CHEMISTRY COMMUNITY

Kuldeep Gill 1H

This makes a lot of sense but what would be considered an appropriate angle?

Kuldeep Gill 1H

Hi, the coordination number is just the number of ligands you have connected to the central atom. So for part A you have 4 Cl forming one bond each for the one Ni so you would have 4 bonds on Ni so the coordination number is 4. An easy way to figure this out is to draw the Lewis structure. Hope that...

Kuldeep Gill 1H

To determine the lowest energy Lewis structure you are going to want to find the formal charge of each atom and add up all the charges. If you find that you can cancel one of the charges out by adding a double bond then do that. You also want to get as close to one charge as possible so either all p...

Kuldeep Gill 1H

That is correct we are not covering naming and he said that we can skip problems that require you to name them.

Kuldeep Gill 1H

Since the dipole moments all cancel out the molecule is non-polar. I initially thought it could be dipole induced dipole but a UA said that we would need to know the other molecule to say that it was dipole induced dipole. So, in this case, it would only be London forces Hope that helps

Kuldeep Gill 1H

Since the top image includes both single and double bonds it would be longer and therefore weaker while the bottom image is composed of double bonds so it would have shorter and stronger bonds. Hope that helps!

Kuldeep Gill 1H

So can a molecule still "have resonance" even if there is one structure that is the most stable form of it? I had the same question and I asked in TA office hours and they said yes. When we are doing resonance we will probably mostly see resonance as moving the double or triple bonds arou...

Kuldeep Gill 1H

Hi, I believe we started Chapter 6 on Friday, May 18th. We are currently on Chapter 4. Hope that helps!

Kuldeep Gill 1H

Hi, I remember hearing in class that it was 10^-18. I'm not sure if you remember but in lecture, we did practice problems where the wavelength was something like 10^-31 and we came to the conclusion that because it was so small we would not be able to detect it. Hope that helps!

Kuldeep Gill 1H

Hi from what I understand Boron does not follow the octet rule. in the case you stated BF3 if you think about it in terms of formal charges boron will have three shared with fluorine one for each of them and then if you were to calculate the formal charge of Boron you would see that it was 0. the sa...

Kuldeep Gill 1H

Hi, you take the number of valence electrons minus half the number of shared electrons minus the number of unshared electrons. This will give you the formal charge of that atom!

Kuldeep Gill 1H

Hi since they said that you were finding the wavelength of an ejected electron and we know that an electron has a mass that is why we know to use the de Broglie wavelength equation. Hope that helps!

Kuldeep Gill 1H

Hi does anyone know if Friday's lecture covered Chapter 3 material or did we begin Chapter 4 material? Thanks!

Kuldeep Gill 1H

Hydrogen cannot be in the middle because it is only able to form one bond.

Kuldeep Gill 1H

The book says Frequency = R(1/n2^2-1/n1^2)

Kuldeep Gill 1H

You use Avogadro's number to find the mass of a sodium ion since sodium ions have mass and in order to find de Broglie wavelength, we need both mass and velocity.

Kuldeep Gill 1H

Since we are looking at n=4 l can be 0,1,2, or 3 and since the 0 is the s orbital its values of ml can be 0 so one orbital 2 electrons then to add to that if l=1 then ml can be -1,0,1 so three orbitals for 6 electrons then if l=2 ml can be -2,-1,0,1,2 so five orbitals for 10 electrons and if l=3 ml ...

Kuldeep Gill 1H

Hi so since the velocity can be + or - .34 we would multiply it by two since that is the amount we can go in two directions this then becomes the uncertainty in velocity. Then you can convert the g into kg and plug into the equation delta-p = mass* delta-v then plug in delta-v and mass into the equa...

Kuldeep Gill 1H

I believe silver's electron configuration is similar to the exception for Copper which is why it would be [KR] 4d^10 5s^1

Kuldeep Gill 1H

I was confused with this question as well. It asks for the wavelength of a sodium ion, does the mean we have to use Avogadro's number to convert from grams to moles or can we solve for the problem by just leaving it as grams. Yes so the way you would go about this is to find the mass of one sodium ...

Kuldeep Gill 1H

Hi so you can use the equation V=R((1/n^2)-(1/n^2)) and plug-in frequency for V and plug-in rydberg's constant and then plug in 4 into the first n^2 since that is the final energy level then you would just solve for the second n value which would be your initial quantum level the other way you could...

Kuldeep Gill 1H

Kara is right! so if they did not give you the l value and just told you n=3 that's when you would include the orbitals in the s and p but since they gave you the l value its a little more specific than just saying all the orbitals in period 3.

Kuldeep Gill 1H

Hi I'm not sure what you mean by thoroughly but n denotes the period on the periodic table so in this case if n= 2 we are looking at period 2 so all of the elements from Lithium to Neon, l can be any value up to the n-1 so if we have n-2 the maximum value of l=1 but also includes 0. A general rule t...

Kuldeep Gill 1H

I was confused with this question as well. It asks for the wavelength of a sodium ion, does the mean we have to use Avogadro's number to convert from grams to moles or can we solve for the problem by just leaving it as grams. Yes so the way you would go about this is to find the mass of one sodium ...

Kuldeep Gill 1H

Yes it is on the constant and equations sheet given on the first page of the exam.

Kuldeep Gill 1H

Hi so what I did was a drew a picture first and labeled all the corresponding parts with what I knew. I think starting with a picture definitely helps you figure out where to begin and what is actually happening. So for A, we are told 3.607*10^-19 J is REQUIRED to remove an electron so since it says...

Kuldeep Gill 1H

Hi I was a little confused when doing the Rydberg equation and finding the energy difference between the two levels I got a negative number which I know means that energy is being emitted to go to a lower energy level so I was wondering if when we state the energy difference would we state it as a p...

Kuldeep Gill 1H

The Balmer series is the visible region of light it ends at the energy level of 2. The Balmer series does not just have red and blue visible regions it is from violet----> red with increasing wavelength so between violet and red is your typical ROYGBIV but just backward. The Lyman series is the UV r...

Kuldeep Gill 1H

On the Chem 14A website, there is a link titled constants and equations.. this is what will be given to us, everything else other than a periodic table is what we would have to know. As far as wavelength I think either is okay as long as the question does not specify but remember in terms of the equ...

Kuldeep Gill 1H

You count how many of each element there is and put that number as the subscript then you do that for all of the elements in that structure that would be the molecular formula and then if you can simplify the subscripts that becomes your empirical formula. Hope that helps!

Kuldeep Gill 1H

The conversion factor from eV to J is given on the cover sheet of the test

Kuldeep Gill 1H

I thought that they were defined by the final energy level not the initial so for the Lyman series it can start at any level but ends at level 1... can anyone clear this up??

Kuldeep Gill 1H

I believe we are still picking problems from Chapter 1!

Kuldeep Gill 1H

Hi, so I figured it out..... I guess we were supposed to know that in the Lyman series ( UV light) that the final energy always goes to 1 so.... after you get 2.923* 10^15 you use the equation V=R*(1/n1^2 - 1/n2^2) and realise that it is final minus initial so set plug in 1 for 1/n1^2 and plug-in 2....

Kuldeep Gill 1H

Hi from my understanding molarity is the concentration.... if they give you the initial volume and end volume as well as one of the concentrations you can use the equation you said.

Kuldeep Gill 1H

I got up to that point as well and plugging it in just does not make sense to me

Kuldeep Gill 1H

My TA suggested doing a formula like setup where you put a blank in front of each one for example ____CH6 + ___O2-----> _____CO2 + ____H20 then you would assign each blank a letter... I'm going to use a,b,c, and d. so now C: 1a=1c H: 6a=2d O: 2b=2c+d then assign one variable equal to one I'm going t...

Kuldeep Gill 1H

We were told that we would eventually have to know how to write a formula given the name.... will we be taught how to do this??

Kuldeep Gill 1H

The part that says it was transferred relates to us knowing that a new volume may have to be obtained which is what the question asks After finding the concentration when transferring it into a new container the concentration should not change so that's why you set the concentration equal to x/.5L N...

Kuldeep Gill 1H

Hi well first you use the formula M1*V1=M2×V2 o find the concentration after it has been diluted after finding the concentration you are going to want to multiple it by 2 since for every nitrogen atoms there are two in the equation then set this concentration equal to moles over .1 L and that will g...

Kuldeep Gill 1H

Hi for G13 you take the dilution equation and setting it up it would be 1*.20=4*M. Solve for M and that you give you the concentration of the diluted solution which would be as you said .05M but then it says you took 100mL of the DILUTED solution and fed it to each plant so for this you would set th...

Kuldeep Gill 1H

I would think you should if it is given to you but I'm not necessarily sure if failing to do so would cause you to lose points!

Kuldeep Gill 1H

Hi, I was just wanted to know... to what extent should we know chemical formulas?

Kuldeep Gill 1H

Okay so the way I understood this is that we start with 2.111g of a solid and we know we have a flask with a volume of 250 mL part a asks what volume of solution should be transferred to get 2.15 mmol of Na+. So to start out with convert 2.111g to moles by dividing by the molar mass of sodium carbon...

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