CHEMISTRY COMMUNITY

Fatemah Yacoub 1F

I believe he will be sending another conformation email with all the logistics.

Fatemah Yacoub 1F

A differential rate law is the rate dependent on concentration while the integrated rate law is the concentration as a function of time.

Fatemah Yacoub 1F

I would be prepared to have to print out a worksheet from online and then scan in your answers and submit it online. There will most likely be a time frame of probably around 3-4 hours.

Fatemah Yacoub 1F

Both methods will ultimately yield the same answer, but if you are only given equilibrium constants then you would use K=k/k'

Fatemah Yacoub 1F

When plotting a second order reaction, the slope should be positive k.

Fatemah Yacoub 1F

There is no way to know for sure the intermediates and all the steps of the reaction based on a single overarching equation.

Fatemah Yacoub 1F

An adiabatic system means that q=0 which does not necessarily connect directly to entropy.

Fatemah Yacoub 1F

It depends on the coefficients, order, and whether they are reactants or products.

Fatemah Yacoub 1F

The pH of an electrode can help you find the concentration of H+, if you are trying to find E cell and need to find Q.

Fatemah Yacoub 1F

The reaction would be 2H20--> O2 +4H+ + 4e-. You would have to look at the table of reductions to figure this out.

Fatemah Yacoub 1F

I thin you switched reactants and products in Q because [H+] should. be in the numerator and 1 should be in the denominator.

Fatemah Yacoub 1F

I think there is a typo in the textbook and it is supposed to be cl2--> 2cl-, otherwise it wouldn't be a redox reaction.

Fatemah Yacoub 1F

The cell potential of Hg2^2+ +2e--> Hg is on the reduction chart and the potential is 0.79 V.

Fatemah Yacoub 1F

I think for the purpose of the class, he only is focusing on drawing a cell diagram with a salt bridge, but you should know the function and application of a porous disc.

Fatemah Yacoub 1F

In the following step they end up multiplying by 2 so I don't think it matters that much which way you attempt it. In the end it the solution is the same.

Fatemah Yacoub 1F

I think you would have to look at the reduction table to figure it out. At least that's the way I approached it.

Fatemah Yacoub 1F

Yes, I believe the test covers the second page of the thermodynamics outline and all of the electrochemistry outline

Fatemah Yacoub 1F

I do not think he will make us reproduce the derivation for an exam

Fatemah Yacoub 1F

Test 2 will cover the second page of thermodynamics as well as all of the electrochemistry outline

Fatemah Yacoub 1F

Can someone explain why n=1 in 6N.7, I thought that N should be 2 because there is a transfer of 2 electrons from H2->2H+

Fatemah Yacoub 1F

In example 6L.2 can someone explain why the Cl- only shows up in the reduction half of the equation and not the oxidation half?

Fatemah Yacoub 1F

Can someone explain the free energy depends on pressure and the equilibrium constant

Fatemah Yacoub 1F

The change in entropy is constant because the difference in entropy between the reactants and products at 2 different temperatures is the same. The entropy is different at two different temperatures, but for Van't Hoff we are using the net change go entry which is constant.

Fatemah Yacoub 1F

The difference in entropy between reactants and products at two different temperatures is the same because despite entropy changing with temperature, you are finding the overall change in entropy, so if reactants entropy goes up then products entropy would do the opposite.

Fatemah Yacoub 1F

Temperature affects entropy because at higher temperature, there are more possible states for particles increasing entropy and vice versa for lower temperatures.

Fatemah Yacoub 1F

Everything given to you in the problem is either a constant or experimentally determined, so when you rearrange the equation to solve for K, you exponentiate the G/-RT, to give you the answer.

Fatemah Yacoub 1F

Can someone explain when one should use the Van't Hoff Factor, and what it tells us?

Fatemah Yacoub 1F

For constant pressure the Cv would actually be switched to CP.

Fatemah Yacoub 1F

I do not believe that Lavelle expects us to know them or what they are used for. He used it as an example to show that an isothermal reaction has an internal energy equal to 0.

Fatemah Yacoub 1F

Pressure is not always used because it is not in the equation, but you can substitute nrt to PV through the ideal gas law.

Fatemah Yacoub 1F

Can someone explain how the cis has 12 possible orientations while the trans only has 3? I don't understand how they got these numbers.

Fatemah Yacoub 1F

Can someone explain why you have to calculate the entropy to heat up the water and then cool it back down. Why can't you just calculate going from 100 degrees to 85 degrees?

Fatemah Yacoub 1F

It is also states that the sign of entropy is always positive as long as the system is above absolute zero. This means that entropy in increase and the system is spontaneous as long as the the system is above absolute zero.

Fatemah Yacoub 1F

If you burn fuel you have a lot of heat lost, meaning the internal energy decrease, but since the energy is lost as heat it does not increase the capacity to do work.

Fatemah Yacoub 1F

I would not focus as much on how to derive the equation, rather focus conceptually on what the equation means and how to manipulate it.

Fatemah Yacoub 1F

You must assume that it is constant pressure because otherwise you a reversible system which requires another set of equations.

Fatemah Yacoub 1F

I believe this is possible, but everything would just equal 0. It is not a realistic scenario.

Fatemah Yacoub 1F

I don't believe we have to know this for the midterm.

Fatemah Yacoub 1F

This is because for reactions with solids or liquids there are no volume changes as compared to a gas that is able to expand and change its volume relatively easily.

Fatemah Yacoub 1F

This would be an isolated system because no heat is able to escape the bomb calorimeter and there would be no matter (such as water) able to escape either.

Fatemah Yacoub 1F

For method two you could assume that all the bonds of the reactants were broken and all the bonds of the products were formed, but the reactant bond enthalpies would be positive and the bond enthalpies of the products would be negative.

Fatemah Yacoub 1F

delta U indicates constant volume while delta H Indicates constant pressure

Fatemah Yacoub 1F

I was also wondering the same thing.

Fatemah Yacoub 1F

Can someone explain when the standard enthalpy of formation is equal to 0?

Fatemah Yacoub 1F

1.0x10^-14= [H3O+][OH-], from this equation you can find the concentration of either hydronium or hydroxide.

Fatemah Yacoub 1F

Changing temperature does not make it a different reaction it simply makes the reaction go more to completion which therefore changes K.

Fatemah Yacoub 1F

Decreasing H2 would cause there to be less reactants meaning the equilibrium will shift back towards the reactants in order to maintain the value of K.

Fatemah Yacoub 1F

The partial pressure and total pressure are different things, but they are related. The partial pressure is the pressure of a specific gas while the total pressure is the sum of all the partial pressures.

Fatemah Yacoub 1F

Can someone explain how the heat curve for water proves steam causes worse burns than water at the same temperature?

Fatemah Yacoub 1F

I got those answers as well!

Fatemah Yacoub 1F

No since solids and liquids are not even used to calculate K they are disregarded in Le Chateliers. Solids and liquids do not affect the equilibrium.

Fatemah Yacoub 1F

You must memorize the strong/weak acids and bases in order to figure it out in a problem.

Fatemah Yacoub 1F

For some problems you must use the quadratic formula, there is no way around it. Even when you ignore K you must do the 5% approximation check.

Fatemah Yacoub 1F

Can someone explain part C of 5I.13? What would the equilibrium constant tell us in terms of stability?

Fatemah Yacoub 1F

I think graphing is a very complex way of showing q vs k which is more visually favorable, but is not necessary to determine q vs k. I think due to time constraints it would not be reasonable to graph q vs k.

Fatemah Yacoub 1F

The activity of pure substances are actually not equal to zero they are equal to one.

Fatemah Yacoub 1F

Can someone explain what he was trying to say in todays lecture when he said that simply comparing the mole ratio would give the right answer but the wrong reasoning? What is the proper reasoning to explain change in pressure?

Fatemah Yacoub 1F

Can someone explain part C of 5G.1. The question is talking about pressure but the answer only addresses amount of reactant and product which is confusing.

Fatemah Yacoub 1F

In 5G.7 why did they write K in terms of Kc if all the products and reactants are gases? Should it be in terms of Kp?

Fatemah Yacoub 1F

I think it is referring to the conjugate seesaw in which a strong acid has a weak conjugate base and vice versa.

Fatemah Yacoub 1F

I think that since these are common acids that you should know them.

Fatemah Yacoub 1F

A Lewis acid simply includes more molecules than a Bronsted acid. They are both proper definitions for acids, but a Lewis acid is more broad and includes more molecules.

Fatemah Yacoub 1F

This is because an electronegative atom has a more stable resulting anion. In H-O-F and H-O-Cl the more EN F pulls the charge away from the O delocalizing the electrons and making the resulting anion more stable and HF a stronger acid.

Fatemah Yacoub 1F

I remember during lecture he said anything with Nitrogen is usually a weak base. So that is a good rule of thumb.

Fatemah Yacoub 1F

Another way of thinking about it is the strength of an acid can be related to electronegativity because in the case of H-O-F and H-O-Cl since F is more electronegative it pulls on the oxygen weakening the O-H bond making HF a stronger acid.

Fatemah Yacoub 1F

Since strong bases fully dissociate a helpful tip is that if Kb is given it is implying a weak base.

Fatemah Yacoub 1F

I don't think we need to know any complicated ones, but you should know the rules and the most common acids/bases.

Fatemah Yacoub 1F

Not a single one is correct to define an acid or base, but you should know the difference between them.

Fatemah Yacoub 1F

For strong acids I had to memorize them as HCl, HBr, HI, HCLO3, HCLO4, HNO3, and H2SO4, but I don't know if Lavelle wants us to know any others.

Fatemah Yacoub 1F

The strong bases are any group I or II oxides and hydroxides and an indication of a weak base is basically anything with N in it.

Fatemah Yacoub 1F

It might refer to where the ligand makes a coordination bond with the central atom.

Fatemah Yacoub 1F

I think you would have to find the oxidation number on the transition metal first and then from there you can determine how many coordination bonds that can form.

Fatemah Yacoub 1F

Can someone explain how cisplatin stops cell division?

Fatemah Yacoub 1F

In class when he was mentioning that energy was increasing he was referring to the original orbitals of 2s and 2p. He drew the arrow of increasing energy next to these orbitals and then put the hybridized orbital in the middle of the 2s and 2p because a hybridized orbital has energy somewhere betwee...

Fatemah Yacoub 1F

As long as you state there is one sigma and 2 pi bonds it doesn't matter how you assign them.

Fatemah Yacoub 1F

A sigma bond almost always forms first so automatically a multiple bond will have one sigma bond. Since pi bonds overlap side by side that is why they form on multiple bonds. Therefore any multiple bond will have one sigma bond and then either 1 or 2 pi bonds.

Fatemah Yacoub 1F

Yes, I believe we are responsible for knowing all of the shapes for the test.

Fatemah Yacoub 1F

It is based on a case by case basis and what shape the lone pairs are affecting. You just have to visualize the effect of the lone pair based on the VSEPR shape you have.

Fatemah Yacoub 1F

It's helpful remember that the polarity should be based off of the VSEPR model and not the Lewis Structure. Since the VSEPR model is the actual shape and often looks a lot different than the lewis structure, the polarity would be based off of the VSEPR model.

Fatemah Yacoub 1F

Sometimes symmetry alludes to polarity but since most of the time the lewis structure isn't the actual representation you have to base it off of the VSEPR model. A symmetric Lewis structure does not automatically mean the molecule is non polar because the VSEPR model is the actual shape of the molec...

Fatemah Yacoub 1F

I think it should be fine either way because bent and angular refer to the same shape.

Fatemah Yacoub 1F

For the purposes of this class they are the same, but in induced dipole moments the molecule that induces the dipole moments has a permanent dipole moment while the same does not occur in London Dispersion Forces.

Fatemah Yacoub 1F

With greater polarizability, the electron cloud is able to be distorted more. Therefore it is easier to form IMFs because since the electron cloud can easily be distorted it is more attracted to partial positive charges.

Fatemah Yacoub 1F

In addition, in dipole-induced-dipole interaction the molecule that induces the dipole moment has a permanent dipole moment while it does not in London Dispersion.

Fatemah Yacoub 1F

It mainly pertains to electrostatic interaction which is the interaction between objects having electrical charges. These forces are what give rise to IMFs.

Fatemah Yacoub 1F

How do you determine if a bond is polar covalent or ionic without looking at ionization energies?

Fatemah Yacoub 1F

There are several transition metals that break the trend in ionization energy, but it depends on a case by case basis. I do not think we are going to be responsible for knowing those.

Fatemah Yacoub 1F

An easy way to remember the trends is electronegativity, ionization energy, and electron affinity all have the same periodic trend.

Fatemah Yacoub 1F

If the atom is in a n=3 shell or higher and has an empty d orbital (elements in the p block) it will be able to expand its octet.

Fatemah Yacoub 1F

The midterm covers all the material up to the end of Outline 3: Chemical Bonds.

Fatemah Yacoub 1F

When computing the formal charges for SO4^2- in class he said that by adding more bonds and making all the oxygens have a 0 formal charge it would make sulfur have a 2- charge and we would rather have the sulfur have a 0 charge than the oxygens? Why would the compound be more stable with oxygen havi...

Fatemah Yacoub 1F

I think you have to draw out all of the possible resonance structures for a compound to portray all the possible positions for the multiple bonds.

Fatemah Yacoub 1F

I was confused as to why P,S, and Cl have an expanded octet if they have no electrons in the d orbital?

Fatemah Yacoub 1F

A resonance structure is specifically related towards elements with double and triple bonds. A resonance structure is basically all of the variations of having a double and triple bond. A Lewis Structure is the basic structure of a compound and while it may contain a double or triple bond it does no...

Fatemah Yacoub 1F

One way to look at it would be that the ground state of an electron would be the normal electron configuration. If the electron configuration is off in any way such as electrons being paired while the orbitals are not yet completely filled, then the electron is considered to be in an excited state.

Fatemah Yacoub 1F

Yes, the second ionization is always higher regardless of whether the electron would gain a full octet or not. The second ionization is always higher because it takes more energy to remove an electron positively charged ion( since one electron has already been removed) rather than a neutral atom.

Fatemah Yacoub 1F

When simply drawing out the configurations once all the oribtals have atleast one electron( at this point they are all parallel to one another) then when more electrons are added, the electrons in the same orbital begin to get paired. All of the orbitals must be filled by at least one electron befor...

Fatemah Yacoub 1F

In addition if electrons only acted like waves then increasing the intensity of light would yield more electron ejection, but that was not the case.

Fatemah Yacoub 1F

Calculating spin state is not necessary as it will always be either +1/2 or -1/2. The positives and negatives simply represent whether the electron is spinning clockwise or counterclockwise

Fatemah Yacoub 1F

S orbitals have no nodal planes simply due to their shape. S orbitals are symmetrical and basically look like a sphere. For this reason they have no nodal planes.

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