CHEMISTRY COMMUNITY

Manpreet Singh 1N

On one of the practice finals, you were able to get credit using either or but it looks more concise when using E and Z for IUPAC naming

Manpreet Singh 1N

More specifically Z is used when the high priority groups are on the same side and E is when the high priority are on the opposite side. The high priority group is determined by atomic number

Manpreet Singh 1N

Dr.Lavelle, I just want to say thank you for all the effort you have put in to make sure we understand chemistry. Because of my lack of a strong chemistry background from high school, the idea of taking chemistry for two years scared me. However since the first day of chem 14A with you, your class h...

Manpreet Singh 1N

Aadding on, we use the carbon count from right to left because it gives us a small number. From left to right the isoporpyl would have been on the 7th carbon, but from right to left it is attached to the 4th carbon. Professor Lavelle said we want the smaller number when writing the name of the compo...

Manpreet Singh 1N

Hello,

I am not completely sure where they exist, but there are reactions that require three half-reactions. Due to this, there are ways to solve for them. Here is a link I found.
http://www.chemteam.info/Redox/Redox-Th ... tions.html

Manpreet Singh 1N

Hello,

We cant use the same process because it 15.27 the reaction is in 1st order but it 35 it is in second order. So it 27, the equation we used is ln[A]=-kt+ln[A0]. But for a second order reaction the equation would be 1/[A]=kt+1/[A0].

Manpreet Singh 1N

Hello,

They used the equation: deltaG=-RTlnK

Manpreet Singh 1N

Hello, For this problem, we want the standard potential for U^4+ +4e- ----> U since we don't have that exact standard potential, we can use the information we have so we have: U^4+ +e- ----> U^3+ E^o=-0.61 U^3+ +3e- -----> U E^o=-1.78 When we add, the U^3+ cancels out because they are on opposite si...

Manpreet Singh 1N

We also multiple the second half-reaction by 5 because we want the electrons to cancel out, so we are left with the only the molecules.

Manpreet Singh 1N

Hello,

My logic for this problem is that the molecule with the greater molar mass has the greatest molar entropy because the more massive the gas molecule, the more disordered it can be.

Hope that helps :)

Manpreet Singh 1N

Hello,

Yes you are correct. You can write the bond enthalpies for both reactants and products, but when you do that, you see that all the C-H bonds cancel out.

Hope that helps :)

Manpreet Singh 1N

Reversible reactions at equilibrium means that once the reactants form the products, the products can form the reactions again using the same amount of energy. A irreversible reaction at equilibrium means that once the reactants make the products, the reactants can not be formed again by the products.

Manpreet Singh 1N

Hello, My logic may be wrong, but the units for molar heat capacity are J*K^-1*mol^-1. I remember Professor Lavelle saying that the more mass something has the more the heat capacity. So since NO2 has one more oxygen (one more atom), there are more bonds that can absorb the energy that is added. Hop...

Manpreet Singh 1N

I am pretty sure, scientists can get very close to having a perfectly isolated system. My TA used the example of a thermos as an isolated system.

Manpreet Singh 1N

Hello, Yes heating and cooling would be only the only way to change energy for a closed system because in a closed system no substance can go in or out, just heat energy can enter and escape. While an open system, because it is open can be changed by substance and by cooling and heating. However, in...

Manpreet Singh 1N

Hello,

I feel like in order to know what bonds are broken and formed there is some extent of knowledge necessary. Adding on, the wording of the question and the chemical formula should tell you which bonds are broken and which are formed.

Hope that helps :)

Manpreet Singh 1N

Manpreet Singh 1N

That is correct. The practice quizzes are in our course reader. Professor Lavelle said that there will not be workbooks this quarter because he has too many students and it would have been very difficult for the TAs to be able to grade them.

Manpreet Singh 1N

A Bronsted Acid is an acid that donates a proton (H+), while a Lewis acid accepts electrons. I like to think that acids like to have a negative charge and therefore either loses a proton or accepts an electron.

Hope that helps :)

Manpreet Singh 1N

Professor Lavelle posted it under announcement on chemistry community. But here it is: Chem 14A Final Exam 3-6pm, Sunday, December 4. Students MUST go to their assigned room. 14A-1, 11am class: Last name A-L in CS50. Last name M-Z in LAKRETZ 110. 14A-3, 1pm class: Last name A-L in CS24. Last name M-...

Manpreet Singh 1N

Adding on, Bronsted Acid and Bases deal with donating or accepting protons, while Lewis acids and bases deal with accepting or donating electrons.

Manpreet Singh 1N

In this question, it talks about the UV spectrum, which falls under the Lyman series. The Lyman series has a n final of 1. On the other hand, the Balmer series (visible light) has a n final of 2.

Hope this helps :)

Manpreet Singh 1N

Hello,

When the percent dissociation is greater than 5%, we can not approximate. We would have to use the quadratic equation to solve for X, to have a more precise concentration.

Hope that helps :)

Manpreet Singh 1N

Hello,
So pH is the measure of hydrogen ion concentration, [H+], while pOH is a measure of the hydroxide ion concentration, [OH-]. The scales for both are 1-14. However in most cases, the question will ask for pH. So remember that pH+pOH=14. So pH=14-pOH.

Manpreet Singh 1N

Hello,

Since acids are proton acceptors, it would gain a proton. Since it is negatively charged to begin with, adding a H+ would neutralize it. So the answer would be H2CO3

Hope this helps :)

Manpreet Singh 1N

Hello,

In lecture, Professor Lavelle said that if the complex has a negative charge(which is rare), the Transition metal will end in ate.

Hope that helps :)

Manpreet Singh 1N

Hello,

My TA did a example of F2 and F2- in our discussion. He said that F2- is weaker because it has one more anti-bonding and because it is paramagnetic. So when we compare two molecules, the one that is paramagnetic is weaker.

Hope that helps:)

Manpreet Singh 1N

Hello, First of all it is not a stupid question. Secondly, professor Lavelle mentioned in lecture than when a complex has a negative charge (which is pretty uncommon) it has a anion. For this case, the complex has a 4- charge. Adding on the CN has a -1 charge. Since there are 6 of them, the charge i...

Manpreet Singh 1N

Hello, The coordination number helps you find out how many ligands are connected to the central atom and helps you determine the overall shape of the specific complex. So in other words, it helps you find how many donor atoms there are. So to answer your question, yes the coordination number is affe...

Manpreet Singh 1N

Hello, The more electronegative atom has lower energy because the row of electronegativity relates to the charge of the nucleus. Therefore, the elements on the right (more charged nucleus) are more electronegative. That means that the more electronegative the atom is, the closer the electrons orbita...

Manpreet Singh 1N

Hello,

O does have a slightly lower ionization energy, but I remember Professor Lavelle saying that O can never be a central atom.

Hope that helps :)

Manpreet Singh 1N

Hello,

For Hf, the atomic number is 72. The lanthanides have atomic numbers 58-71, which is before 71. Therefore the f orbital is full and the electron configurations for Hf=[Xe]4f^14 5d^2 6s^2

Hope that helps :)

Manpreet Singh 1N

Hello,

For the d-shell, l=2, the notation for ml is -2,-1,0,1,2

Hope this helps :)

Manpreet Singh 1N

Hello, In class Professor Lavelle said that the lewis structure that has the most of the elements that have a formal charge of 0 will make the greater contribution because it is the most stable. So it part a) Xe and F have a formal charge of 0 in the first one, while the second one has Xe with -1 an...

Manpreet Singh 1N

Hello, So basically for this question, we are told to find a M (a metal) with the corresponding configurations. So for part a of #9. it is [Ar]3d^7. To begin we look at the noble gas [Ar], that is our starting point. Then we look at the next part, which is 3d^7. The first row of the d series is 3d. ...

Manpreet Singh 1N

Hello, The question tells us that a line at 102.6nm was observed in the ultraviolet spectrum. This brings in the Balmer and Lyman series. We know that Balmer is visible light with n1=2, while Lyman is UV with n1=1. I think it is something we have to remember. we can solve this problem by using v=R(1...

Manpreet Singh 1N

To begin the problem we need to find the energy of the light. So we use the E=(hc)/wavelength so energy= ((6.626x10^-34 Js)(3x10^8m/s))/(420x10^9m) I changed the units of wavelength from nm to m, so that is why I multiplied 10^-9 to the given wavelength. We get the energy to be about 4.7x10^-19J. I ...

Manpreet Singh 1N

To begin the problem we need to find the energy of the light. So we use the E=(hc)/wavelength so energy= ((6.626x10^-34 Js)(3x10^8m/s))/(420x10^9m) I changed the units of wavelength from nm to m, so that is why I multiplied 10^-9 to the given wavelength. We get the energy to be about 4.7x10^-19J. I ...

Manpreet Singh 1N

Hello, h is a constant. However in this answer manual it uses h bar, which is equal to h/(2pi) when you do the math (6.626x10^-34)/(2pi)=1.0546x10^-34. just to make it clear (h bar)/2 is the same as h/(4pi). in the solution manual it has 1/2 separated. But if you do it with h/4pi instead, you should...

Manpreet Singh 1N

To begin the problem we need to find the energy of the light. So we use the E=(hc)/wavelength so energy= ((6.626x10^-34 Js)(3x10^8m/s))/(420x10^9m) I changed the units of wavelenth from nm to m, so that is why I multiplied 10^-9 to the given wavelength. We get the energy to be about 4.7x10^-19J. I l...

Manpreet Singh 1N

I am stuck on this question from the post assessment. Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1. Answer the following three questions. A. What is the kinetic energy of the ejected electron? I think I...

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