accepting reaction. Write one reaction for it acting as an acid (donate proton to water) and one for it acting as a base (accept proton from water).

Chem_Mod

Amphiprotic means molecule can both accept and donate a proton. This makes it necessarily amphoteric as well, which is a molecule that can act as both an acid and a base. Acids and bases by Lewis definition do not have to donate or accept protons

Chem_Mod

You will not be asked to draw the entire structure of EDTA. However it may be helpful to know which atoms/lone pairs are involved in forming coordinate covalent bonds

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