CHEMISTRY COMMUNITY

Ashish Verma 2I

Being given the activation energy is a good hint to use it.

Ashish Verma 2I

Why is it that the lower the standard reduction potential, the higher the reducing power, and the higher the standard reduction potential, the stronger the oxidizing agent? I believe that is the trend, if I'm not mistaken and I have that committed to memory but I don't understand why that is the cas...

Ashish Verma 2I

There will probably be a higher emphasis on stuff after both the Midterm and Test 2, (kinetics especially) as those are topics we have yet to be tested on at all.

Ashish Verma 2I

Thank you so much for your prompt reply, appreciate it!

Ashish Verma 2I

Dr. Lavelle, I've been a freshman student in both your 14A and 14B class this year and I just wanted to say thank you so much for going above and beyond with the amount of effort you've poured into all of us making sure that we always have more resources than we could ever ask for and always seeming...

Ashish Verma 2I

Can we always use Platinum as an inert electrode for our cell diagrams when one is needed or are there others that must be used in certain cases?

Ashish Verma 2I

Also I don't have a mnemonic for it but be sure to remember the degeneracy equation (W=X^n) as I don't believe that one is given to us on our formula sheet.

Ashish Verma 2I

Why is it that for First Order Reactions the time required for the reaction to go to one fourth remaining is double the half life, but this is not true for second order reactions, as in problem 7B13 from the 7th edition?

Ashish Verma 2I

Wmax is directly proportional to the Gibbs free energy so it will be at its highest when Delta G is at its highest as well.

Ashish Verma 2I

Using standard reduction potentials. I believe a lower standard reduction potential indicates a higher reducing power and thus a higher standard reduction potential indicates a higher oxidizing power.

Ashish Verma 2I

I believe you are heating the liquid water to its boiling point, vaporizing the liquid water at its boiling point into a gas , then cooling the gaseous to the desired temperature, while adding together the entropy for each of these separate stages. Though I can't be sure as I can't seem to find the ...

Ashish Verma 2I

I believe in lecture he said that Mercury will only be used as an inert conductor if it present in the reaction initially but I"m not entirely sure, sorry. But I believe Platinum should be used if an inert conductor is needed unless Mercury is part of the reaction.

Ashish Verma 2I

I don't believe the 5N section is part of our syllabus?

Ashish Verma 2I

The integrated rate laws for zero order, first order, and second order reactions all include the rate constant and time. The differential form of these rate laws expresses teh rate of reaction. And by substituting the final concentration for half the initial in the integrated rate laws, we can solve...

Ashish Verma 2I

Could you be more specific? What kind of a problem are you talking about?

Ashish Verma 2I

Well they aren't present in the final equation but are in the initial step so I suppose they could be in the rate law for one specific step but not the overall rate law? But I'm not sure honestly, sorry.

Ashish Verma 2I

I believe you would use the definition of an intermediate and a catalyst to determine that. By definition, an intermediate is not present at the start or end of a reaction, it is created and used up within the process of the reaction. In this case none of the components fit that definition. On the c...

Ashish Verma 2I

pH 7 is neutral at STP but not for all temperatures as Kw is temperature dependent.

Ashish Verma 2I

It is not an intermediate as it is present at the beginning of the reaction. Intermediates are not present at the start or end of a reaction and are created and used up within the process of the reaction. HBrO is present at the beginning, thus making it a potential catalyst I believe.

Ashish Verma 2I

Yes I believe that's true; you reverse a reaction in order to have the highest resulting cell potential when you add the two together.

Ashish Verma 2I

No if something is an intermediate it is created and used up within the reaction steps and therefore I do not believe it should be present in a rate law.

Ashish Verma 2I

I believe if something has a low reduction potential it has strong reducing power and a high reduction potential signifies strong oxidizing power.

Ashish Verma 2I

It's E_cell sometimes written as E⁰_cell

Ashish Verma 2I

LEO says GER for Redox reactions:
Losing Electrons = Oxidation
Gaining Electrons = Reduction

Ashish Verma 2I

You don't need to use an equation just compare standard reduction potentials.

Ashish Verma 2I

There are two reactants in this equation, hydogen cation and hydroxide anion. Analyzing both their reduction potentials, we see that hydrogen cation has the stronger standard reduction potential, thus it must be reduced and the hydroxide reaction must be reversed and it must be oxidized. By balancin...

Ashish Verma 2I

Based on lecture, I believe it is because the goal of a cell diagram is to provide directions for how to conduct the described cell and to build that solid state cell hydroxide must be present in the form of a solid compound as the solid state cell could not be built using just aqueous hydroxide, bu...

Ashish Verma 2I

You can convert between partial pressure and molar concentrations using the Ideal Gas Law:
$PV=nRT$
$P=\left ( \frac{n}{V} \right )RT$
$P=\left ( concentration \right )RT$
$concentration=\frac{P}{RT}$

Ashish Verma 2I

An ideal gas is one which perfectly obeys the Ideal Gas Law: PV = nRT.
All collisions between molecules are elastic, kinetic energy and momentum are conserved, and the law is obeyed exactly, without any discrepancies.

Ashish Verma 2I

Temperature is directly proportional to the average kinetic energy of molecules of a gas. By increasing the pressure these molecules are placed in a more compact setting, causing them to bounce off each other more frequently, and thus move faster. As these molecules move faster, the temperature incr...

Ashish Verma 2I

Electronegativity increases as you go up the periodic table as the energy level is lower so electrons are held closer to the nucleus with lower atomic radii and an increased effective nuclear charge. As you go across the periodic table electronegativity also increases as the energy level remains the...

Ashish Verma 2I

In cisplatin the molecule is organized so that both binding sites are on the same side of the molecule so it can bind twice to the same strand of DNA stopping cell division.

Ashish Verma 2I

Do you mean the equilibrium constant? That represents the ratio of products over reactants in a reaction at equilibrium.

Ashish Verma 2I

Salts that contain the conjugate base of a weak acid will raise pH by removing protons from water to generate hydroxide. Sodium acetate is a good example of this.

Ashish Verma 2I

Perfect, thank you so much!

Ashish Verma 2I

The answer book is correct. The formula should be AX3E as there are three atoms bonded to the central atom, one oxygen atom and two chlorine atoms.

Ashish Verma 2I

1 sig fig

Ashish Verma 2I

Because it donates a proton to become a stable anion (Br-)

Ashish Verma 2I

I think he just wanted us to memorize them but generally speaking many strong acids are either polyprotic or the conjugate acid of a halogen anion (with HF being a notable exception).

Ashish Verma 2I

Ionic bonding usually involves a metal cation and a non metal anion and a transference of charge.
Covalent bonding usually involves two nonmetal atoms and a sharing of electrons in a bond, (no charge only potential dipoles)

Ashish Verma 2I

Pages 490-491 in the 7th Edition have a small blurb on this.

Ashish Verma 2I

I don't think you need to know anything specifically about acid rain just the concepts involved in that example from the book regarding the reaction of carbonic acid (or a similar strong acid) and water.

Ashish Verma 2I

The question focuses on using this quantum equation for hydrogen: E_{n}=\frac{hR_{h}}{n^2} Manipulating the equation with the information given, (the energy released as the electron went from one energy level to another) you get: E=\frac{hR_{h}}{n_{f}^2}-\frac{hR_{h}}{n_{i}^2} Where: E = The energy ...

Ashish Verma 2I

It seems to be asking for they hybridization of the second carbon in the chain. That carbon is attached to two other carbons, the first in the chain with a single bond and the third with a triple bond. As such it has two regions of electron density and thus exhibits a hybridization of sp.

Ashish Verma 2I

An amphoteric substance can act as either an acid or a base. An amphiprotic substance is a more specific type of amphoteric substance that can easily accept or donate hydrogen ions.

Ashish Verma 2I

I would assume their orientation as in bond angles and shape and VSEPR formulas. How they would experience repulsion and orient themselves far apart from each other such as in a square planar shape.

Ashish Verma 2I

There is an informational blurb with some example equations in section 6.E.4 on pages 490-491 of the seventh edition.

Ashish Verma 2I

Keep in mind the hydronium ion has a 1+ charge as H30+ but yes that is correct. Also note that both HSO4- and H20 are amphoteric. But yes you are correct. :)

Ashish Verma 2I

What is the difference between polarizability and polarizing power?

Ashish Verma 2I

Less than 120, also in the case of T-Shape versus seesaw both sets of bonds you've referred to are less than 90 degrees but indicating which is specifically lower is not usually important as long as you state that they are each lower than 90.

Ashish Verma 2I

You're right it would dissociate into the K+ and F- ions; however, F- is the conjugate base of hydofluoric acid, HF, which is a WEAK acid not a strong acid, and therefore does still affect pH.

Ashish Verma 2I

I'm not sure but I believe that this is outside the scope of our course as it is from section 2.G.1 in the textbook and the entire 2.G section is absent from the syllabus.

Ashish Verma 2I

Bromine is in Group 4 of the Periodic Table while Fluorine is in Group 2. Based on periodic trends and atomic mass, Bromine is a heavier atom with a much larger atomic radius. As such, the single bond in HBr is much longer than the single bond in HF. Longer bonds are weaker so hydrobromic acid has a...

Ashish Verma 2I

I had a question about one of the parts of the Photoelectric Effect Module Post Assessment. Question 30 specifically. The given information is as follows: Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 10^5 m/s. The work function for sodium is 150.6 kJ/mol. The ...

Ashish Verma 2I

I believe we are still on the quantum section for homework since we haven't covered anything further so we aren't prepared for homework on sections that have never been discussed in lecture. Also my TA allowed us to do homework from the high school chem review section for both Week 1 and Week 2 sinc...

Ashish Verma 2I

I'm confused as well and am not entirely sure but I think the equation is more conceptually used to prove that at this scale the means of measurement affects the outcome like Dr. Lavelle emphasized. I think it just proves that by measuring either momentum or distance of an electron in this type of a...

Ashish Verma 2I

One cheap mnemonic device to remember these is: Never Have Fear Of Ice CoLd Bread
Never: $N_2$ Nitrogen
Have: $H_2$ Hydrogen
Fear: $F_2$ Fluorine
Of: $O_2$ Oxygen
Ice: $I_2$ Iodine
CoLd: $Cl_2$ Chlorine
Bread: $Br_2$ Bromine

Ashish Verma 2I

I have the seventh edition book and solutions manual and I had the exact same issue as you. My third equation in the book is also listed as FeBr_2+Na_2CO_3->NaBr+CO_2+Fe_3O_4 But since the solutions manual uses Fe_3Br_8 I'm assuming that's a typo like you said and just using that as the first reacta...

Ashish Verma 2I

Question E.21 (b) asks you to calculate the moles and number of molecules in 25.92 mg of hydrogen fluoride. I tried to solve the question but ended up getting an answer with the incorrect number of significant figures and I can't explain why: \frac{25.92 mg HF}{1}*\frac{1gHF}{1000mgHF}*\frac{1molHF}...

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