CHEMISTRY COMMUNITY

chemboi

In general, I think Dr. Lavelle said so. If another commenter could clarify, that would be great!

chemboi

I agree! Thank you to all the team for their hard work, and also Dr. Lavelle for being understanding and alleviating stress :3

chemboi

Are you asking how to set this up? If the reaction is COBr2 -> CO + Br2, then it's an ICE table setup.

chemboi

I think its B because while you have equal number of reactant molecules, the reaction will be taking place in less volume of solution, so any temperature change will be more pronounced.

chemboi

if what you're adding will dissociate into a proton, it will make the solution more acidic. HCl is an immediate candidate. I believe NH4+ is considered a weak acid, so C seems like a good answer.

chemboi

pOH + pH = 14

chemboi

One liter is 1/1000th of a cubic meter. way of thinking about that: 10cm^3 is 1000cm^3 is a liter. 10cm is 0.1(meter). 0.1^3 (for 3 side lengths) is 0.001, or 1/1000th

chemboi

In a situation where you'd be solving 1 equation with 2 unknowns without knowing A, I'd think the problem would say the value.

chemboi

Cell potentials, I believe, are reported as reduction potentials. An oxidizing agent allows oxidization of another species by itself being reduced, and a reducing agent similarly allows something else to be reduced by itself being oxidized.

chemboi

Q is [product]/[reactant], use in the appropriate place in the expression for the Nernst equation (will be given on eq. sheet)

chemboi

Are those concentrations given in the form of experiments increasing the concentration of one species, then seeing the change in overall reaction rate? If so, those let you figure out the reaction order with regard to each reactant, which can inform the overall instantaneous initial rate.

chemboi

If you know the half reactions, I think adding (in the appropriate manner with regard to signs) their standard reduction potentials could give you think the cell potential/

chemboi

I'm pretty sure that's if the two species are in the same phase

chemboi

Hm perhaps because you'll have a pool of electrons available to move down a gradient and do work?

chemboi

It relates gibbs free energy to the cell potential, and is useful when you know the concentrations of your reactants and products.

chemboi

Splitting the net reaction into the two half reactions that occur at each electrode respectively make it clearer to work through the problem and see what's going on at each electrode.

chemboi

What gets reduced depends on the reduction potential of the metal vs. that of water; the question probably wants us to know this instead of blindly assuming the metal will be reduced.

chemboi

That should account for all of the reactants' contribution to the total reaction order

chemboi

Unsure about the first part of your question, but as for why you reverse one of them, its to yield the overall net reaction as given in the problem.

chemboi

The connection lies in the Nernst equation: more reactant than product gives a more positive Q term, which makes it more likely for E(cell) to be positive (spontaneous)

chemboi

Transition metals can occur in different oxidation states, so especially in their case it helps to clarify. In general for any species, I think it is helpful to clarify however (not just transition metals)

chemboi

I recommend reading over "disproportionatoin reactions" from here: https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidation_State/Oxidation_States_II I don't know why this occurs, but it does :/ I th...

chemboi

Spontaneity is determined by the sign of G, where G = H - T*S. For exothermic reactions., H is negative, so it is more common for G to be negative as well (S is positive, or negative T*S term is lower in absolute value than H)

chemboi

Your final equation is balanced, and I really like how you did each step, made it easy to follow :) The e- are indeed on the correct side, because this is a reduction half reaction, not oxidation. Cr is 6+ in Cr2O7(2-) and ends up as 3+, so it has gained electrons and its oxidation number has... red...

chemboi

To clarify the good previous response, make sure you have for your final division operation's numerator: [H+] minus the initial conc. from starting pH conditions
Using [H+] alone would overestimate the dissociation of the acid.

chemboi

For hydrogen, it depends on what kind of atom the hydrogen is bonded to, could make the charge of H 1+ or 1-
Oxygen has lots of exception situations such as in peroxides.

chemboi

To add on to previous responses, if you're given an unbalanced (stiochiometrically and unbalanced charge) starting equation, it's much easier to balance the half reactions individually than the net reaction right off the bat.

chemboi

Delta S should remain zero I think. W depends on what you've got in your system :) unless you've got atoms at abs. zero, W is likely >0

chemboi

Individual tests aren't curved but the class grade distribution will be at the end of the quarter if the pre-curve average falls short of what he wants for the final average.

chemboi

I think they equal each other when there is no work being done on the system. Enthalpy is the change in heat of a system at constant pressure, and yes it is a state function vs. heat which depends on the path of the reaction.

chemboi

Whichever factor allows for more "free movement" of the atoms entails more entropy, because the atoms can exist in more states.

chemboi

if its negative, it is spontaneous (will proceed forward as written)

chemboi

at constant volume, q = delta U because no work is being done in expansion, so w is zero. q+w = delta U, so only q is left over. If someone could clarify the second part, that would be great because I'm not sure :(

chemboi

These values, especially for gases, are likely to be given. However, the previous poster makes a good point that we could be expected to derive them for a given situation if other appropriate values are known. Could show up as more of a conceptual type question.

chemboi

The subscript describes the conditions: Cp is for constant pressure, Cv is for constant volume.

chemboi

The 4 in front of the TNT molecule in the given equation indicates this.

chemboi

To expand on the previous response, Dr. Lavelle seems to be devoting lots of time to showing the related nature of the equations during the derivation process. This seems like prime material for a conceptual type question, so it may be prudent to understand the relation of the equations.

chemboi

This ties in with the formula deltaS = (q reversible) / T , so with higher T the change in S is reduced.

chemboi

Typically if combustion enthalpies are given, it implies using the product-reactant sum formula.

chemboi

We won't be expected to have these memorized. Perhaps only for water would it be expected that we know, but that's not a gas. So for your question, no :)

chemboi

Perhaps knowing the structures is advantageous for knowing how many of which bond types are broken. It seems in relevant problems that the molecules are given in an expanded format, which implies the relevant bonds.

chemboi

One important one to remember is that the w = -P(deltaV) equation only applies for constant external pressure situations.

chemboi

integrals may be used for reversible changes of volume in a gas producing reaction. when you integrate from v1 to v2 of an external pressure, you get P*deltaV

chemboi

for work, use w = -P(deltaV). Make sure to keep track of units: converting atm to pascals, and converting cm to meters, will give pascals*meters&3 which is joules. change in internal energy is w if no heat transfer occurs.

chemboi

The basis of the question gets to the heart of w = -P deltaV, so I would think it's fair game.

chemboi

While another poster may want to break it down super in depth, I think throughout interpretation of explanations you should keep in mind the differences between the heat energy of the water and its temperature. For steam, increasing temperature requires imparting more heat.

chemboi

I thought that the reasoning for formic acid being slightly weaker is the CH3 group in acetic acid does have that inductive behavior, but to the effect of donating electrons and decreasing the polarization of the total molecule.

chemboi

The given value of K = 41 has only 2 sig figs, so by the sig fig rules your answer must only have 2.

chemboi

Preserve decimal points during intermediate steps, and only cut off decimals as part of adhering to sig fig rules for your final answer.

chemboi

Here, Cl- and NO3- anions are the conjugate bases of strong acids. Thus, they act as exceptionally weak bases in solution, and can be considered practically neutral ions in solution. Any change in pH will not come from these anions, but may come from the metal cations. We see that Al3+ and Cu2+ are ...

chemboi

Both are valid, cyanido is in correspondence with new IUPAC naming conventions. Textbook uses this new convention, but in lecture Dr. Lavelle has used the old system. I guess whichever you choose to use, stay consistent with it.

chemboi

Depends on the strengths of the contributing acid, in this case citric vs. acetic acid.

chemboi

Some molecules like HCl remain "intact" in other states liek gas phase, but in water (aqueous), they dissociate. It's an important distinction to keep track of.

chemboi

I believe that the hybridization of any region of electron density around the central atom can be described in terms of the central atom's hybridization. Oxygen is sp3 hybridized, so all bonds and lone pairs are also sp3 (lone pairs count as a region of electron density).

chemboi

I believe that electron promotion is not necessary for hybridization. For example, it does not occur in NH3, which has hybridization.

chemboi

My interpretation is the AXnEm form of central atom, # surrounding atoms, and # lone pairs. If anyone else knows for sure, that would be greatly appreciated!

chemboi

I found this resource (and teh associated part 2 page) very helpful for oxidation states: https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Oxidation_State Knowing oxidation states lets you figure out the net cha...

chemboi

If atoms with the lone pairs are separated by two "spacer" atoms and can rotate, or are in the same plane, then that is a configuration conducive to polydentate behavior.

chemboi

I think that the extent of naming we have covered would be sufficient to justify a question that asks us to write out the formula of a compound such as EDTA given the expanded name.

chemboi

I don't think the method to calculate such values has been covered explicitly so far in lecture

chemboi

I don't think we have learned how to calculate actual values. We should know relative values, however.

chemboi

The lewis structures should not look the same. trigonal planar lewis structure will have a central atom with 3 regions of electron density, whereas trigonal pyramidal has 4 regions: 3 bonds and 1 lone pair

chemboi

The nitrogen is sp2 hybridized, with trigonal planar geometry which has bond angles of 120 degrees.

chemboi

NOF bound to H

chemboi

Bond angles of certain configurations you must memorize, ie 90, 120, 109.5 and as lone pairs come into play, you should be able to reason that the electron repulsion from the lone pairs would cause the bond angles to decrease slightly relative to their normal angle.

chemboi

I found this website very helpful in pointing out their differences: https://www.majordifferences.com/2014/1 ... -bond.html

chemboi

I found this website helpful: https://www.majordifferences.com/2014/1 ... -bond.html

chemboi

XeF2 is nonpolar. The three lone pairs are in the equatorial plane, while the two F form the :poles" at 180 degrees from each other.

chemboi

They are experimentally determined.

chemboi

To elaborate on the previous response, a valid lewis structure for NH3 would have the 3 H drawn perpendicularly to eachother.

chemboi

That follows the general convention of cutoffs of 0.4 and 1.7 electronegativity differences to determine if a bond is nonpolar (0-0.4), polar (0.4-1.7), or ionic (1.7+)

chemboi

Presence of asymmetric lone pairs, and asymmetric distribution of atoms with electronegativity differences are indicators of polarity

chemboi

Also keep in mind which atoms must obey the octet rule, and which can support an expanded octet. These considerations will help guide you to a valid structure.

chemboi

To elaborate on step 2, the central atoms is usually the one with the lowest ionization energy. Additionally, keep in mind situations in which an expanded octet is valid.

chemboi

What is the reasoning behind only having one double bond, instead of two or three?

chemboi

To clarify the previous response, many covalent compounds are polar. For example, H2O and NH3 possess covalent bonds but are polar and thus soluble in water.

chemboi

To expand on the first response, we refer to bonds as being on a spectrum between covalent and ionic character because a given bond may not be perfectly covalent nor ionic; while a bond with an electronegativity difference of 0.1 has high covalent character, it nonetheless is not a perfectly equal s...

chemboi

To elaborate on the previous answer, it is useful and clear to think of the strengths of dipoles as proportional to the electronegativity differences of involved atoms.

chemboi

In my searching I haven't found much more than a description of the duration as "an instant". Probably an arbitrarily brief moment of time that does not bear relevance for topics covered in this class.

chemboi

As this (https://www.quora.com/Why-does-titanium-have-4-valence-electrons-if-it-is-in-group-3) answer to a similar question explains, even though the 3d electrons are not in the n=4 level, their energy is so close to that of the 4s electrons that all are considered valence electrons that can partici...

chemboi

Valence electrons includes all e- not part of the "noble gas core", so you count the 5 ei in the d state in addition to the 2 e- in the s state.

chemboi

So far it seems that questions deal with first ionization energy. Perhaps it is good to keep in mind that the second IE will always be greater than the first. Also, the difference between 1st IE and second IE may be drastically different between elements: the difference is much more pronounced in Al...

chemboi

It's a known value, likely to be provided on a test.

chemboi

I think the spirit of this questions is enforcing the concept that the Balmer series is visible spectrum, and Lyman is UV.

chemboi

I don't believe we are expected to know the equations that are the wave functions. Worth checking with your TA though.

chemboi

The half filling or ful filling (5 or 10 electrons) of 3d orbital is a lower energy state for the electrons to be in, with 4s having 1 electron, than 4s2 and 3d4 or 3d9. see here for explanation: https://chemistry.stackexchange.com/questions/53246/why-are-full-and-half-filled-orbitals-the-most-stabl...

chemboi

I don't think its possible to pin down individual elements if you are only given n, l, m. Rather, these describe orbitals in elements of that n

chemboi

I think Dr. Lavelle said that it's arbitrary. Worth clarifying with him or your TA. Also, wouldn't that number of electrons correspond to oxygen?

chemboi

It's a weird exception which intuition cannot help you out with very much as far as I know. Some rare earth metals are similarly bizarre. See this article for the in depth explanation: http://www.madsci.org/posts/archives/2000-02/951518136.Ch.r.html Points 3 and 4 explain tungsten: "As you go d...

chemboi

In the cases of these atoms, a lower energy configuration is for each of the five 3d orbitals to be half-filled (Cr) or filled (Cu) and the 4s to have 1 electron rather than 2. I believe these energy level differences are quite small, so it is difficult to have an intuition for these exceptions beyo...

chemboi

I believe that single electrons parallel spins occupying empty orbitals is a lower energy (more favorable) than pairing first. I'm not 100% sure why pairing-first filling is not favorable, but I think it has to do with the involved electron-electron repulsion in such a scenario.

chemboi

I believe the setup for the experiment was using silver atoms, passed through a magnetic field. These neutral silver atoms have an unpaired electron, in the 5s state I believe. If there was no spin on the electrons to interact with the magnetic field, one would expect a majority of the silver atoms ...

chemboi

To add on to the first response, the Bohr model works for atoms or ions with only one electron, so hydrogen or a rare C 5+ ion or He+ ion.

chemboi

To add on to the previous response, as far as theoretical applications, the photoelectric experiment also indicated that the wave model of light alone could not explain all observed behaviors (such as the photoelectric effect).

chemboi

chemboi

Im not sure what the greatest observed wavelength is, but https://en.wikipedia.org/wiki/Extremely_low_frequency lists some EM waves as having wavelengths in the ballpark of 100,000 km!

chemboi

To elaborate on the previous answer, the differential wavelength absorption of different elements leads to different atomic spectra (that rainbow spectrum looking rectangle with black lines) which is what molecular spectroscopy deals with

chemboi

From http://chemistry.bd.psu.edu/jircitano/sigfigs.html

"Non-zero digits are always significant.
Any zeros between two significant digits are significant.
A final zero or trailing zeros in the decimal portion ONLY are significant"

chemboi

Some calculations, such as those regarding wavelength, operate with units of meters, so watch out for those. In that case, if you are given some measure in centimeters, convert that value to meters to then use in your calculation.

chemboi

Rather than rounding in intermediate steps, preserve your decimal precision. Once you have your final answer, perhaps with many decimal places, express it according to sig fig restrictions present in the original problem.

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