CHEMISTRY COMMUNITY

Simon Dionson 4I

I actually use H+ to help balance my basic equations, except an extra step would have to be taken. When I have H+, I convert that to OH- by adding OH- to both sides. H+ and OH- neutralize to H2O so H+ never appears in my final equation I learned this method in this video: https://www.youtube.com/wat...

Simon Dionson 4I

Test 2 is available in your TA's mailbox in Young Hall.

Simon Dionson 4I

along with the responses above, I also try to consider which assignment of the cathode or anode will result in a positive Eº cell value

Simon Dionson 4I

Elements with positive oxidation numbers will generally gain electrons whereas elements with negative oxidation numbers will generally lose electrons. This can change depending on if you're dealing with a reduction or oxidation reaction.

Simon Dionson 4I

Does the presence of a catalyst change the value of rate constants?

Simon Dionson 4I

I think we should be able to determine the rate law based off of what the axes are labeled and what gives us a linear line or not.

Simon Dionson 4I

Intermediates are included in the elementary rate laws but not the overall rate-law since they cancel out.

Simon Dionson 4I

I believe Dr. Lavelle said that we will never have to determine the intermediates ourselves since they are experimentally proposed/recorded

Simon Dionson 4I

I believe you include whatever reactant is lost/decreased.

Simon Dionson 4I

I initially did the same thing, but I found that by using the second-order half-life to find k and using that in the integrated rate law, you could find t by using [A]0 and [A] as whatever amount of [A]0 is remaining.

Simon Dionson 4I

I don't believe you include the stoichiometric coefficients when writing the cell diagram. As for H+ being included, I have always written it in my cell diagrams if it is involved in the half-reactions and that's been pretty consistent with my homework answers and the book.

Simon Dionson 4I

Adding on to the answers above, you use an inert electrode if there is no solid METAL. An example is Iodine, I2(s). You would still have to use an electrode for half-reactions involving solid iodine since it is technically a gas/nonmetal.

Simon Dionson 4I

If given Eº values, you can decide which ones are the cathode and anode if Eº is positive when subtracting them (if you aren't given a cell diagram)

Simon Dionson 4I

You first start off with
O2 --> H+

balance with oxygen with water

O2 --> H+ + 2H2O

balance hydrogen with protons

5H+ + O2 --> H+ + 2H2O

cancel protons

4H+ + O2 --> 2H2O

balance charges

4H+ + 4e- + O2 --> 2H2O

To balance OH-, you use the basic method.

Simon Dionson 4I

For predicting the voltage of galvanic cells for b: Ag(s)|AgI(s)|I-(aq)||Cl-(aq)|AgCl(s)|Ag(s) The anode reaction is: AgI + e- --> Ag + I- The cathode reaction is: Cl- + Ag --> AgCl + e- I reversed these reactions and assigned them negative values since oxidation occurs at the anode and reduction oc...

Simon Dionson 4I

For the galvanic cell diagram Pt(s)|I2(s)|2I-(aq)||Ce4+(aq) ,Ce3+(aq)|Pt(s), why is platinum included at the anode even though the iodine being oxidized is a solid?

Simon Dionson 4I

Jessica Esparza 2H wrote:Do you add the same amount of OH-?

Yes, the OH- will neutralize the H+ and leave you with water and OH-

Simon Dionson 4I

Start by identifying the oxidized and reduced species. The oxidation number for Cr changes from +6 to +3, indicating reduction (gain of 3 electrons). The oxidation number for C changes from -2 to -1, indicating oxidation (loss of 1 electron). You now have the skeleton for your half-reactions: (Cr2O...

Simon Dionson 4I

Protons aren't transferred, only electrons. Protons are used to balance redox reactions in acidic solutions, however.

Simon Dionson 4I

Oxidation numbers are the theoretical charge of the molecule. Many follow patterns: group 1 is +1, group 2 is +2. H is 0 or +1 and O is 0 or -2. A molecule that breaks this rule is H2O2 where the oxygens are actually -1

Simon Dionson 4I

The overall charge is 0 on both but the charge of C changes.

Simon Dionson 4I

Why do molecules and atoms with greater mass have more molar entropy?

Simon Dionson 4I

When counting positions on a complex molecule, would we count lone pairs as their own position? For example, would NH3 have 4 possible positions?

Simon Dionson 4I

I'm confused as to why neither (3R/2) or (5R/2) are used for nitrogen; they just used R by itself as N2's ideal heat capacity.

Another general question I have is when to use 3R/2 and 5R/2 for heat capacities when calculating entropy.

Simon Dionson 4I

Will the entropy equations for isothermal expansion/compression, pressure change, and heat change be on the midterm?

Simon Dionson 4I

Hey everyone, I left my black Ray bans in Bunche 3211 in the back row of the classroom for discussion 4I (Friday’s 12-12:50) on Friday 2/7. If you have them, would you mind returning them? My name is Simon Dionson and I have lectures MWF from 3-3:50.

Simon Dionson 4I

Sean Cheah 1E wrote:Because the process is isothermal, q = -w. Decreasing pressure must mean that volume is increasing, therefore work is negative. Then q is positive and delta S = q/T is also positive.

Oh, that makes sense. Thank You! I think I interpreted decreasing pressure as decreasing volume.

Simon Dionson 4I

If the pressure of a gas in a container is decreased isothermally, wouldn't the entropy change be negative? The answer key is saying that it is positive.

Simon Dionson 4I

How do we know to use 5/2R and 3/2R for constant pressure and volume?

Simon Dionson 4I

Is it incorrect to report the entropy change in Celcius rather than kelvin?

Simon Dionson 4I

Is it possible for a heat capacity to be negative?

Simon Dionson 4I

Why is there a negative in: q = -C(cal)ΔT. What is this indicating/what is losing heat and gaining heat?

Simon Dionson 4I

Elizabeth Harty 1A wrote:What is the difference between enthalpy and entropy?

Entropy is the disorder of a system, Enthalpy is the measure of heat/energy related to a substance when it is changed.

Simon Dionson 4I

Benzene has a special value for bond enthalpy that is noted in the table; I also made the same mistake.

Simon Dionson 4I

Phase changes are constant to change intermolecular forces. Sloped lines can indicate heat capacities.

Simon Dionson 4I

When calculating a reaction using bond enthalpies, will the answer always be negative (exothermic)?

Simon Dionson 4I

How do we know if q is negative or positive for q(system) when analyzing a bomb calorimeter?

Simon Dionson 4I

If we took Chem 14A with Lavelle last quarter does anyone know when the final is handed out/if it was already?

Simon Dionson 4I

At 700k less ammonia is formed since the K value is smaller. The K value being smaller indicates that there is more reactant that product at equilibrium

Simon Dionson 4I

Increasing pressure will favor the side with fewer mols of gas

B) 1mol of H2O vs 1mol of H2 + 1mol CO --> reactants favored
D) 2mol of HD vs 1 mol of H2 + 1mol of D2 --> neither favored since both have 2 mols of gases

Simon Dionson 4I

the -log means that the higher the number, the lower the value. Higher numbers result in stronger acids/bases for pka and pkb.

Simon Dionson 4I

In my experience, I usually use this rule in weak acid/base equilibria.

Simon Dionson 4I

Since reactants are represented at the bottom, if the number is smaller than 10^-3 then more reactants are present at equilibrium

Simon Dionson 4I

Water is only not included if it is a solvent (liquid) or a solid.

Simon Dionson 4I

Does anyone have the answers for both parts?

For part a, I calculated: K(a) = 0.09, pKa = 1.0

Part b, I calculated: K(b) = 5.7*10^-4, pKb = 3.2

I can't find the solutions so I wanna make sure these are right

Simon Dionson 4I

When can we approximate x when expressing K(a) or K(b)

i.e. when (x)(0.10+x)/(0.10 - x)

can be approximated to

(x)(0.10 + x)/(0.10)

Simon Dionson 4I

Do strong acids have K(a) values like weak acids do? Or is there even a distinction between K and K(a)?

Simon Dionson 4I

Tracy Tolentino_2E wrote:Do we not have to calculate anything?

I didn't do any math, I used the K values to base my answers. Since K at the lower temperature was smaller, I assumed more products were formed at 600K vs 700K.

Simon Dionson 4I

Removal of a product will cause the equilibrium to shift to the right, meaning there will be more reactants consumed to reach equilibrium again.

Simon Dionson 4I

If p(NH3) is decreased will O2 increase?

3NH3 + 5O2 = 4NO + 6H2O

My initial answer is yes because the reaction will shift to the left, is this explanation correct?

Simon Dionson 4I

For H2 + I2 <=> 2HI

If 60% of 0.133M H2 has reacted at equilibrium, do I represent the change as - 60% of H2 and the equilibrium concentration as the remaining 40%?

Simon Dionson 4I

To set up the quadratic equation, there would actually be no denominator correct?

NH4HS(s) = NH3(g) + H2S(g)

We would only use NH3 and H2S in the ICE table?

Simon Dionson 4I

You can also look at the product formed. F2 -> 2F has more atoms formed than Cl2 -> 2Cl, F atoms form more than Cl atoms at the same conditions, so they are more stable.

Simon Dionson 4I

If we were calculating a reaction with multiple phases (i.e. solid, liquid, aqueous, gas), will calculating P(q) and P(k) only deal with the gases?

Simon Dionson 4I

When calculating Q, do we apply the same rules of not including the concentrations of solids and solvents?

Simon Dionson 4I

I don't see the answer key for the marshmallow practice problems so can someone tell me if the answer to 34 is a pH of 2.1? I'm not sure if it's 2.1 or 2.3. CaO + H2O --> Ca(OH)2. There are 2 hydroxides produced for every 1 CaO. If you double the moles of OH-, the answer comes out to 2.276 vs 2.1 i...

Simon Dionson 4I

It will be provided on the front sheet along with other constants and equations.

Simon Dionson 4I

The only possible l values are 0, 1, 2, 3 (s, p, d, f respectively)

Simon Dionson 4I

I think as long as you know the values, you should be fine. I also don't think they'll explicitly mark you down if you assign a random ml value to an atom since they can be in any orientation in a given moment.

Simon Dionson 4I

If light is absorbed, a system is receiving an input of energy, so the value is positive. If light is emitted, a system is releasing energy (essentially losing it), so the value is negative.

Simon Dionson 4I

If we were to dissolve Na2CO3 in water, would CO3 2- be completely or partially protonated by the water?

i.e.

CO3 2- + 2H2O -> H2CO3 + 2OH-

vs

CO3 2- + H2O -> HCO3- + OH-

Could the same be said for bases with a negative charge greater than 1? thanks.

Simon Dionson 4I

water is amphoteric since it can form H3O+ or OH- if it accepts or donates a proton respectively. Another amphoteric molecule is HCO3- (bicarbonate)

It can accept a proton to form H2CO3 (carbonic acid)
It can also donate a proton to form CO3 2- (carbonate ion)

Simon Dionson 4I

Arrhenius acid: a compound that supplies H+ in water
Arrhenius base: a compound that supplies OH- in water

Brønsted acid: a compound that donates protons
Brønsted base: a compound that accepts protons

Lewis acid: an electron pair acceptor (i.e. H+)
Lewis base: an electron pair donor

Simon Dionson 4I

Why is acetic acid considered a weaker acid than formic acid? I thought carbon is more electronegative than hydrogen.

Simon Dionson 4I

Why is Ag+ considered a Lewis acid and not a Lewis base? Doesn't it have several electron pairs it can donate?

Simon Dionson 4I

Is [PtCl2(en)2]2+ coordination number 4 or 6? Does en form a chelate that binds to the platinum more than once?

Simon Dionson 4I

Electrons with high electronegativity are likely to form polar bonds which interact with partial charges in water molecules.

Simon Dionson 4I

I'm guessing resonance structures still apply to the idea that they're hybrids of the resonance structures, so the number of signature and pi bonds don't change based on what resonance structure you use.

Simon Dionson 4I

Yes, but the atom would have to have a partially positive on the hydrogen and there would have to be a strongly electronegative atom, like N, O, or F

Simon Dionson 4I

Stronger bonds result in higher melting points, so if bonds were broken, the melting point would decrease

Simon Dionson 4I

Visually, it's similar to a trigonal bipyramidal but you would have to remove one bonding pair. So it would be 90, 180, 120

Simon Dionson 4I

He explicitly told us on Friday's lecture that only hybridization wouldn't be on the test. I think it would be safe to assume this will be on the test.

Simon Dionson 4I

Are there any shortcuts or quick ways of finding an atom's hybridized orbitals?

Simon Dionson 4I

Polarising power increases with an atom with fewer electrons and is more positively charged (up and left). Polarizability increases with an atom with more electrons and a more negatively charged nucleus (down and right)

Simon Dionson 4I

∂+ and ∂- represent partial charges due to difference in electronegavities. A dipole is indicated by ∂+ --> ∂-

Simon Dionson 4I

A double bond is stronger than a single bond due to its stability and distance between the bonding atoms. Shorter bond lengths allow attractive forces between atoms to be stronger.

Simon Dionson 4I

Polarizability is how likely an electron cloud can be distorted in shape. Polarizability also increases with the size of atoms and how many electrons they have. If an atom has more electrons, there is a higher likelihood electrons can gather on one side and create an induced dipole.

Simon Dionson 4I

The more electrons a molecule has, the stronger its induced dipole interactions are. If a molecule/atom has more electrons, they have increased probability of electrons creating a partial charge by gathering on one side of the electron cloud.

Simon Dionson 4I

What is the relationship between the strength of intermolecular forces and vapor pressure?

Simon Dionson 4I

Polarizability increases with the size of the electron and how many electrons it has. For example, Iodine has high polarizability because it is a big atom with many electrons compared to something like sodium.

Simon Dionson 4I

The orbitals of oxygen have one more electron in any of the p orbitals. These create electron-electron repulsions that make it easier to remove an electron than nitrogen.

Simon Dionson 4I

An ionic bond would be denoted as the cation being [M]+ with the anion having the donated electron(s) surrounding it with a charge [M]-. A covalent bond has a line for each bond its sharing electrons make. H2 is H-H because its electrons make a single bond

Simon Dionson 4I

Is there a relationship demonstrated between a particle in a box and nodes in an atom? If so, how can it be explained?

Simon Dionson 4I

To find the kinetic energy of the electron, you actually have to use 2 different equations. Use de Broglie's to find that lambda = h/mv and solve for v. Using the v from de Broglie's equation, find the kinetic energy using E(k) = 1/2(m)(v)^2. Note that m in both these equations is the mass of an ele...

Simon Dionson 4I

In the 2p orbital, all nitrogen's are half-filled with electrons whereas oxygen has one full orbital and two half-filled orbitals. This first full orbital in oxygen has greater electron repulsion than the first half-full orbital in nitrogen, which makes it easier to remove an electron

Simon Dionson 4I

Threshold energy is the minimum energy needed to remove an electron from a metal surface (photoelectric effect). Ionization energy is the energy needed to remove an electron from an element.

Simon Dionson 4I

Use E = hc/lambda to find the individual energies of the lasers. Convert the work function from eV to joules. Choose the laser that is greater the work function and from there you have the laser and can calculate the kinetic energy from E = (work function) - (kinetic energy).

Simon Dionson 4I

How can you tell exactly what pair of elements exhibit a diagonal relationship? Does it have to do with what ions they can readily form?

Simon Dionson 4I

If an electron is removed from Mn, why is the electron removed from the 4s orbital instead of 3d orbital?

Simon Dionson 4I

Would someone explain to me the answers of b and c? When m(l) in part b is given, is that talking about a specific orbital making the answer 2? For part c, I understand that 2p and 2s can exist, so are the orbitals of those two added together?

Simon Dionson 4I

This question is asking for the number of orbitals present in a given subshell. The answers in the answer guide are given in electrons, but wouldn't the answers be based on the value of 2l and 2l+1?

Simon Dionson 4I

We can tell the de Broglie equation must be used because it's relating wavelength, mass, and velocity. The de Broglie uses SI units so you'll have to convert the values given.

Simon Dionson 4I

Emission spectrums represent the energy an electron requires for it to jump up from a lower level to a higher level and back down (when it jumps back down, it releases energy as a photon, which is the line on the spectrum). Where there isn't a line, think of those regions as radiation not having eno...

Simon Dionson 4I

Electron spins have a charge (denoted by + or - 1/2). If they spin in the same direction, they have similar charges and repel each other.

Simon Dionson 4I

The threshold energy (work function) can be thought of as the energy needed to eject an electron from a metal. Any energy radiated that is higher than this will result in a higher kinetic energy. (That's what the equation E = (work function) + E(k) is is relating) In energy levels, electrons have to...

Simon Dionson 4I

I had a general question on what it means when the velocity is given as
5.00 + or - 5.0 m*s^-1. If we are using p=mv for momentum, how could I relate the + or - interval into ΔP?

Simon Dionson 4I

It's also important to note that Planck's constant, J * S, cancel out the bottom units in wavelength = h/mv (J = kg*m^2*s^-2) (mass = kg) (velocity = m*s^-1)

So wavelength = (kg)(m^2)(s^-2)(s)/(kg)(m^s)(s^-1) = wavelength in meters.

That's how I see it in the context of units like mass and seconds.

Simon Dionson 4I

I'm having a hard time understanding what the wavy lines represent. Is it the path of an electron? Or is it representing a wavelength of an electron? Is it representing an electron as a wavelength at a certain energy level? (i.e. n=1, etc.)

Simon Dionson 4I

Dr. Lavelle's example problem on the 10/9/19 lecture is actually an alternative to way to use the Rydberg equation if both the initial n and final n are known and its possible to use the difference of the calculated energies. Unfortunately for this problem, I'm not sure it's possible since the final...

Simon Dionson 4I

I need some clarifying on the concept of: as n approaches infinity, E = 0. From what I understand it's because the difference between energy levels becomes smaller, but is the electron becoming less bound to the atom also a valid reason why E = 0?

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