CHEMISTRY COMMUNITY

Anthony Hatashita 4H

When do we use Pt in a cell diagram?

Anthony Hatashita 4H

It's possible given that it would be able to come into contact with more at one time, but for this class I'm pretty sure just the stoichiometric coefficients matter.

Anthony Hatashita 4H

Half lives are useful for lots of real world applications. If asked for in a question it would ask for it explicitly and the equations should be on the equation sheet.

Anthony Hatashita 4H

Rate constants are always positive because they are based off time and concentrations which are always positive.

Anthony Hatashita 4H

I had the same question, hopefully it isn't too complicated. If it's take home I'm assuming we have to take pictures and submit that way? I'm not sure.

Anthony Hatashita 4H

He never went over it in class and the topic has an asterisk next to it meaning we don't need to know this.

Anthony Hatashita 4H

Q is anode/cathode usually. You can write out the full reaction equation to verify on a case to case basis.

Anthony Hatashita 4H

You can use either, the equation sheet lists both so whichever you are most comfortable using is best.

Anthony Hatashita 4H

Adding water will decrease the concentration of everyone else, changing the value of Q which changes cell potential.

Anthony Hatashita 4H

Q is the concentration of product divided by the concentration of reactant. [product]/[reactant]. Lots of questions will give you the concentrations and you'll be able to solve from there.

Anthony Hatashita 4H

What's the difference between these two?

Anthony Hatashita 4H

The equations should be as follows:
12OH- + Br2 → 2BrO3- + 6H2O + 10e-
5 x (Br2 + 2e- → 2Br-)
I'm not sure where you got the H from when you did your calculations, maybe you forgot that this was in a basic solution? You use H2O and OH- to balance the H on each side.

Anthony Hatashita 4H

You have to look at the charge of Mn, not the entire compound. In this case, Mn goes from +7 to +2 and is being reduced.

Anthony Hatashita 4H

I think it's supposed to be Cl2 -> Cl-. I ran into that problem as well but the answer book shows Cl- and I got the right answer doing this.

Anthony Hatashita 4H

Because it's a basic solution, in order to add more H to one side you need to add H2O to that side and OH- to the other, resulting in one more H per H2O molecule on that side while leaving the O balanced.

Anthony Hatashita 4H

I saw this and thought the same thing and I think you're right. The answer key has Cl- in it while the question itself does not.

Anthony Hatashita 4H

Anodes are more negative and cathodes are more positive, so the electrons naturally go from anode to cathode. Cathodes get reduced and anodes get oxidized.

Anthony Hatashita 4H

What is the purpose of galvanic cells? Is it just to do work on something or is there more than that?

Anthony Hatashita 4H

So the order of a cell diagram would be solid|gas|aqueous || aqueous|gas|solid. The two lines in the middle represent the salt bridge and the left side represents the anode and the right side represents the cathode. A cell bridge is just another way to represent the transfer of electrons from an an...

Anthony Hatashita 4H

I think that's the main thing that we need to know about salt bridges, but you should probably also know how to recognize a salt bridge in a cell diagram.

Anthony Hatashita 4H

When electrons move, the salt bridge helps balance the charges of the two solutions.

Anthony Hatashita 4H

ΔS is a state function, so they are the same either way. In irreversible however, ΔSsurr=0.

Anthony Hatashita 4H

Pressure and volume always have an inverse relationship, so it's P1/P2 instead of vice versa.

Anthony Hatashita 4H

For the surroundings it's 0, the system is W=-PDeltaV I believe. I'm not sure about the second part of your question though, sorry about that.

Anthony Hatashita 4H

For reversible, the reaction is able to go in the reverse direction afterwards which means work has to be done on it by the surroundings to return, or at least that's how I think of it, which means that ΔSsys=-ΔSsurr.

Anthony Hatashita 4H

For reversible, delta S of the system is equal to -delta S of the surroundings.

Anthony Hatashita 4H

You can think of this as two parts, first heating up the kettle, then heating up the water. You just need to add the two values you get for each part to get the final answer.

Anthony Hatashita 4H

The equations we need will be on the equations and constants sheet given to us for exams.

Anthony Hatashita 4H

If the system is doing work, it is losing energy, so it is negative. Think of it with yourself as the system, if you open a door you are transferring energy to the door, losing energy yourself (-).

Anthony Hatashita 4H

Considering the first law equation, work and heat would have to equal 0. For an isothermal process, heat is 0, so if there is a reaction that is isothermal that does no work, delta U should be 0.

Anthony Hatashita 4H

1 Torr = 0.001316 atm, so you can convert to atm by multiplying 750 by 0.001316, giving 0.987 atm.

Anthony Hatashita 4H

Adiabatic means there is no heat exchange, however there can already be heat within the system.

Anthony Hatashita 4H

Think of this problem as having two parts: first you heat up the copper, then the water. You add up the change in energy for both of them to find the total.

Anthony Hatashita 4H

Increasing energy in an open system can be done by adding or removing a substance, heating or cooling the system, doing work on the system or vice versa.

Anthony Hatashita 4H

With an isolated system you can still observe changes happening within without interruption or influence from outside the system.

Anthony Hatashita 4H

Isolated systems cannot exchange energy with their surroundings. Think of a thermos retaining heat, it counts as an isolated system because the energy transfer from within to outside is negligible.

Anthony Hatashita 4H

My solution manual says 28J, so I think you may have misread something.

Anthony Hatashita 4H

When I first did this problem I thought that a) would be closed since it can't be completely isolated, but it's close enough to completely isolated to count.

Anthony Hatashita 4H

Think of it as two problems, heating up the copper and then heating up the water. You add the two together to get the total.

Anthony Hatashita 4H

There can be reactions that happen within the isolated system, but no energy can leave or enter the system.

Anthony Hatashita 4H

I don't think there's a way for something to be completely isolated, but if it's close enough the difference is negligible.

Anthony Hatashita 4H

If they're the same then why do we have different names for them? Just to clarify?

Anthony Hatashita 4H

They don't affect either because they are pure and don't change in reactions.

Anthony Hatashita 4H

H+ and H30+ are interchangable for us I believe. H+ is just H30+ without the water.

Anthony Hatashita 4H

The percent protonization or deprotonization needs to be less than 5% for the x value to be valid; it makes it acceptable to use approximations.

Anthony Hatashita 4H

I think we use Kp when all of the reactants and products are gases.

Anthony Hatashita 4H

I think if you increase the amount of gas or increase the temperature it also increases temperature.

Anthony Hatashita 4H

Just make sure the coefficients for specific molecules remain on the same molecules as a power when calculating K.

Anthony Hatashita 4H

It's a constant and should be given with any exams so I don't think we need to worry too much about it.

Anthony Hatashita 4H

If we are given partial pressure but need concentration or vice versa it helps us convert between the two.

Anthony Hatashita 4H

Large K value means that the reaction shifts to the right, small K value means the opposite.

Anthony Hatashita 4H

how do you get 33b?

Anthony Hatashita 4H

The textbook says that we only use the first deprotonation equilibrium for polyprotic acids except for sulfuric acid, so do we just calculate the pH of these acids just as we would for monoprotic acids?

Anthony Hatashita 4H

I don't think this is explicitly stated anywhere but I would do alphabetical order just in case, don't wanna lose free points on the final

Anthony Hatashita 4H

Yes, for example cobalt is cobalate and copper is ferrate.

Anthony Hatashita 4H

Don't use the prefixes when listing in alphabetical order

Anthony Hatashita 4H

I think it would say if there was an overall charge, indicated by brackets

Anthony Hatashita 4H

They are interchangable and should give the same points on a test

Anthony Hatashita 4H

There's a table in the textbook that shows the most common polydendate ligands

Anthony Hatashita 4H

I wasn't sure on this one either, but I think it has to do with being polydendate? I'm not sure though.

Anthony Hatashita 4H

Cuprate for copper I believe

Anthony Hatashita 4H

It's probably a good idea to know them just in case, you never know what's gonna be on the final

Anthony Hatashita 4H

Why do we use "cyano" instead of "cyanido?" Does it have to do with the bonds on the central atom for hexacyanoferrate(II)?

Anthony Hatashita 4H

Lone pairs are equatorial usually, but for something like AX4E2 they are axial so they are farthest apart.

Anthony Hatashita 4H

They should all be linear, two atoms together can only make a straight line between.

Anthony Hatashita 4H

Dipoles have to cancel out or it will be polar. I think it helps to look for symmetry in molecules and atoms within molecules to find if something is polar.

Anthony Hatashita 4H

Does anyone have any advice about studying intermolecular forces?

Anthony Hatashita 4H

What do we need to know for test 2? Everything covered after the midterm?

Anthony Hatashita 4H

I wouldn't imagine it being any different from previous tests.

Anthony Hatashita 4H

Bond angles are something you have to memorize, and I'm pretty sure the angles are determined by electron densities, not atomic attraction, but I'm not positive there.

Anthony Hatashita 4H

If there is a lone pair on the central atom, it makes the overall shape trigonal pyramidal, if there are no lone pairs on the central atom it's planar pretty sure.

Anthony Hatashita 4H

I think the answer is 1 because the bonds are the same angle, but I'm not sure.

Anthony Hatashita 4H

When drawing the lewis structure, make sure to remember formal charges!

Anthony Hatashita 4H

Element in the third period or further can have an expanded octet because the electrons and go into the d-orbital.

Anthony Hatashita 4H

Anything in the third period or past that because they have an empty d shell that can be expanded into

Anthony Hatashita 4H

O is more electronegative than P so it receives the negative charge, and you want the charges to be spread out, not on one atom usually

Anthony Hatashita 4H

Yes, this is testable with calculating the formal charges of the atoms

Anthony Hatashita 4H

I think it's the first elements where they simply don't have enough electrons to fill an octet and anything after the third period where expanded octets can occur.

Anthony Hatashita 4H

I understand the trends on the periodic table, but what does a high electronegativity mean?

Anthony Hatashita 4H

All of them have no unpaired electrons because the shells are filled, but there are some cases where 3d is filled with 8 electrons, leaving 2 unpaired electrons due to Hund's rule I believe.

Anthony Hatashita 4H

For Ga+3 I believe the answer is [Ar]3d^10 because the 3d shell fills before the 4s does. You write the "largest" noble gas filled with electrons in brackets followed by the extra electrons.

Anthony Hatashita 4H

The question asks about how many unpaired electrons Sn^4+ has, which is 0 because the entire 4d shell is full.

Anthony Hatashita 4H

Mn has a ground state of [Ar]3d^7, hence the 7 valence electrons while Sb is [Kr]4d^10 5s^2 5p^3 with 5 valence electrons in the higher energy state.

Anthony Hatashita 4H

It doesn't matter, just make sure you have the right amount of them.

Anthony Hatashita 4H

Yeah you're right, I think it also has to do with whether or not the electrons are shared between the atoms or just taken from one to the other as well.

Anthony Hatashita 4H

I'm not sure to be honest, but I think it has to do with 4s, 4p, and 4d being filled, leaving 5s and 5p electrons.

Anthony Hatashita 4H

They can't form an octet and never would in nature because they would never receive electrons for that, they are stable at lower values.

Anthony Hatashita 4H

I think that means the number of covalent bonds an atom has, which is tightly related to the amount of valence electrons it has.

Anthony Hatashita 4H

2.998x10^8 is most accurate, but for some solutions in the manual it uses even more precise numbers. I think for the tests we're given the constants to use though so we should all be fine!

Anthony Hatashita 4H

Any EMR has the same speed, we just call it the speed of light (c)

Anthony Hatashita 4H

On the class website there's a link that says Constants and Equations, that should help you :)

Anthony Hatashita 4H

64 total Joules are emitted over 2 seconds, and we have all the information we need we need to find E=ch/(wavelength). After you find E, you can divide 64 by E to find out how many photons there are, and then you can divide that number by Avogadro's number, 6.022x10^23 to find the moles of photons.

Anthony Hatashita 4H

So do we have to convert keV to J for all problems or is it specific to this one?

Anthony Hatashita 4H

You can eliminate a b and d through the process of elimination. I first looked at b but that's the opposite of the right answer. When the frequency decreases, the wavelength increases, so c is the correct answer as the change in the electrical field is lower at any point when the wavelength increases.

Anthony Hatashita 4H

From figure 1a.10 you know that the line in question is part of the Lyman series. From this you know that n1 is 1, and from there you find out that 1/n2^2 is around .111 or 1/9.

Anthony Hatashita 4H

In the spectrum of atomic hydrogen, several lines are generally classified together as belonging to a series (for example, Balmer series or Lyman series, as shown in Fig. 1A.10). What is common to the lines within a series that makes grouping them together logical? I'm pretty sure the answer is beca...

Anthony Hatashita 4H

Do we have to memorize equations or are we given a list of equations?

Anthony Hatashita 4H

As I first did this problem I immediately looked for a "the wavelength increases" answer, but there was none. I also used the process of elimination to eliminate a b and d, leaving only c as the possible answer. From the correct answer you can learn that higher wavelength results in less c...

Anthony Hatashita 4H

From my understanding, yes you are correct. Using whole numbers just makes things easier to use or read for lab experiments or simple calculations.

Anthony Hatashita 4H

The molar mass of the metal hydroxide M(OH)₂ is 74.10 g.mol. What is the molar mass of the sulfide of this metal?
What does it mean by the sulfide of this metal?

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