CHEMISTRY COMMUNITY

Chris Freking 2G

15.39 Determine the time required for each of the following second-order reactions to take place: (a) 2 A -> B + C, for the concentration of A to decrease from 0.10 mol/L to 0.080 mol/L, given that k = 0.015 L-1 mol min for the rate law expressed in terms of the loss of A. In the 6th edition solutio...

Chris Freking 2G

You would have to refer to the standard reduction potential chart. The more positive the value of E⁰, the stronger the oxidation power.

Chris Freking 2G

In 14.23a, the galvanic cell is constructed in the solutions manual as: Hg(l) | Hg 2 2+ || NO 3 -(aq), H+(aq) | NO(g) | Pt(s) I understand that liquids may be used as conductors. Can a liquid in the anode/cathode always be used as a conductor? If not, how can it be determined whether it can be used ...

Chris Freking 2G

https://lavelle.chem.ucla.edu/wp-conten ... ns_14B.pdf

Chris Freking 2G

Why would the cold/heat pack activation be considered spontaneous?

Chris Freking 2G

What would happen if you have 2 slow steps? Is the slower step still the determining rate? Is having more than 2 slow steps possible? You can still have multiple "slow" reactants, but ultimately the slowest one will still determine the overall rate law. (An extremely small bottleneck will...

Chris Freking 2G

The only things you should have to remember from 14A are any "fundamentals" that are applied throughout various sections in 14B (i.e. balancing a reaction)

Chris Freking 2G

Use the formula deltaGr⁰ = -(RT)lnK; solving for K gives K = e^-(deltaGr⁰/RT) Solve for the first step reaction: K1 = 10K2 = e^-(G1⁰/RT) = e^(2.00*10^5/RT) For the second step reaction: K2 = e^-(G2⁰/RT) Combining the two steps gives the reaction 10 = e^(deltaG1⁰+deltaG2⁰/RT) = e^(2.00*10^5+deltaG2⁰/...

Chris Freking 2G

Conducting solids (electrodes) should be on the outside of the cell, and aqueous should be on the inside next to the salt bridge.

See viewtopic.php?f=140&t=43085

Chris Freking 2G

Why is it favorable for the ions to flow through the salt bridge to the anode from the cathode? Is it because the electron flow from the anode to the cathode makes the cathode negatively charged the cathode negatively charged and the anode positively charged, thus making the salt bridge ion flow fav...

Chris Freking 2G

Chris Freking 2G

Tony Ong 3K wrote:does anyone know if we can argue our midterm scores?

Try to contact your TA about it

Chris Freking 2G

Athena L 1B wrote:What was the problem for this one?

This was question 8A for the Fall 2018 Chem 14A Midterm

Chris Freking 2G

You do flip the enthalpies when the equation is flipped. The final equation should be (2*-345kJ) + (2*21kJ) + (150kJ), where the first equation is multiplied by 2, the second equation is flipped and multiplied by 2, and the final equation is flipped as to give the final reaction 2GingerAle + 5RootBe...

Chris Freking 2G

Madeline Ho 1C wrote:Can someone explain how to do #3?
I got deltaHrxn= 1mol(1560 kJ/mol) - [1mol(-1300 kJ/mol)+2mol(-286 kJ/mol)] = 3432 kJ

Using Hess' Law will give the reaction
deltaHrxn = (1mol)(-1300kJ/mol) + (2mol)(-286kJ) + (1mol)(1560kJ) = -312kJ

Chris Freking 2G

It is given that the gas mixture has 2.250 mol of helium gas and 1.492 mol of krypton gas. The temperature of the system is 348.15K and the volume of the system is 11.0L. To calculate the total pressure of the system, do we first calculate the pressure using mol of helium gas, then calculate the pre...

Chris Freking 2G

According to Lavelle's important midterm information document on his website, the midterm will cover "Chemical Equilibrium, Acid and Base Equilibria, Thermochemistry, Thermodynamics to the end of entropy."

Chris Freking 2G

Some examples:
Open - An unsealed flask of liquid; an open soda can
Closed - A sealed flask of liquid, an unopened soda can
Isolated - The universe; A high-quality sealed thermos*; a bomb calorimeter*

* (these are not 100% isolated, but for our reference the heat transfer is approximated to 0)

Chris Freking 2G

Chem14BTestSchedule.pdf: (132.76 KiB) Downloaded 77 times

Chris Freking 2G

The bomb calorimeter is also an example of an "isolated system." There is no truly 100% isolated system other than the universe (arguably). But we refer to a high-quality sealed thermos or a bomb calorimeter as an isolated system because the heat transfer is so slow that it can be approxim...

Chris Freking 2G

From the lecture on 1/25/19:

N₂(g) + O₂(g) --> 2NO(g) (ΔH_rxn = 180kJ)
2NO(g) + O₂(g) --> 2NO₂(g) (ΔH_rxn = -112 kJ)

Adding the above reactions gives the net reaction:
N₂(g) + 2O₂(g) --> 2NO₂(g) (H_net = 68kJ)

Chris Freking 2G

@504909207

Probably Outline 3 (Thermochemistry and The First Law of Thermodynamics) since we have already started covering that material and the 1st and 2nd outlines have already been tested on

Chris Freking 2G

It can also be seen in this carbon phase diagram that carbon is in the form of granite at standard conditions

Chris Freking 2G

Lavelle emphasizes to do the homework as a way to prepare for the tests.

Chris Freking 2G

If 18.3% of the BrCl gas remains at equilibrium and we have the initial concentration of BrCl, it is possible to calculate the amount of BrCl left at equilibrium. 0.183 * (1.84 * 10^-4) = 3.36 * 10^-5 M of BrCl at equilibrium. Then use the ICE box and plug in this value: 2BrCl(g) ⇌ Br2 (g) + Cl2(g) ...

Chris Freking 2G

pH can be negative if the concentration of H+ ions have a molarity over 1.

See viewtopic.php?t=24564

Chris Freking 2G

K is the equilibrium constant, which is derived from dividing the concentration of product by the concentration of reactant. (K=[P]/[R]). When a reaction is at equilibrium, products and reactants are being formed at equal rates, so the concentration of either remains constant.

Chris Freking 2G

In part C, K = (Partial pressure of X)^2 / (Partial Pressure of X2)

Plug in the mole fractions of each molecule (X = 12 molecules/17 molecules) and (X2 = 5 molecules/17 molecules)
Plugging those fractions to K and multiplying by the initial pressure of X2 (0.10 bar) will give the answer 0.17 .

Chris Freking 2G

K refers to the [P]/[R] ratio at equilibrium. Although Q also refers to the [P]/[R] ratio, it can be measured at any point in time in a reaction--including equilibrium. If Q<K, then the forward reaction (products forming) is favored. If Q>K, then the reverse reaction (reactants forming) is favored. ...

Chris Freking 2G

Reese - Dis 1G wrote:Can someone explain 27? I thought you only added -ate when the compound was negative.

If K₂ has a +2 charge, then [Ni(CN)₄] must have a -2 charge since the overall compound is neutral. It's called "nickelate" because the compound in the brackets has the -2 charge.

Chris Freking 2G

The electron gains energy when it jumps from a lower energy level to a higher energy level.

Chris Freking 2G

You are also already told initially that H2Se has a higher boiling point than H2S, so you don't have to worry about determining which one has the higher boiling point. All you have to do is explain why H2Se is the one with the higher boiling point, and in this case the only possible way for H2Se to ...

Chris Freking 2G

I understand that all amphiprotic compounds can be amphoteric compounds. But what is an example of an amphoteric compound that is not amphiprotic?

Chris Freking 2G

Lewis Acids accept electrons while Lewis Bases donate electrons.

This is not to be confused with Bronsted:
Bronsted acids donate protons while Bronsted Bases accept protons.

Chris Freking 2G

Hydronium ions trigger taste buds that send a signal of "sourness" to the brain.

Hydroxide ions break down fatty acids and oils on the skin which reduces friction on your skin, which is why they feel soapy.

Chris Freking 2G

ZrF₇^3- and HfF₇³ also have pentagonal bipyramidal shape.

Chris Freking 2G

SF₄ has the AXE notation AX₄E. Therefore it has the see-saw molecular shape which is asymmetric and the dipole moments do not cancel out.

: Screen Shot 2018-11-24 at 3.35.29 PM.png (9.24 KiB) Viewed 735 times

Chris Freking 2G

Aluminum can also sometimes have an incomplete octet for molecules like AlCl₃. However, AlCl₃ only exists at high temperatures--in normal temperatures, two molecules of AlCl₃ combine to form Al₂Cl₆ where both Al atoms obey the octet rule.

Chris Freking 2G

In acrylonitrile, yes, the C-N bond is a triple bond. However not every C-N bond has to be a triple bond (eg. Methylamine)

Chris Freking 2G

Chris Freking 2G

Chris Freking 2G

The 3 lone electrons form a trigonal shape in the equatorial plane as it is the most favorable in terms of electron repulsion, so the two other bonded atoms will take on a linear shape.

Chris Freking 2G

Chris Freking 2G

A dipole is a molecule with opposite charges on both ends (positive on one side, negative on the other side). For example, H2O is a dipole as the hydrogens have a partial positive charge and the oxygen has a partial negative charge. An induced dipole is the result of two molecules interacting and ca...

Chris Freking 2G

Be and Br have similar electronegativities. Since the electronegativity values of Be and Br are so close, they both pull the electron with nearly the same strength, meaning the electron is closer to the middle and is shared between both Be and Br

Chris Freking 2G

The electron configuration of Ru is [Kr] 4d^7 5s^1.

Why is the electron configuration not [Kr] 4d^6 5s^2? Wouldn't it make more sense to have a filled 5s and a partially filled 4d rather than a half-filled 5s and partially filled 4d?

Chris Freking 2G

Ag has an electron configuration of [Kr] 4d^10 5s^1 because the element is much more stable when the electrons are configured in the element to have a full d shell and a half-full s shell. When Ag is ionized (Ag+), the electron is removed from the outermost shell 5s. Therefore, the electron configur...

Chris Freking 2G

The correct Lewis structure is the one with as many atoms having a formal charge of 0. Sometimes for compounds like SO₃ 2- there are variations of the Lewis structure that all have the same overall formal charge which is where resonance comes into play (multiple possible lewis structures)

Chris Freking 2G

l refers to the subshells of n. The allowed values of l depend on the value of n and range from 0 to n-1. For example, if n=3, then l can = 0, 1, or 2.

Chris Freking 2G

In the heavier transition-metal elements, the energy levels of the subshells get so close that their energy levels actually depend on how many electrons are occupying them. For example, 4s is filled before 3d because 4s has a lower energy level (rather than 3d filling before 4s like we might normall...

Chris Freking 2G

The Rydberg equation does not take into account electron-electron repulsion, which is why it's okay to use for Hydrogen and H-spectrum (only one electron to worry about). It will not work for atoms with multiple electrons because of electron-electron repulsion.

Chris Freking 2G

Electrostatic attraction refers to the attraction between a positively charged nucleus and a negatively charged electron.

Chris Freking 2G

When delta E is negative, it means that energy has been released. Positive delta E refers to the absorption of energy.

Chris Freking 2G

Handwriting notes is very helpful because I feel that I comprehend the material more, but it's all a matter of personal preference.

Chris Freking 2G

Both are commonly used as symbols for frequency. Perhaps you could use whichever one is given to you in the problem but ultimately it should not matter.

Chris Freking 2G

Use the formula M = n/V. In the context of this problem, solve for V.

V = n/M = (4.50 * 10^-3 mol)/(0.278 mol*L^-1) = 1.62 * 10^-2 L, or 16.2 mL

Chris Freking 2G

The crucible and product together weigh 28.35g. Subtract the mass of the crucible (26.45g) from the mass of the crucible and the product (28.35g) to get the mass of the product (1.900g). Since we know we started with 1.500g of Sn, you can subtract that mass from the mass of the product (1.900g) to f...

Chris Freking 2G

To clear up any confusion, it's referring to the oxygen itself, not oxygen gas molecules. To find the fraction of the total mass of the sample due to oxygen, divide the molar mass of the oxygen by the molar mass of the entire compound. (4 * 15.999 g mol^-1) / (206.53 g mol^-1) (The 4 comes from the ...

Chris Freking 2G

Since only 20 mL of the first solution is diluted, you should first find the molarity of the first solution and using that molarity and 20 mL volume as your M1V1 later on. Start by finding the molarity of the original solution. Convert the 5.00g of KMnO 4 into moles by dividing it by the molar mass....

Chris Freking 2G

I will typically use the exact atomic weight listed on the periodic table and round my final answer if needed. Future problems should supposedly have given atomic weights

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