CHEMISTRY COMMUNITY

Gwen Casillan 3E

No, it's not on the equation sheet, but we do have to know it so either know how to derive it or just memorize it.

Gwen Casillan 3E

I'm stressed and struggling too, so you're definitely not alone. I recommend attending UA sessions as they're super helpful, as well as checking your confidence/understanding of concepts. Remember, grades don't define you, so don't stress too much and remember to pace yourself and not be too hard on...

Gwen Casillan 3E

They'll probably be updated soon, so don't stress too much about them not being accurate at the moment.

Gwen Casillan 3E

Any good advice or tips for studying this week? I'm trying not to stress too much.

Gwen Casillan 3E

When delta G is negative, the forward reaction and products are favored, and when delta G is positive, the reactants and reverse reaction is favored.

Gwen Casillan 3E

I'm hoping to take it either next quarter or over the summer.

Gwen Casillan 3E

We mostly never ignore n, as it stands for moles. Also, pay attention to the units the question is asking for.

Gwen Casillan 3E

The values are given to us on exams, so most likely you don't need to memorize them.

Gwen Casillan 3E

Hospital Playlist is great! I am also hoping to rewatch Reply 1988 one of these days.

Gwen Casillan 3E

Gº is under standard conditions, whereas G isn't.

Gwen Casillan 3E

Meet with your TA during their office hours!

Gwen Casillan 3E

Thank you so much for posting this. It really helped me examine my state of mind and how hard on myself I sometimes get. Exams are always stressful for me, so this provided some insight on my worries.

Gwen Casillan 3E

This mostly depends on what units the question is asking for, but most of the time, convert to Kelvins.

Gwen Casillan 3E

A solution is neutral when the concentration of H3O+ is 1.0x10^-7, which is important because when [H3O+] = 1.0x10^-7, the pH is 7, and [OH-] = 1.0x10^-7, and pOH = 7. Kw is the concentration constant of neutral water at 25º C, which is 1.0x10^-14, which can only be true when concentrations of [H3O+...

Gwen Casillan 3E

Percent ionization is found by dividing the concentration of [H+] at equilibrium by the initial concentration of the acid and multiplying by 100 (to get it in %).

Gwen Casillan 3E

I agree with what John said. I believe Dr. Lavelle referred to standard states to explain how when something is not in its standard state, one needs to take into account the phase change as well when calculating change in enthalpy.

Gwen Casillan 3E

Definitely attending UA sessions and Step-up sessions, as well as practicing textbook problems to just solidify your understanding of the topics covered. I personally go over problems that I have difficulty with and practicing with similar problems in the textbook until I feel comfortable with the c...

Gwen Casillan 3E

You can generally assume x is negligible if the K value is 1000x (10^3) smaller than the initial concentration given, or if the K value is extremely small, like the example Lavelle showed in class where the K value was 10^-30-something. However, it's always a good idea to check if x is really neglig...

Gwen Casillan 3E

I think Ka and Kb values would be given in the problems. I don't believe we need to memorize any specific values. However, it would be good to know strong acids and bases, as well as what a larger or smaller Ka value relates to how acidic a compound is.

Gwen Casillan 3E

So first, find K and set up an ICE table. K is found from [NO]^2/[N2][O2], so it would be [0.400]^2/[0.300][0.300] = 16/9 Products [NO] are added, so reaction shifts toward reactants. Setting up an ICE table you get: [N2]=[O2]=(0.300 + x) and [NO]=(0.700-2x) *keep in mind that change in NO is -2x be...

Gwen Casillan 3E

The change in x would be +x for both N2 and O2, while for 2NO the change in x would be -2x. This is because adding 2NO to the reaction at equilibrium causes the reaction to shift towards the reactants according to Le Chatelier's Principle. The molar coefficients for N2 and O2 are both 1 so the chang...

Gwen Casillan 3E

Basically, you want to look at what exactly is changing/being added to the reaction in order to figure out the change in molarity. The change will not always be positive for products and negative for reactants. Looking at a reaction one would generally ask, does the reaction move towards making more...

Gwen Casillan 3E

Yes, Q is the reaction quotient, and K is the equilibrium constant. Q indicates that there is a shift, while K indicates equilibrium. We use Q and compare it to K to see whether the reaction is at equilibrium (Q=K), favors reactants (Q>K), or favors products (Q<K).

Gwen Casillan 3E

I definitely love watching Pride and Prejudice, Your Name, and Howl's Moving Castle. I love romances and any Ghibli films are always a good watch for feelings of nostalgia and comfort.

Gwen Casillan 3E

It would be the inverse of K, so 1/K.

Gwen Casillan 3E

When combining equations, keep in mind that the K value of a reverse reaction will be the inverse of the original K value: (1/K), while the K value of a reaction that has a coefficient (c) will be the original reaction's K value raised to the power of the coefficient: K^c. Then, once the new K value...

Gwen Casillan 3E

The textbook can definitely be helpful supplemental material to support learning. I generally us it to reaffirm knowledge of concepts and getting more practice and comfortable with problems.

Gwen Casillan 3E

Definitely doing practice problems help, as well as making sure one understands a concept and is comfortable enough to solve problems before moving on.

Gwen Casillan 3E

Devan Nathu - 2H wrote:Just to clarify, will we always be able to assume that the reaction is at equilibrium when K=Q? Thanks!!

Yes, since K is the equilibrium constant, when Q (the reaction quotient) is equal to K, the reaction is at equilibrium.

Gwen Casillan 3E

I really enjoy chemistry jokes, periodically, that is.

Gwen Casillan 3E

This is really inspiring to read about, and I am so grateful to be able to read your advice and experiences.

Gwen Casillan 3E

So if the central atom has lone pairs in addition to the bods, then the molecular shape will be bent due to the additional repulsion that the lone pairs have on the other atoms/bonded pair electrons.

Gwen Casillan 3E

So you’d find the charge on the cation by calculating the charge of the anion and figuring out what the charge should be on the cation to get the total charge of the ligand. In this case the total charge is 2-, and CN has a -1 charge (and there are 5 CN- anions, adding up to a -5 charge). Therefore,...

Gwen Casillan 3E

They are prefixes used for naming ligand when the name for a ligand already includes di- (for example, one wouldn’t say di-di-ethylene, but rather bis-diethylene), tri-, or tetra-, and also when a ligand is polydentate (like en/ethylenediamine).

Gwen Casillan 3E

Bond angles were covered in lecture and can also be found in the textbook. Bond angles are affected by repulsion between electrons, with stronger repulsions between lone pair and lone pair electrons and weaker repulsions between bonded pair and bonded pair electrons.

Gwen Casillan 3E

Lone pair electrons are further from the positively-charged nucleus of an atom compared to bonding pairs of electrons and are therefore bound with less electrostatic force, causing them to have stronger repulsive forces than bonding pairs of electrons. Lone pairs have a larger range of negativity co...

Gwen Casillan 3E

The 3 hydrocarbons at room temperature show that the greater amount of atoms there are, the stronger dispersion forces are, increasing the attractive forces between them, explaining why octadecane (C 18 H 38 ) is closer to a waxy solid than pentane (C 5 H 12 ), which is a mobile fluid, even though t...

Gwen Casillan 3E

Yes, I believe there should be a double bond, but as VSEPR primarily looks at electron densities and the position of atoms, the lewis structure depicting the single bond between Sb and O allows one to predict the shape by primarily looking at bonded pairs of electrons and lone pairs of electrons.

Gwen Casillan 3E

Some examples of dipole-induced dipoles are HCl and N₂, H₂O and O₂, and H₂O and Br₂

Gwen Casillan 3E

The repulsion forces in order of least to most strong are:
bonding-bonding pr < lone-bonding pr < lone-lone pr

Gwen Casillan 3E

It's best to apply significant figures at the end in order to retain as much accuracy as possible. Rounding early on or in the middle of calculations may initially seem convenient and more practical, but the dropping of a few decimal places/numbers can be severely detrimental to the final result, as...

Gwen Casillan 3E

There was only one reactant for M.3 and therefore, you did not need to find a limiting reactant/there was none. If there is more than one reactant, you would use the given moles/masses and molar masses to determine the limiting reactant and use that information to calculate the theoretical yield.

Gwen Casillan 3E

Just use the stoichiometric steps you used to calculate the the 12 mols of the ClO2F product to determine the moles of the other product (Br2).

Gwen Casillan 3E

Yes, you would need to consider the different masses of the isotopes because they have different percentages of abundance, both of which you would then use to calculate the mass of the average lithium atom by adding them together: (7.42/100)(9.988 * 10^-24 g) + (92.58/100)(1.165 * 10^-24 g) = the av...

Gwen Casillan 3E

Yes, you could technically leave the volume in mL since the molarity would cancel, but remember to keep in mind that otherwise, it would be wiser to leave it in liters.

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