CHEMISTRY COMMUNITY

Hovik Mike Mkryan 2I

Hello,
Make it that concentration you found squared over x and equal it to the ka you found and solve for x. Hope this helps!

Hovik Mike Mkryan 2I

Hello,
It was mentioned previously that for the tests we should not approximate and use the quadratic formula unless stated otherwise. Hope this helps!

Hovik Mike Mkryan 2I

The final should include all material throughout the quarter, but I believe there will be more problems revolved around material after the midterm. Hope this helps!

Hovik Mike Mkryan 2I

Hello, Professor Lavelle touched on this after elementary reactions when we spoke about catalysts. We just need to know that if the catalyst is in the same phase as the reactant it is homogeneous and if it is different , than it is heterogeneous. Therefore, it would not make an impact on outcomes of...

Hovik Mike Mkryan 2I

Hello,
What I was told was that the aqueous solution in a cell diagram mostly always comes after the sold bridge and the solid components go on the outer most. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I believe that this term refers to that in a balanced equation and that forward and reverse rates are equal. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I always use the term oil rig to help me remember which is oxidation and reduction. Oil= oxidation is losing electrons during the reaction and rig= reduction is gaining electrons. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I believe the average for the midterm was a 73%. Hope this information helps!

Hovik Mike Mkryan 2I

Hello,
The test 2 average for my discussion section was a 38-39. Hope this information helps!

Hovik Mike Mkryan 2I

Hello,
During office hours I was informed that (aq) solutions should be nearest the salt bridge and the solids follow with the I separating them. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
Water melting will be spontaneous at its melting point. The scenario has to make clear under what conditions the water is present. The water melting will be spontaneous at 0 Celsius , but it won't be below that temperature. Hope this helped!

Hovik Mike Mkryan 2I

Coefficients have to be taken into account when calculating the standard free energy with the use enthalpy, entropy. and temperature. Trying to figure out what the Delta G is what will define if it is spontaneous or not. Even for Delta G products - Delta G reactants calculation the coefficients need...

Hovik Mike Mkryan 2I

Hello,
H+ is required but H2O is not. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
At office hours today I was told that looking at the reduction potentials and allowing them to be positive and knowing the bigger one is the cathode is how you can tell them a part and that is how you know which one needs to be reversed. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
Max work is in regards to the relation with the E potential. That is why wmax and delta G both equal -nFEcell. Hope this helps!

Hovik Mike Mkryan 2I

Hello, if I am understanding your question correctly, the molar entropy increases because as a solid goes into two aqueous solutions, which contain more particles and disorder. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I believe professor Lavelle mentioned in class that at room temperature the most stable is graphite and at higher temperatures diamond can become more stable. Hope this helped!

Hovik Mike Mkryan 2I

Hello, if delta G is negative it results in the reaction being spontaneous because the forward reaction is favorable and happens naturally without extra needed energy. Hope this helped!

Hovik Mike Mkryan 2I

Hello, to calculate degeneracy one has to know how many molecules are present and how many orientations it can form. The # of orientations can be denoted x and the number of molecules can be denoted n. Therefore x^n will give you the degeneracy. Hope this helped!

Hovik Mike Mkryan 2I

Hello, I believe for part A it is positive because it is the entropy for the surroundings which are gaining the heat given off the body. Hope this helped!

Hovik Mike Mkryan 2I

Does anyone have the solutions?

Hovik Mike Mkryan 2I

Hello,
Basically it means that if the calorimeter is gaining heat that means that the rxn is losing the heat therefore in order to convert the heat for the calorimeter into the heat for the rxn you have to add a negative sign. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I believe that for this problem both the heat capacity for ice and liquid water is used because it is required to melt the ice first than raise the temperature from 0Celsius to the desired temperature. Hope this helped!

Hovik Mike Mkryan 2I

Thank you very much Joyce!

Hovik Mike Mkryan 2I

Hello,
I believe the constants and equations on Dr. Lavelle's website is the one that is going to be supplied on the mid-term and it should suffice. Hope this helped!

Hovik Mike Mkryan 2I

Hello, I believe a bomb calorimeter is a closed system because it has constant-volume and is sealed so that no gas can escape and allows heat exchange. Hope this helped!

Hovik Mike Mkryan 2I

Hello, to solve 2a and 2b in the first worksheet I would go about setting realizing the differences of the problems with the initial one with the given KC. If you square the KC you get the answer for 2a and if you inverse the KC you get the answer for 2b. I would go about setting the products/reacta...

Hovik Mike Mkryan 2I

Hello, the way I remember is by that irreversible expansion has constant pressure, therefore you can take out -P and multiply with Vfinal-Vinitial. For reversible reactions the pressure is not constant and the second equation takes that into account. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I believe that work is not a state function because it is proportional to the distance an object is moved. The path taken to get to final is needed to calculate amount of work, but state functions depend only on the state it is in not how it got to that state. Hope this helped!

Hovik Mike Mkryan 2I

Hello, For the ICE table the amount that changes to reach equilibrium is denoted by x for both the products and reactants. If the sample problem was 2A + 3B ----> 3C and we were given the initial concentrations for A and B at both 0.1, then we would do 0.1-2x for A and 0.1-3x for B. For C it's initi...

Hovik Mike Mkryan 2I

Hello,
I attended a TA's office hours today and he told me that it would be good to know the strong acids and bases for the test. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I also think it would be great if the solutions for the worksheets would have the work, because I myself sometimes can't get the provided answer. Thanks in advance for whoever can make that happen!

Hovik Mike Mkryan 2I

Hello,
I believe that in today's lecture Dr. Lavelle said that enthalpy is a state property and that the value is determined by its current state. He mentioned that qsubp is the same as Delta H, therefore a state property. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I also had this same question and how the balanced equation was derived.

Hovik Mike Mkryan 2I

Hello,
I memorized the most common six strong acids and bases which is not that tough. You can find them here: https://opentextbc.ca/chemistry/chapter ... and-bases/. Hope this helped!

Hovik Mike Mkryan 2I

Hello, Usually if the water is not at room temperature the different value of temperature is given, but the concentrations are already given at that specific temperature so it is just additional information to help us understand the different concentration when and if it is compared to room temperat...

Hovik Mike Mkryan 2I

Hello,
As mentioned by Professor Lavelle in class, an inert gas that was never present in our reaction will have no effect on the equilibrium when added inside the system. Hope this helped!

Hovik Mike Mkryan 2I

Hello, I believe that the potassium onion is present on both sides alone, therefore they have no affect on the concentration of the pH or POH and they cancel as mentioned. But the HNO2 is present only on the left and the deprotonated NO2 is on the right. There isn't another separate NO2 molecule on ...

Hovik Mike Mkryan 2I

Hello,
I was stuck on the same problem than realized that either the book problem forgot to add the 10^-3 for the concentration for part C or the solutions manual added the 10^-3. I believe it is a problem error. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I believe calculating concentration for a reaction in all gas phase is used with the PV=nRT formula.Hope this helps!

Hovik Mike Mkryan 2I

Hello, I believe in the last lecture Professor Lavelle said we can give an answer in partial pressure or concentration if the question does not specify, but if the question wants the Kc we need to give it in concentration, but we can use partial pressure to find the concentration with P=(conc) RT. H...

Hovik Mike Mkryan 2I

Hello, I believe when adding liquid to a system the K will not change because all the other components are in gas phase. The concentrations are not changing because no moles of gas is being added to to the system. If there was (aq) solutions in the system then it would have effect on the concentrati...

Hovik Mike Mkryan 2I

Thank you very much!

Hovik Mike Mkryan 2I

Hello, You calculate the mass percentage by dividing the weight each element has present in the entire molecule and multiplying by 100%. For example in CH3, the entire molecule weighs around 15.4 g. To find the mass percentage of hydrogen you do (1.008g) x3 divided by 15.4g times 100% to get the mas...

Hovik Mike Mkryan 2I

Hello,
I believe knowing that O2 is required for a combustion reaction and that it always produces H2O and C02 are the basic knowledge needed for this course. Hope this helped,

Hovik Mike Mkryan 2I

Hello,
I believe that the lone pair on the nitrogen will cause the displacement with h2O. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
I had the same question and my TA told me that they both are the same thing and can be used interchangeably. One is the old way of saying it and one is the new. Hope this helped!

Hovik Mike Mkryan 2I

Hello, Kw is representing the concentration of both the hydroxide and hydronium concentrations. Kw= 1.0 x 10^-7 multiplied with another 1.0 x 10^-7. Each 1.0 x 10^-7 represents the concentration for both hydroxide and hydronium. Therefore, kw can equal to both of the answers you stated. Hope this he...

Hovik Mike Mkryan 2I

Hello,
I believe you are talking about calculating pH levels, if so the formula is pH =-log [H3O+]. For example calculating the hydroxide concentration in a solution that has pOH 6.34.
On a calculator you would just put 10^-6.34.

Hovik Mike Mkryan 2I

Hello,
To my knowledge knowing nomenclature is not expected in this class, but just to be safe I would know the basic ones like ammonia, ammonium, carbon dioxide, water, Hydrate, and hydroxide.

Hovik Mike Mkryan 2I

Hello,
Since the actual structure is not shown, I predict that the abbreviation shown is to represent both atoms at the end of each sigma bond. This would be true if the sigma bond was connected to two Carbon atoms.

Hovik Mike Mkryan 2I

Hello,
During lecture it was mentioned that the first four elements do not follow the octet rule and sometimes Boron satisfies its octet with 6 electrons. Also, all elements after at the quantum level 3 and after can occupy more than 8 electrons and do not follow an octet rule. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
An easy way I use to find out the atomic orbitals is squaring the principal quantum number. For example, if quantum number is n=2, the number of atomic orbitals would be 4.

Hovik Mike Mkryan 2I

Hello,
To my understanding the intermolecular hydrogen bonds in the water are attached with the O and H bonds and do not separate and surround the other ions that are present in the water. The O bonds with the positively charged ions while the H bonds with the negatively charged ion. Hope this helped!

Hovik Mike Mkryan 2I

Hello,
Yes, if there are no lone pairs present there will not be the lone pair repulsion effecting the arrangement. Therefore, both the arrangement and molecular shape will be the same.

Hovik Mike Mkryan 2I

Hello,
I spoke to my TA this week and she notified me that the test will include hybridization, shape, the two types of forces and polarization. Hope this helps!

Hovik Mike Mkryan 2I

Hello, I believe the CHI3 will have a higher melting point because the Iodine atoms are bigger and have many more electrons than the Fluorine atoms. Both are polar covalent bonds and both have London Forces, but the bigger and more packed electrons will take more energy to melt. Hope I answered your...

Hovik Mike Mkryan 2I

Hello, Unfortunately you need an access code to access sapling learning and it can be purchased through the UCLA bookstore or from the website itself. https://www.saplinglearning.com/ibiscms/login/. Also, during the beginning of this course Dr. Lavelle sent out an email with the link to purchase the...

Hovik Mike Mkryan 2I

Hello,
When I asked a UA last week they said we do not need to know oxidation numbers yet, maybe in the future we will discuss them. Hope this helped.

Hovik Mike Mkryan 2I

Hello, I believe it is not tirgonal planar because of the two lone pairs that lie on the central Iodine atom. 3 Cl atoms have 21 electrons and 1 I atom has 7. Resulting in 28 electrons. It would take 24 electrons to fulfill the Cl octets with I only having 6 electrons on it (the three bonding pairs)...

Hovik Mike Mkryan 2I

Hello,
Dipoles are formed because of partial positive and partial negative charges within a molecule. The charges between the polar (having unequal charges) molecules creates the attraction and bond. Hope this helped.

Hovik Mike Mkryan 2I

Hello,
The O in N=O=N gives O a +2 charge and the N with a -1 charge. Basically, the way I look at it is making sure I have formal charges closest to 0. Resulting in N being the central atom because of formal charges.

Hovik Mike Mkryan 2I

Hello,
I also got confused on this question, but I believe the question is asking which is the most ionic from all three. Hope this helps!

Hovik Mike Mkryan 2I

Hello,
Electronegativity values for nobles gases are not 0 and follow the trends like the other elements. As you go up the group electronegativity increases. Noble gases are not considered reactive because they already have filled octets.

Hovik Mike Mkryan 2I

Hello, I believe knowing that intramolecular bonds are stronger than intermolcular bonds and the sequence of the bonds in each category should suffice. In intra it is metallic>ionic>polar covalent> and lastly non-polar covalent. After comes the intermolecular forces which are Hydrogen bonding> dipol...

Hovik Mike Mkryan 2I

Hello,
I believe it is always negative because of the conservation of energy law. Work is being done against the gravitational pull.

Hovik Mike Mkryan 2I

Hello,
the exceptions discussed were the atoms Cr and Cu for the 3d orbitals. Cr prefers 3d^5 4s^1 instead of 3d^4 4s^2 and Cu prefers 3d^10 4s^1 instead of 3d^9 4s^2. Hope this helped.

Hovik Mike Mkryan 2I

Having the resonance structure with formal charges closest to 0 (especially central atom) makes it the the structure with highest contribution. The f(c)=0 makes the molecule the most likely to react.

Hovik Mike Mkryan 2I

Hello,
As I recall, during Thursdays review session we were told that we are not going to be tested on nomenclature.

Hovik Mike Mkryan 2I

Hello, a ground state electron configuration is writing out the element in a neutral state with no charge. If a given element loses or gains an electron(s) then the element is considered to be in its excited state. For example the atom Fluorine has 9 electrons in its ground state(neutral), but F^- h...

Hovik Mike Mkryan 2I

Hello, for this problem I believe resonance structures do not need to be stated because we are trying to find an unpaired electron to know if the species is a radical one or not. Changed around the area of double bond will not result in losing an electron.

Hovik Mike Mkryan 2I

Thank you very much for the feedback guys.

Hovik Mike Mkryan 2I

Hey guys, I wasn't able to make today's review sessions. I was wondering if there were any handouts given and if someone could post pictures or send them via email? Thank you!

Hovik Mike Mkryan 2I

Hello,
I believe it is taken out of the s orbital because it is the outermost shell and would require the least amount of energy.

Hovik Mike Mkryan 2I

Hello, Dr. Lavelle mentioned in his lecture to choose the atom with the lowest ionization energy as the central atom.

Hovik Mike Mkryan 2I

Hello, I usually consider the same reasoning for any type of radius for atoms. The trends I have noticed is that as you go down a group the radius increases. As you go from right to left in a period the radius increases. Also, radius increases as you go from the top right corner to the bottom left. ...

Hovik Mike Mkryan 2I

Hello, I usually consider the same reasoning for any type of radius for atoms. The trends I have noticed is that as you go down a group the radius increases. As you go from right to left in a period the radius increases. Also, radius increases as you go from the top right corner to the bottom left. ...

Hovik Mike Mkryan 2I

Hello, I usually retake a module if I got questions wrong and I would specifically refocus on the numbers I got wrong until I got them all right. I do not think there is the key to the modules.

Hovik Mike Mkryan 2I

Hello, To my understanding, there are 3 orbitals for 3p and they are labelled as x,y and z. n=3 gives us the third shell. l=1 gives us that it is respect to sub-shell p and m-sub(l) values are -1,0,1. The problems specifically only writes the m sub(l)=0. This means that it is specifically talking ab...

Hovik Mike Mkryan 2I

Hello, yes the Rydberg constant will be supplied for test 2 and also many more constants. You can see them all under the "constants and equations" section on https://lavelle.chem.ucla.edu/class-websites/chem14a/. Hope this helps!

Hovik Mike Mkryan 2I

Hello, I believe that the atom will excite itself to n=2 from n=1 and the excess energy will be given off as kinetic energy.

Hovik Mike Mkryan 2I

Hello, Dr. Lavelle noted on Friday that the upcoming test will include material from Chapter 1 and up to Fridays lecture.

Hovik Mike Mkryan 2I

Hello, I believe for this class if we refer to the table Dr. Lavelle showed us in class for the wavelengths should suffice. It starts from the y-rays all the way up to long radio waves.

Hovik Mike Mkryan 2I

Hello, I believe it will not be because testing for quantum world starts Oct. 23.

Hovik Mike Mkryan 2I

Hello,
As far as I understand every Sunday the website takes into account how many posts you have made that week and gives points accordingly(maximum 3). Therefore, I believe every Monday everyone starts off with 0 posts.

Hovik Mike Mkryan 2I

Hello,
I believe that they are because when we are talking about the frequency, we are talking about the frequency of the incoming light.

Hovik Mike Mkryan 2I

Hello, I have been using the Sapling homework online problems. I purchased the 6-month access. The extra problems help me solidify my understandings and what I need to work more on.

Hovik Mike Mkryan 2I

Hello,
According to my notes from Mondays lecture, I noted that light acts as both waves and particles. This is due to the fact that at such a small scale it is possible to have both properties. Hope this cleared up your question!

Hovik Mike Mkryan 2I

Hello,
Tin is one of the eight elements that can possess more than one type of charge. Tin has two charges, therefore you will see Tin (II) representing Sn2+ and Tin (IV) representing Sn 4+ to specify which charge of tin is in that compound. Hope I helped!

Hovik Mike Mkryan 2I

Hello,
Yes you do not need to use Avagadro's number to solve any of the parts in this question, just need to convert the molar masses to moles.

Hovik Mike Mkryan 2I

Hello, For this equation knowing M(molarity)=mol/ volume (L) is the first step. With the given information, converting the 2.111g Na2CO3 to moles and dividing by 0.250 to get the molarity(0.07967) will help to answer the rest of the question. For a) convert 2.15mmol to mol with 1000mmol=1mol and div...

Hovik Mike Mkryan 2I

Hello,
I remember Dr. Lavelle told us in class to not round off to significant figures up until the final answer. I hope this helped!

Hovik Mike Mkryan 2I

Hello,
I did this problem by individually dividing the elements by the total mass (161.20g) and getting the percent composition of each element of the compound. Using 4 significant figures I got the percentage composition to be 52.15%C, 9.379%H, 8.691% N, and 29.78%O. I hope this helped!

Hovik Mike Mkryan 2I

Hello, I solved this problem by labeling each compound as x and y. Compound x (NaNO3) contains 27.05% sodium, while compound y (Na2SO4) contains about 32.37% sodium. I used two equations, the equation x+y=5.37 in order to get one variable alone and the second equation 0.2705x+0.3237y= 1.61g to calcu...

Hovik Mike Mkryan 2I

Hello, I was too stuck on this problem and figured out that the equation was not fully balanced. There needs to be a 3 stoichiometric coefficient in front of product AgCl(s). After balancing the equation, one has to find out the limiting reactant. After converting both 0.750g of C6H9Cl3 and 1000g (r...

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