CHEMISTRY COMMUNITY

Kevin Liu 1J

Lone pairs are not necessarily all that is considered. As said, formal charge should be zero or whatever the original charge on the molecule is.

Kevin Liu 1J

Kind of off this, but on the final were the questions asking to identify the type of compound referring to identification of amphoteric or amphiprotic?

Kevin Liu 1J

Based on the number of electrons bonded, the cyanide should have a triple bond between carbon and nitrogen along with a lone pair on each atom.

Kevin Liu 1J

It was more just based on relative ideas on pH. Yes, it said titration but just knowing which one was the stronger acid/base could help you determine if the pH was greater than, less than, or equal to 7.

Kevin Liu 1J

The marshmallow review had a similar question. The main thing was that the nitrogen in cyanide is attached as the heme complex to the central iron.

Kevin Liu 1J

H+ concentration is the key to differentiating acids and bases as this provides the basis to determining the pH of the substance.
pH < 7 is acidic
pH > 7 is basic

Kevin Liu 1J

The organic chemistry tutor is good ^ as well as khan academy videos on the subject. Chemistry libre texts is also a good online resource too.

Kevin Liu 1J

Chapter 6 covers everything with regards to acids and bases.

Kevin Liu 1J

In terms of bonds, the simples way to remember sigma and pi bonds is to differentiate by bond length:
single bond = 1 sigma bond
double bond = 1 sigma bond, 1 pi bond
triple bond = 1 sigma bond, 2 pi bonds

Kevin Liu 1J

At least one tests, applying cis and trans should be either given in the question itself if drawing the structure or identifiable based on the structures given.

Kevin Liu 1J

Since hydrogen bonding requries partial positive/negative bonding between Hydrogen (partial positive) and N, O, F (partial negative, it is usually polar.

Kevin Liu 1J

I'm not too sure, but single bonds have lower melting points than like triple bonds that have higher melting point.

Kevin Liu 1J

I'm not too sure what the connection is between these and hybridization. The most direct connection hybridization has is with molecular shape/structure.

Kevin Liu 1J

Hybridization does represent the blend of the resonance structures and refers to their molecular shapes (i.e. sp, sp2, sp3, etc.)

Kevin Liu 1J

sigma and pi bonds are simply labeled based on the bond type (single = 1 sigma, double = 1 sigma, 1 pi, triple = 1 sigma, 2 pi)

Kevin Liu 1J

I'm pretty sure we just need to know the conceptual basis behind sigma and pi bonds, not necessarily the hybridization.

Kevin Liu 1J

Sigma and pi bonds both involve the overlap of atomic orbitals in different ways, but to put it simply, pi bonds are weaker due to the fact that there is less overlap present normally than there is in sigma bonds.

Kevin Liu 1J

We should not need to based on our current knowledge of hybridization.

Kevin Liu 1J

Hybridization should not be included, but we will be expected to know all the various molecular shapes (linear, tetrahedral, etc.) along with bond angles that do relate to the hybridization

Kevin Liu 1J

Hybridized orbitals are basically a mix (or hybrid so to speak) of atomic orbitals.

Kevin Liu 1J

I'm not too sure what you mean, but we are sometimes asked to indicate partial positive and negative charges, which is an indication of whether or not the bond is polar.

Kevin Liu 1J

It's because oxygen actually has 8 protons in its nucleus compared to nitrogen's 7, making oxygen have a greater attractive force towards a bonding pair.

Kevin Liu 1J

Usually resonance structures are differentiated by a difference in bonds (single, double, triple) that can be rotated among the similar atoms to form the resonance structures.

Kevin Liu 1J

I think we need to write it all out when it explicitly says to write out all resonance. Otherwise bracketed should be fine.

Kevin Liu 1J

I believe so. That is what most of us were told to do in reference to formal charges

Kevin Liu 1J

Expanded octet often is restrained to those with odd number of valence electrons

Kevin Liu 1J

Noble gases are not included in the trend for electronegativity.

Kevin Liu 1J

I think we just need to know polarizability in terms of its relation to anions and polarizing power; basically periodic trends. Hope this helps

Kevin Liu 1J

A more symmetrical distribution contributes to the overall stability of the molecule, indicating why formal charges closer to zero are more stable.

Kevin Liu 1J

Formal charge is represented by the formula: FC = Ve- - (bonded + s/2) and is used for determining the most plausible resonance structures among other purposes.

Kevin Liu 1J

Both ionization energy and electronegativity are related with ionization energy referring to how much energy is needed to remove an electron, while electronegativity refers to the relative strength of the pull of shared electrons unequally within a bond.

Kevin Liu 1J

Ionic bonds are where electrons are fully transferred from one atom to another, while covalent bonds are strictly shared between atoms.

Kevin Liu 1J

Particles: λ = h / (mv)

Light: c = λ v ---> E = hc/λ (derived from E = hv)

Kevin Liu 1J

Usually, I just use 2.998 x 10^8 to improve accuracy and just round off for significant figures only after the final calculation.

Kevin Liu 1J

p is momentum

m is mass in kg

v is velocity in m/s

p is measured in kg * (m/s)

Kevin Liu 1J

speed of light is 2.998 x 10^8 m/s and this is utilized as waves are part of the light spectrum. Hope this helps!

Kevin Liu 1J

c = lambda (v)

E = hv ---> E = hc / lambda

Hope this helps!

Kevin Liu 1J

Hertz refers to s^-1. This is justified by the fact that c = v λ; c is in units of m/s, λ is in units of m, therefore leaving v in units of s^-1.

Kevin Liu 1J

The location of impact does not affect the number of photons, rather the energy of the photons (blue/violet in comparison to red/green). Hope this helps!

Kevin Liu 1J

For the most part, it shouldn't be difficult to differentiate between which equation to use as you determine which value you are given in the problem then utilize the respective equation (either E = hv or E = pc). Hope this helps!

Kevin Liu 1J

Not too sure, but our TA said to know at the very least the values that divide the ultraviolet region, visible light region, and infrared region since these are the most commonly utilized in calculations.

Kevin Liu 1J

I'm not too sure what you mean by p-block metals, since p orbitals normally refer to the gaseous elements, but the s-block metals normally are reactive (become cations) in the sense that they are willing to bond with p-block elements to form a complete octet. Hope this helps!

Kevin Liu 1J

There is a direct relationship between these two aspects that causes one to increase/decrease in relation to the corresponding increase/decrease of the other factor.

Kevin Liu 1J

s, p, d, and f refer to specific types of orbitals that are found in elements. So these numbers simply represent the maximum amount of electrons that can occupy a space. Hope this helps!

Kevin Liu 1J

This is basically the same, but simply put, it is the same equation but is utilized based on the values given. If the wavelength is given, E = hc / lambda is ued. If the velocity is given, E = hv is used. Hope this helps!

Kevin Liu 1J

In terms of periodic trends, uhedlund described it pretty well. Just to be more specific, the trend displays that electronegativity increases from the bottom left corner of the periodic table to the upper right hand corner. This means that elements like Fluorine and Chlorine have an extremely high e...

Kevin Liu 1J

I'm not too sure, but x-ray crystallography is the utilization of a crystalline structure to diffract x-rays in a beam and these help determine the three-dimensional structure. Hope this helps!

Kevin Liu 1J

You need to include the answer achieved in part a in the part b calculation.

So, 1 / (1.2 x 10^-14 mol) represents the time in seconds.

All that needs to be done after is a conversion to time in years. Hope this helps!

Kevin Liu 1J

I think for the sake of not complicating it, the book's intention is to just draw a distinction between those mixtures strictly including gases and those strictly including liquids. Gaseous solution seems contradictory in nature like Priscilla said.

Kevin Liu 1J

So, we are given that the final moar concentration is 0.10 M and that you will have a volume of 10 mL diluted solution. We need to first calculate the number of molecules present in such a solution: 10 mL (0.10 mol / 1000 mL) (6.022 x 10^23 molecules / 1 mol) = 6.0 x 10^20 mlcls From here, all you n...

Kevin Liu 1J

I would balance the equation first: Ca Mg(Co3)2 --> MgO + CaO + 2 CO2 Then find the amount of moles of CO2 using the mass (10.04 g - 5.12 g = 4.92 g). Use the moles of CO2 to calculate the moles of MgCO3. Finally, use the moles of MgCO3 to calculate the mass of MgCO3 that remains after the decomposi...

Kevin Liu 1J

In terms of rounding, I usually write down a rounded percentage for the step, but keep the original unrounded number in my calculator. Then using these original calculated values in the final calculation, the results seem to be more accurate and add up to a value much closer to 100%.

Kevin Liu 1J

I believe once you figure out which reagent is the limiting reagent, you must use the whole amount given in the problem for the limiting reagent calculation. So yes, the theoretical yield would be the same as the amount of the limiting reagent.

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