CHEMISTRY COMMUNITY

Rajshree 1F

rate constant is specific to the given temperature

Rajshree 1F

a is the stoichiometric coefficient in front of the reactant a. we were given the reaction A --> P therefore a = 1

Rajshree 1F

it tells u the relationship between rate and concentration. rate = k[A]^n. if its a zero order reaction, then the concentration of reactant does not affect the rate. a first order reaction (rate = k[A]^1) means that the concentration of reactant and the rate are directly related. if u double the con...

Rajshree 1F

0.693 is the ln(2)

Rajshree 1F

given the concentration of ur reactant at different time intervals, u can create a graph. if u have a straight line when u plot [reactant] vs time u have a zero order reactant. if u have a straight line when you plot ln[reactant] vs time u have a first order reactant. if u have a straight line when ...

Rajshree 1F

OIL RIG is the one I use. Oxidation Is Losing and Reducing Is Gaining

Rajshree 1F

water is added to half reactions where the oxygens are not balanced.

Rajshree 1F

i think for this class, we have learned that cathodes are typically drawn on the right. regardless you can tell which side is the cathode or anode knowing the reduction potentials of the half reactions.

Rajshree 1F

i try to remember these terms as actual agents. so an oxidizing agent is tasked with oxidizing another molecule and in the process, it gets reduced. a reducing agent is tasked with reducing another molecule and in the process, it gets oxidized. when given an equation, you should first identify the o...

Rajshree 1F

When trying to balance out the H and O, an acidic equation should only have H+ and H2O while basic solutions should only have OH- and H2O.

Rajshree 1F

the integral is a way to derive the equation you do not need to use the integral to solce and i think when it comes to the work equation, it is only applicable during reversible expansion.

Rajshree 1F

they are the same equation once factor out -1 from the brackets in the textbook equation

Rajshree 1F

Use Kelvin because the R value also uses Kelvin

Rajshree 1F

"i" represents the current aka the electron flow therefore as i approaches 0, there is max difference in potential (since no electrons are flowing)

Rajshree 1F

salt bridges allow the flow of spectator ions that affect neutrality but not the redox reactions. neutrality is important for redox reactions to continuously occur.

Rajshree 1F

pertaining to this class, i think you can use them interchangeably

Rajshree 1F

you would use the external pressure for that equation because the equation solves for the amount of work needed to be done on the surroundings given the external pressure.

Rajshree 1F

the integral is a step in deriving in the final reversible equation for work. understanding the concept behind it should be sufficient for this class i do not think we need to calculate the integrals for any work related problem. we can instead use w = -nRTlnV2/V1

Rajshree 1F

i would make a list of equations and understand when what equation is used (understanding the units can be a good first step for doing so).

Rajshree 1F

when speaking of work specifically, yes if the system is being compressed then the work value will be positive and if the system is being expanded then the work value will be negative. however when discussing internal energy, the presence and exchange of heat can change sign of the final answer.

Rajshree 1F

the problem will specify if the problem is Cp or Cv

Rajshree 1F

compression = work is being done ON the system therefore it is positive
expansion = work is being done BY the system therefore it is negative

Rajshree 1F

what is the difference between temperature and enthalpy and where does heat fit into these definitions? thanks!

Rajshree 1F

N represents the molecules in the system!

Rajshree 1F

for this question, you would use q = n*Cv*deltaT. while n is given and deltaT can be easily solved for, the Cv value you would use is based upon the molecule Xenon is. since it is an ideal gas and it not bonded to anything, you would use 3/2 R. hope this helps!

Rajshree 1F

it seems you are making a simple error when u add 346 to 6265. it should equal 6611 which is then set up to 6611 = c/(mass*deltaT). hope this helps

Rajshree 1F

you use w = -nRTln(Vf-Vi).

Rajshree 1F

i think it doesnt incorporate the C-H bonds because in both structures those bonds have not changed. the question asks about the difference in C-C and C-.C bond enthalpy

Rajshree 1F

yes, you would add the enthalpy of sublimation or vaporization (depending on if its a solid or liquid) to ur final answer.

Rajshree 1F

H2SO4 is a strong acid so your original thought process is correct however it is also a polyprotic acid meaning it can deprotonate twice. you regard hso4- as a weak acid and solve the ice box as you typically would except you would add x to the 0.15M (the amount of hydronium ions formed from the dis...

Rajshree 1F

the first deprotonation of H2SO4 disassociates completely making it a strong acid but the second deprotonation should be regarded as a weak acid and the concentration of hydronium ions or SO4(2-) should still be calculated using the ICE box. After calculating x, add that to the 0.15 and find the pH.

Rajshree 1F

endothermic reactions use heat as a reactant therefore if u add a reactant, the product increases which raises the K value i believe

Rajshree 1F

noble gases have no effect because they do not react with the reactants or products

Rajshree 1F

with most of these problems, there will be typically be one reactant or product that starts with an initial value of 0. knowing that, the side with the zero should have +x and the other side should have -x. if all initial concentrations are given then Q can be calculated and compared to K to see whe...

Rajshree 1F

because NH3 is also a product as is H2S, both of them is 0.2+x. it is misleading because you think since you are starting with NH3, the reaction should lose the concentration but that is not necessarily the case. we still want to make more H2S which means more NH3 as well.

Rajshree 1F

p in a name in chemistry typically means the negative log of the value so pKa is the negative log of Ka.

Rajshree 1F

the equilibrium constant is the ratio of products to reactants in any reaction and can help us determine different concentrations of products and reactants formed when given limited information because despite the change concentration or pressure, Kc does not change.

Rajshree 1F

i am already feeling a little lost after this lecture, could anyone explain to me the difference between enthalpy and temperature and why would steam cause severe burns compared to water at the same temperature? thank u!

Rajshree 1F

for acids you calculate %ionization because it donates a proton. for bases u calculate %protonation because it accepts a proton. the math for both is almost identical but depending on whether u are solving for an acid or a base, the term changes.

Rajshree 1F

Kc is the ratio of products and reactants when the system is at equilibrium. Q represents a ratio of a system not yet at equilibrium

Rajshree 1F

kelvin i am pretty sure

Rajshree 1F

During lecture 6, Lavelle talks about how a weak acid could have a [h3o+] of __ x 10^-10 but in that case we should take into consideration the autoprotolysis of water and add that conc to 1.0 x 10^-7. i was wondering why we didn't do that for the CH3COOH example because the pH came to be 2.89 which...

Rajshree 1F

dividing by 0.5 is the same as multiplying by 2. your work is still correct, the textbook just did it differently but the answers should be the same.

Rajshree 1F

if the product formed is H3O+ then Ka should be used and if OH- is formed then Kb should be used.

Rajshree 1F

part d talks about the equilibrium concentrations not the equilibrium constant. therefore though the number of products does increase, the ratio stays the same.

Rajshree 1F

Someone asked this question in the module and I remember Lavelle said we are better off doing the math because it would be more accurate.

Rajshree 1F

yes you are correct it is added to the ratio because it is in the gaseous state

Rajshree 1F

yes, Q is the ratio of [P]/[R] when the system is not at equilibrium (K)

Rajshree 1F

concentration = n/V and since V is constant in this equation but n is different, you would get different values for the pressure of the gases (P = (n/V)*RT).

Rajshree 1F

for c, only temperature affects the equilibrium constant. for d, the concentration of reactants would lead to an increase in products but the equilibrium constant (K) would stay the same. for example whether u have Kc = 5/1 or 10/2 it still equals 5. hope this helps.

Rajshree 1F

you would use the ideal gas law. in lecture lavelle manipulates PV = nRT into P = n/V*RT and says n/V is the equation for molarity (concentration) and simplifies it to P = conc*RT. depending on whether you are given Kp or Kc you can plug it in and since R is a constant and temperature is always give...

Rajshree 1F

i believe with the way the equatorial axis is arranged, the electron repulsions will cancel out and the shape would be linear with a bond angle of 180

Rajshree 1F

The most important structure is the one with the lowest FC. For that specific structure, while there might be many resonance structures, there is one structure that has a FC of 0 and is therefore the most important one in order to derive the hybridizations of each atom.

Rajshree 1F

I think you would place the lone pairs on the equatorial axis because the bonds would be further away from the lone pair vs if you were to put it on the axial axis. One LP would give a seesaw shape, 2 LP would give a T-shape, and 3 LP would give you linear (only the axial axis having bonds)

Rajshree 1F

Hi! I think carbonato and sulfito generally act as monodentates (in rare cases, however, they may act as bidentates). I believe the only polydentates you really need to know are oxalato (a bidentate), EDTA (a hexadentate), and en (a bidentate). Hope this helps! I think we also need to know that die...

Rajshree 1F

The latter half of the list is very similar to the names of the polygons with the same number so it shouldn't be hard to memorize if u make that association. However, in our exercises we have not seen more than hexa- so I think knowing until at least hexa is crucial.

Rajshree 1F

I think knowing polyatomic ions of strong acids and bases as well as for the ligands for coordination compounds.

Rajshree 1F

I believe HFO is a weak acid because of how electronegative it is, it initiates other chemical reactions in water and with itself. I do not think we need to know this for the course

Rajshree 1F

I think bond length and bond strength are inversely related so I think those are the two factors to consider. There is also stability of the anion that can affect acid strength when bond length is not enough to conclude the strength of the acid.

Rajshree 1F

We learned that metal oxides turn into bases and since neither Cl, Br, or I are metals, they would not form bases but instead acids,

Rajshree 1F

Your body will typically tell you if something is damaging your skin so if the chip residue does not cause any stinging I am sure your hands are fine

Rajshree 1F

I don't think hexagonal planar or pentagonal pyramidal are very common geometric complexes therefore I believe it is only octahedral.

Rajshree 1F

I believe so, it is a trigonal bipyramidal with two of the equatorial bonds replaced with lone pairs.

Rajshree 1F

I think the way the formula is written gives a hint to how the molecule is structured since there is H2 at the front of the C, you can infer both hydrogens are bonded to the first C and the same can be done with the last C and H2. From there you complete the octet rule of the middle carbons. This is...

Rajshree 1F

Structures that have resonance hybrids because the electrons are delocalized therefore so is the pi bond.

Rajshree 1F

Yes the ring would only contain carbons because hydrogens can't form two bonds which is needed to make a ring.

Rajshree 1F

I am not sure either but it definitely has to do with pi bonds. Also why does an even number of internal carbons mean the hydrogens are in the same plane?

Rajshree 1F

Yes I believe sp3d hybridized orbitals are a thing and since Lavelle has not brought up exceptions yet, I don't think there are any.

Rajshree 1F

Because the Cl are on opposite sides of the carbons, they cancel each other out because both dipole moments are opposite in direction and the same in magnitude.

Rajshree 1F

You would just put AXE without the ½ subscript

Rajshree 1F

For VSEPR, I think we only take into account the lone pairs on the central atom therefore HCN would be linear.

Rajshree 1F

If it isn't in the outline for this section, I don't think we will need to know it. If it is apart of the textbook readings then to be safe I would just understand the concept of it or what it means.

Rajshree 1F

I think of it as connecting a linear molecule with a trigonal planar molecule. the bonds that create a 180 are the axial and the bonds that create a 120 are equatorial. I find staring at the models Lavelle holds up during lecture or finding a website that gives a 3D simulation is the best way to vis...

Rajshree 1F

You can't draw SO₃^-2 without the lone pair because the molecule has a charge of -2. only the sulfur atom can have an expanded octet therefore the shape Lavelle gave the molecule during lecture is the only correct one.

Rajshree 1F

I think the key concept being tested through this question is that when removing electrons from a transition metal, the electrons in the outermost s subshell go first. Hope this helps!

Rajshree 1F

Yes you are exactly right. It is easier to lose 3 electrons than gain 5. I think Group 14 and transition metals are where things can get tricky but for that question, it is a rule of thumb that Ga would be become a cation.

Rajshree 1F

Since the midterm is multiple choice I would suspect we would need to choose between some Lewis structures given to us. However, the structure with the least amount of atoms with a non-zero formal charge is usually the most accurate Lewis structure.

Rajshree 1F

BeCl₂ is a Lewis acid because the central electron can still accept electrons because of its Lewis structure. Be is bonded to Cl on both sides but there are still empty spaces where Be can accept electrons.

Rajshree 1F

the sub-shells are the variable l or the angular momentum quantum number aka s, p, d, or f. the orbitals are the ml or the magnetic quantum number so for s, there is one orbital for p, there are 3 orbitals, etc.

Rajshree 1F

I think the textbook just gave the numbers for context, we only need to know trends for this course, not actual values of each element. Hope this helps!

Rajshree 1F

covalent molecules can have ionic character - ionic being that there are charged areas of the molecules due to one atom holding onto the shared electrons more closely. therefore like ions, the molecule has delta positive and delta negative portions. salts can be covalent character when the anion is ...

Rajshree 1F

i think it has to do with the graph Lavelle showed in Week 5 Friday lecture where some different molecules have different ionic or covalent characters based on the anion distortion and attraction to the cation

Rajshree 1F

I always understood the octet rule where the atom is most stable in that form. However, after learning about d-orbitals and exceptions to the octet rules, do these atoms have a different number that is the most stable or does it work differently?

Rajshree 1F

the amount distortion is based upon the size of the anion. the bigger the atom and therefore the number of electrons it has, the more distorted it will be when attracted to a cation.

Rajshree 1F

I think the equation is a fool proof way of finding the formal charge of each atom and therefore the overall charge of the molecule.

Rajshree 1F

I think the best way to remember which elements are metalloids are by memorizing them. Since there are only seven of them on our periodic table, I don't think it would be too hard to memorize them with a mnemonic such as: Boring Silly Goats Are At The Party (Boron, Silicon, Germanium, Arsenic, Anti...

Rajshree 1F

Lavelle explains it really well in his lecture 11 diagram with each n level being the average of their orbital energy levels. 3d and 4s are so close that 4s sometimes can have a lower energy than 3d. for elements after atomic number 20, 4s is a higher state than 3d.

Rajshree 1F

i think the electron affinity equation stays in the same format no matter energy getting released or absorbed but i may be wrong. since Lavelle didn't mention the switching of the two values I am going to assume it stays the same.

Rajshree 1F

i think it has to do with the structures of the p orbitals in the carbon atom. hopefully this picture will help better visualize.

Rajshree 1F

based on the n, you can know what l (orbitals) are possible. for example n = 3 means l values can be 0, 1, and (n=1) aka 2. l = 0 is the s-orbital. l=1 is the p orbital. l=2 is the d orbital. depending on which element in the level you are working with, you can deduce which l value it has based on w...

Rajshree 1F

I think we are supposed to know the electromagnetic spectrum aka different waves and their ranges. Lyman series are ultraviolet waves being emitted and Balmer waves are visible light. By knowing the spectrum range, you should have enough information to tackle any question on the exam.

Rajshree 1F

Sometimes the chemical reactions coefficients can get pretty big and I feel really lost on how big is too big. Any tips or tricks as to not get carried away making me coefficients too big?

Rajshree 1F

I think there are certain rules about how to order elements in a compound that we have not learned yet. Regardless, no matter what way you write it, it is still the same compound

Rajshree 1F

Oh my gosh thank you so much!! I have been looking for some more extra practice since I am still confused a little.

Rajshree 1F

based on what you are given, both equations can be used to get to the unknown variable you are looking for

Rajshree 1F

i think it has to do with effective nuclear charge and how 2p is better shielded than an electron in 2s orbital

Rajshree 1F

Yes! The scientists hypothesized that making the light brighter would give the electrons more energy however increasing the number of photons (aka brightness) had no such effect but increasing frequency did.

Rajshree 1F

How in depth do we need to know black body radiation? Lavelle said during lecture just the concept however there is an equation attached to this topic so now I am worried.

Rajshree 1F

I don't think we will need to know that until we have covered acids and bases.

Rajshree 1F

I think the homework problems are extra practice to help prepare for the exam however I do not think we will need to know the conversion outside of the metric system. From my understanding it is more the concept and the problem solving that is the focus of these extra practice problems!

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