CHEMISTRY COMMUNITY

Jeremy_Guiman2E

Thank you for a wonderful quarter Dr. Lavelle! Your plentiful resources have been very helpful in me improving my work in this class, and I feel prepared for the rigor of future Chemistry courses. All the best!

Jeremy_Guiman2E

Enzymes are catalysts because they are not consumed in reactions but can accelerate reactions by lowering activation energies.

Jeremy_Guiman2E

When ions dissociate in solution, they are considered aqueous and are therefore not considered solids/liquids.

Jeremy_Guiman2E

Ideal gases do not take up space and don't interact. They obey gas laws.

Jeremy_Guiman2E

Thank you for an excellent quarter and for handling the current situation well! Best of luck to everyone.

Jeremy_Guiman2E

Galvanic cells transforms energy from spontaneous redox reactions into electrical energy. Electrolytic cells convert electrical energy into chemical energy.

Jeremy_Guiman2E

Rate laws would look at reactants because we are using initial concentrations of reactants to determine rate into product creation.

Jeremy_Guiman2E

You'd use the conc of B to solve for the new concentration of A.

So [A]t = (0.153 mol A/L) - [(2 molA/1 mol B)(0.034molB/L) (this is the new concentration of A)

Then, solve for rate constant using the initial/new concentrations with 115s as time.

Jeremy_Guiman2E

We know that the reaction would take the format _A -> _B + _C because it talks about the decomposition of A. We see also that A decreases about 10 increments of partial pressure and correspondingly, C increases by that amount while B only increases by half that amount. Because B is produced at half ...

Jeremy_Guiman2E

005384106 wrote:At what point do you add elections to the half reaction?

you would add electrons at the end to balance charge on both sides.

Jeremy_Guiman2E

Electrodes are conductors that make contact with electrolytes. Electrolytes involve substances that dissolve to give an electrically conducting solution (NaCl is an example of a strong electrolyte).

Jeremy_Guiman2E

You need only look at the number of electrons involved in the reaction.

Jeremy_Guiman2E

I think you'd use "simpler" situation where the element would need to gain less electrons. So you'd use Fe2+ as opposed to Fe3+. Not 100% sure though!

Jeremy_Guiman2E

As stated above, you'd just compare the stoichiometric coefficients (for purposes of simplicity you can assign mol values to each coefficient). Because the reaction is N2 (g) + 3 H2 (g) -> 2NH3 (g), you can use this to answer what they ask about rate: So, you can see that N2 is consumed at 1/3 the r...

Jeremy_Guiman2E

1 atm and 273 K.

Jeremy_Guiman2E

Amperes measure electric current (coulombs in a second) whereas coulombs measure electric charge (amp * second). Amperes thus measure how fast the amount of charge is moving.

Jeremy_Guiman2E

Positive SRP correlates to a spontaneous reduction!

Jeremy_Guiman2E

W(max) is the maximum amount of work possible. It appears that by definition G is the maximum non-expansion work that can be done under constant T and P.

Moreover, see this: https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=5141

Jeremy_Guiman2E

You will balance charges (using electrons) and amount of elements (often with the addition of H20 or H+).

Jeremy_Guiman2E

6K.1 The following redox reaction is used in acidic solution in the Breathalyzer test to determine the level of alcohol in blood: H + (aq) + Cr 2 O 7 2- (aq) + C 2 H 5 OH(aq) --> Cr 3+ (aq) + C 2 H 4 O(aq) + H 2 O(l) Identify the elements undergoing oxidation or reduction and indicate their initial ...

Jeremy_Guiman2E

I believe workshops are an opportunity for students to get more practice via problems (they understand a majority of the material) whereas step-up sessions involves step by step guidance as to how to approach problems.

Jeremy_Guiman2E

Would we be able to determine Magnesium's charge first? Or is it always easier to start with elements like Oxygen first when determining molecular charges?

Jeremy_Guiman2E

Nathan Rothschild_2D wrote:The answer to this might be obvious but how does this connect to batteries?

At a basic level, batteries are storage places for chemical reactions to occur. These chemical reactions can be used to generate electrical energy for our usage.

Jeremy_Guiman2E

Anything done to prepare for the midterm is probably ok, though more specifically Thermochem/Thermodynamics problems as those have been our most recent topics of discussion as we just started Electrochemistry.

Jeremy_Guiman2E

Electrochemistry is understanding chemical applications involving electricity. In class we looked at oxidation and reduction - redox - reactions as examples (loss or gain of electrons, respectively). Chemical changes are often seen with changes in electrical energy which we will be looking at.

Jeremy_Guiman2E

Hi, I'm currently ill and wasn't able to make it to class this past Monday (2/10). Would someone be able to either post their notes here or send them to me at jguiman39@ucla.edu ? Thanks!

Jeremy_Guiman2E

Basically you compare the molecule(s) before and after (it would thus be very helpful to draw them out) and identify (based off of your drawing) which bonds are present before that are not present after AND which bonds are present after that were not there before.

Jeremy_Guiman2E

The cis form would have the higher residual entropy because it has many more possible orientations than the trans form. If we used the Boltzmann entropy equations, we can see that the cis form would have a greater entropy. Residual entropy is a particular type of entropy - it is "the nonzero en...

Jeremy_Guiman2E

12B) Suppose a researcher finds that delta H(rxn)=-2756 kJ for the reaction at 200. degrees C. Assuming all heat capacities are constant, calculate delta H(rxn) at the temperature of the human body, 37 degrees C. Hint: since enthalpy is a state function, the process can be divided into three steps. ...

Jeremy_Guiman2E

Why do molecules at 0 K have entropy still? Are there circumstances under which molecules could have no entropy?

Jeremy_Guiman2E

A will be greater than B or C because it has a greater amount of individual particles (and therefore greater entropy). C will be greater than B because its atoms are vibrationally active as opposed to B's which are not. Thus, we get A > C > B.

Jeremy_Guiman2E

I think we use the integral when we're trying to sum an infinite number of steps (in volume changes?), but can someone explain when to use it and how/why we need to?

Jeremy_Guiman2E

The first step is to find the heat capacity that is calibrated by the calorimeter. To do this use the eqaution C=q(calorimeter)/change in T. Essentially you would do C=(3.50kJ)/(7.32K) = 0.478kJ/K. Then you can use this heat capacity to calculate the heat released by the reaction. So set up the equ...

Jeremy_Guiman2E

4A.13 A constant-volume calorimeter was calibrated by carrying out a reaction known to release 3.50 kJ of heat in 0.200 L of solution in the calorimeter (q= -3.50 kJ), resulting in a temperature rise of 7.32 C. In a subsequent experiment, 100.0 mL of 0.200 m HBr(aq) and 100.0 mL of 0.200 m KOH(aq) w...

Jeremy_Guiman2E

You do use w = P * A * D!

w = 2 atm * (0.015 m)² * 0.2 m * pi
w = 2.8274 * 10^-4 m³ * atm

Then, you can multiply the answer by 101,325 Pa to convert from atm to Pa. Once you have your answer in m³ * Pa, it is equal to that amount in joules (J).

Jeremy_Guiman2E

Why must you convert the answer to pascals when the units (J) does not include pascals?

You must convert the answer to Pascals because the textbook gives a conversion factor that converts meters cubed x Pascals into joules (1 Pa x m³ = 1 J). Remember that 1 atm = 101,325 Pa.

Jeremy_Guiman2E

I'd be prepared to use all of them. However, Professor Lavelle noted that the examples he gave in class were very simple and that the textbook has both simpler and more complex examples.

Jeremy_Guiman2E

Method 3 uses the standard enthalpies of formation for both products and reactants. It is different than Method 2 which looks specifically at the bonds of the reactants and products. You can take the standard enthalpy of formation for the products and subtract the standard enthalpy of formation for ...

Jeremy_Guiman2E

To add on to the above post, attached is the specific example Dr. Lavelle used:

Jeremy_Guiman2E

The midterm will be Wednesday, February 12. This is Week 6.

Jeremy_Guiman2E

Method 2 involves using specific bond enthalpies to calculate delta H of the reaction. Basically, for the reactants, bonds would need to be broken and therefore energy is required (positive delta H). Conversely, for the products, new bonds are formed, meaning that energy is released (negative delta ...

Jeremy_Guiman2E

Homework 3 should probably be from the Acids and Bases Equilibria section given that we finished covering it last week (and we already completed the Chemical Equilibrium section as well).

Jeremy_Guiman2E

Strong acids by definition completely (or nearly completely) dissociate in solution. Conversely, weak acids would dissociate much less. Less dissociation as with weak acids means that the molecules are still held together and would therefore have stronger, shorter bonds.

Jeremy_Guiman2E

Increasing temperature adds heat to the reaction. Because the reverse reaction is endothermic, adding heat means it is inputted into the system and will favor the reverse reaction.

Jeremy_Guiman2E

This is how I did it, and your process looks good!

Jeremy_Guiman2E

Thank you Brian, this was really helpful!

Jeremy_Guiman2E

how are stoichiometric coefficients involved in Equilibrium? Why when they are the same, its still in equilibrium? Stoichiometric coefficients are involved in the equilibrium equation as they are the exponents used for the concentrations. It is important to ensure the equation is balanced first. To ...

Jeremy_Guiman2E

Pure liquids and solids don't affect the reactant amount as their concentrations stay the same.

Jeremy_Guiman2E

Yes, we use initial concentrations of given compounds/substances to determine their final concentrations at equilibrium. These final concentrations are part of the equilibrium composition.

Jeremy_Guiman2E

K values will most likely be given as they are unique values depending on the equations.

Jeremy_Guiman2E

5I.13 (a) In an experiment, 2.0 mmol Cl2(g) was sealed into a reaction vessel of volume 2.0 L and heated to 1000. K to study its dissociation into Cl atoms. Use the information in Table 5G.2 to calculate the equilibrium composition of the mixture. (b) If 2.0 mmol F2 was placed into the reaction vess...

Jeremy_Guiman2E

This seems to be the correct electron dot diagram: https://qph.fs.quoracdn.net/main-qimg-d14aef43ff5b2ed2d29871b3296d8571 The outer I's have a formal charge of 0, and I believe the formal charge of -1 is ok on the central I atom because of the fact that I 3 - is a negatively charged molecule, with a...

Jeremy_Guiman2E

Yes, essentially anion radius > neutral atom radius > cation radius. Atomic radius increases down a group (as amount of electrons/shells increases) and decreases from left to right (as more protons draw in electron shells more).

Jeremy_Guiman2E

This chart also sums up the point made above. The presence of lone pairs typically alters the bond angles to less/greater than expected, except in the two cases seen (steric number 5, linear shape and steric number 6, square planar shape) as these electrons can be placed anywhere on the axis, so the...

Jeremy_Guiman2E

17.37 With the help of Table 17.4, determine the coordination number of the metal ion in each of the following complexes:

(a) [NiCl₄]^2-
(b) [Ag(NH₃)₂]⁺
(c) [PtCl₂(en)₂]²⁺
(d) [Cr(edta)]^-

How would you determine the coordination number for c and d?

Jeremy_Guiman2E

If something can produce/donate an H⁺ ion, it is probably a Bronsted acid. But, if it can accept one, it is a Bronsted base. The base is often OH^-, but as you can seen, ammonia (NH₃) can also act as a Bronsted base (proton acceptor) to produce ammonium ion, NH₄⁺.

Jeremy_Guiman2E

Recall that Bronsted acids are proton (or H + ) donors, and that Bronsted bases are proton (or H + ) acceptors. I believe that H 2 SO 3 can dissolve into sulfite ion and H + ions, while Ca(OH) 2 can dissolve into Ca 2+ and OH - . Thus, because H 2 SO 3 can donate protons, it is a Lewis acid. And, be...

Jeremy_Guiman2E

Is the partial positive on the central O and the partial negative on the outer O just due to a greater amount of lone pairs on the outer O, or are there more/other reasons?

Jeremy_Guiman2E

Dipoles would cancel each other out when the polar bonds in a molecule are equal and opposite. Molecules that have symmetry thus often have dipole cancellation. This can be seen in CO 2 as it is a linear shape. In the CO 2 molecule, the partial negative oxygens will have a dipole moment outward, eff...

Jeremy_Guiman2E

https://pubchem.ncbi.nlm.nih.gov/image/imgsrv.fcgi?cid=1089&t=l I used the above for reference. The 2 hydrogens are bonded to their respective oxygens. This is enough for hydrogen bonding because the hydrogens will get their partial positive as O is the most electronegative element present. All...

Jeremy_Guiman2E

They're the same thing. You can use them interchangeably.

Jeremy_Guiman2E

Would a molecule like nitrogen trifluoride (NF₃) have an electronegativity difference (partial positive/partial negative) suitable for hydrogen bonding (i.e. with water)? I'm just not sure if the fact that these are 2 of the most electronegative elements affects this.

Jeremy_Guiman2E

Why does repulsion strength occur in this way: lone-lone pair > lone-bonding pair > bonding-bonding pair?

Jeremy_Guiman2E

Yes, you'd be determining the bond angle for the ClO₂⁺ ion.

The textbook calls it the OClO bond angle as it's just saying that the bond exists in this way (O - Cl - O), suggesting that Cl is the central atom with an O on both sides. Instead of saying angle ABC, it's saying Angle OClO.

Jeremy_Guiman2E

Lone pairs on neighboring atoms will repel each other. This causes the bonds to weaken as the electron pairs try to move farther apart from one another.

Jeremy_Guiman2E

The roman numerals after these elements indicate the ion being used. Because the element given has a roman numeral after it, you know it has multiple ions (i.e. Fe 2+ and Fe 3+ could be iron(II) and iron (III). For example, indium (III) sulfide will form In 2 S 3 from In 3+ and S 2- . Thus, you know...

Jeremy_Guiman2E

So for a and b of this problem, we're given: (a) [Ar]3d 6 ; (b) [Ar]3d 5 (and that these are M 3+ where M is a metal). This means that to find the neutral atom (so we know which element we're looking for) we'd need to look for the electron configuration with 3 more electrons than what we were given ...

Jeremy_Guiman2E

No problem!

Jeremy_Guiman2E

Essentially, your set up would look like this: E = (5.00 x 10^-3 g Na) (1 mol Na / 22.99 g Na) (6.022 x 10²³ atoms/mol) (3.37 x 10^-19 J/atom) = 44.1 J

The 3.37 x 10^-19 J/atom comes from your answer in part a. Hope this helps!

Jeremy_Guiman2E

You would just use the mass of an electron, which is m = 9.109 383 × 10⁻³¹ kg. Hope this helps!

Jeremy_Guiman2E

The Rydberg constant is 3.29 x 10 15 when using frequency = R (1/n 1 2 - 1/n 2 2 ). Because c = wavelength x frequency, there is another form of the equation. Plugging in c/wavelength for what was frequency and rearranging the equation, we'd get: 1/wavelength = R/c x (1/n 1 2 - 1/n 2 2 ). The R/c va...

Jeremy_Guiman2E

Yeah, as Chem Mod said above, "uncertainty in velocity is the difference between the highest and lowest possible values for the velocity". So given 3.24 +/- 0.06 m/s, our highest possible value is 3.30 and our lowest is 3.18. The difference of the above values is 0.12 m/s which is what del...

Jeremy_Guiman2E

For part a, the frequency = 3600cm -1 * c So, frequency = 3600cm -1 * 2.998 x 8 8 m/s = 3600cm -1 * 2.998 x 10 8 cm/s (Here, I converted everything to cm because it doesn't really matter, but you can convert to meters too.) This gives you about 1.1 x 10 14 s -1 when rounded and using correct sig fig...

Jeremy_Guiman2E

They want you to find the initial and final values of "n", so you know you'd need the Rydberg equation. But, in order to use the Rydberg equation, we'd use the frequency. So, because we're given 102.6 nm, that equals 102.6 x 10 9 m. And plugging it in to: c = lambda * nu OR speed of light ...

Jeremy_Guiman2E

Yeah, test 2 starts Tuesday, October 23 in your discussion section. See:

https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Chem14ATestSchedule.pdf

Jeremy_Guiman2E

I believe he uses 6.63 X 10^-34 when there are only a maximum of three significant figures are given. Otherwise, if given four or more, he will use 6.626 X 10^-34.

Regardless, I would agree though that using 6.626 X 10^-34 consistently will give the most accurate results.

Jeremy_Guiman2E

Significant figures are not affected by constants similar to how measured values like 1m = 100 cm are not used as sig figs. You would just need to look at what numbers you are given in the actual question itself as constants do not affect the sig figs.

Jeremy_Guiman2E

Yes, because Avogadro's number is used as a conversion factor, it is reversible. So, going from "x" atoms to moles, you would say "x" atoms times (1 mol/6.022 x 10 23 atoms) = y mol. Let's say you have 15 atoms. 15 atoms x (1 mol/6.022 x 10 23 atoms) = 2.5 x 10 -23 mol. Alternati...

Jeremy_Guiman2E

It looks like you are talking about the CaC 2 + 2H 2 0 --> Ca(OH) 2 + C 2 H 2 problem. In class, he calculated the molar mass of C 2 H 2 but not Ca(OH) 2 . He didn't calculate the molar mass of calcium hydroxide (Ca(OH) 2 ) because it's not relevant to answering the question. Certainly, for purposes...

Jeremy_Guiman2E

Ok, so we are given 5.15 g of epsom salts, and know the formula unit would be MgSO 4 + 7H 2 O. We will use the molar mass (246.48 g/mol) to determine how many total moles of epsom salts we have. So, 5.15 g / 246.48 g/mol = 0.0209 mol epsom salts. However, because there are 7 water molecules in each ...

Jeremy_Guiman2E

In this problem, we are given several mass percentages as well as the molar mass of the compound, L-Dopa, tasked to find its molecular formula. First, we will assume that we have 100 g of the substance because our mass percentages total to 100% . So, we have 54.82g C, 5.62g H, 7.10g N, and 32.46g O....

Jeremy_Guiman2E

So, we are given that there are 4 moles of butane, C 4 H 10 , as well as the chemical equation with varying stoichiometric coefficients. What we know so far, thus, is: 4 C 4 H 10 (g) + ___O 2 (g) --> ___CO 2 (g) + ___H 2 0(g) Typically, we can balance chemical equations using the CHO technique (Carb...

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