CHEMISTRY COMMUNITY

andrewcj 2C

Look at the rate constants of the forward and reverse reactions. The equilibrium constant K is equivalent to the forward rate constant divided by the reverse rate constant.

andrewcj 2C

You use them in the appropriate situations, so if a system does expansion work at constant temperature, then you probably have to use the volume change equation. If the system changes temperature at constant volume, it's likely you have to use the temperature change equation.

andrewcj 2C

I believe they pertained to kinetic energy.

andrewcj 2C

As long as the electrode is still in the solution and has not completely degraded, it shouldn't affect voltage of a galvanic cell.

andrewcj 2C

It's important to consider whether the reaction is exothermic or endothermic when trying to determine the outcome of changing temperature on a reaction. If the reaction is is exothermic, raising temperature will favor the reverse reaction, which means it will increase the rate of the reverse reactio...

andrewcj 2C

k' is the rate constant for the reverse of a reaction. It can be used in conjunction with k to calculate the equilibrium constant by setting the two rates equal to each other.

andrewcj 2C

I think the most important thing pertaining to graphs is that you understand the relationship between the variables. For example, the graph of a 0 order reaction is a linear relationship between reactant concentration and time, whereas the graph of a 1st order reaction is a linear relationship betwe...

andrewcj 2C

The Arrhenius equation relates the rate constant to temperature and activation energy.

andrewcj 2C

When finding n though, make sure you don't accidentally double the value. You only need to count the electrons on one side of the redox reaction, as they are being transferred. So if 2 electrons are being transferred from Fe to Cl for example, make sure you don't count 4 from the 2 being transferred...

andrewcj 2C

You include liquids if they are involved in the reactions in the voltaic cell. However, if one electrode has a chemical in aqueous solution, but water is not part of any of the reactions in the voltaic cell, you need not write water into the cell diagram. However, the chemical in aqueous solution sh...

andrewcj 2C

My TA told me that test 2 covers only material after the midterm, so that the thermochemistry on it would only be delta G stuff. Is this true, or should we still go over entropy material?

andrewcj 2C

In the book, there are several problems in which a species acts as both an oxidizing agent and a reducing agent. Are we going to be responsible for knowing this for the test? I don't remember going over this in lecture.

andrewcj 2C

When looking at redox reactions or electrochemical cells, I find it helpful to label each element with its own charge, as it clears up any confusion that may arise, as you stated. If you label Mn with +7 in MnO 4 - , it becomes clear that it is being reduced, as it goes from a +7 charge to a +2 char...

andrewcj 2C

Concentration cells are galvanic cells in which both the anode and cathode have the same element.

andrewcj 2C

You can also examine the reactions happening at the electrodes. The oxidation reaction happens at the anode and the reduction reaction happens at the cathode. If it is still unclear, you can also check reduction potential, in the sense that E has to be positive for the galvanic cell to work spontane...

andrewcj 2C

In acidic solution, you add water to balance out the oxygen in the half reaction.
In basic solution, you do the same as above but you also add OH^- to the side with H⁺ in order to neutralize the acidity of that side, which makes water.

andrewcj 2C

I'm not entirely sure, but how I understand it is that with constant temperature and pressure, the energy change is equivalent to the work done, assuming no heat is released. This would be w(max), or the maximum amount of work possible.

andrewcj 2C

The salt bridge allows for positive charges to be equalized in the anode half cell and negative charges to be equalized in the cathode half cell. This prevention of charge buildup allows the galvanic cell to continue working.

andrewcj 2C

This is because if you use the normal method, you are saying there is a bunch of H+ in a basic solution, which makes no sense. By changing the H+ to H 2 O by combining it with OH - and adding OH - to the other side, the reaction equation now states that the reaction occurs with a bunch of OH - in ba...

andrewcj 2C

The work is done by the flow of electrons. So if there were some electronic gadget on the connection between the cathode and anode, it would be powered by the cell.

andrewcj 2C

From what I can gather, the midterm was quite difficult for most people, so there's a chance a few points may be added to normalize the grades of the class.

andrewcj 2C

It shouldn't happen, I don't think. In redox reactions we tend to focus on individual atoms being reduced or oxidized, so it shouldn't share electrons.

andrewcj 2C

Something I found helpful on the amphetamines problem was to list what you know. It helped me get insight on how to approach the problem, even when the solution wasn't obvious.

andrewcj 2C

It won't be cumulative. I believe it picks up from Gibbs free energy.

andrewcj 2C

delta G naught is under standard conditions. if conditions are not standard, use delta G.

andrewcj 2C

When the question says "its", it is referring to the entire system, which includes the kettle and the water.

andrewcj 2C

I don't think you need to memorize this. This is just saying the heat of combustion for x moles of ethane = moles of ethane * (heat of combustion/1 mole of ethane).

andrewcj 2C

In an irreversible process, heat is transferred out of the system, which increases the entropy of the universe. However, in reversible processes, theoretically, no heat is lost, which means the entropy of the universe does not change.

andrewcj 2C

The transfer of heat only stops when both objects are at the same temperature. This means the ice will continue to gain heat until it is melted and at the same temp as the water, or the water has frozen over and is the same temp as the ice. Since I'm assuming the water won't freeze, energy will be c...

andrewcj 2C

Yes, if you add matter to a system, the entropy will increase. More moles means more molecules which means more microstates for the molecules to be in.

andrewcj 2C

I assume you're referring to calculations using equilibrium constants. If the equilibrium constant you are using is less than 10 -3 , you can usually assume that subtracting x does not change a constant very much. However, this does not mean that x is zero. For example, if given 10 -5 = x 2 /(0.15-x...

andrewcj 2C

Make sure you only count gaseous moles though, as liquids and solids are not noticeably affected by pressure changes to the equilibrium.

andrewcj 2C

In a bomb calorimeter, the item being burned is considered part of the system. An isolated system cannot exchange matter and energy with its surroundings, but it can exchange matter and energy within itself.

andrewcj 2C

The heat from the burned skin would be transferred to the ice cube by conduction, and the ice cube would begin to melt. However, if it were applied after the skin was already burned, there would already be tissue damage, and the ice cube wouldn't reverse the burn.

andrewcj 2C

It is important to distinguish between breaking bonds and breaking the intermolecular attractions between molecules in a substance. During phase changes, the chemical bonds in the molecules of the substance aren't actually being broken. Rather, the energy added is being used to break the attractions...

andrewcj 2C

How is method 4 different than method 3?

andrewcj 2C

Using Le'Chatlier's principle, the system will work to counteract the changes brought upon it. Therefore, heating the reaction favors the reactants, as this uses up the heat. Conversely, cooling the reaction favors the products, as it would release heat, restoring the initial state.

andrewcj 2C

Something to remember is that strong acids don't have K values. In the k expression, the denominator would be essentailly 0, which is an invalid expression.

andrewcj 2C

To account for non-standard conditions, you just add the enthalpy changes required to get the reactants into their respective states in the conditions given.

andrewcj 2C

The second way Dr. Lavelle suggested to measure the enthalpy of a reaction was to use bond enthalpies. Essentially, you would find how much energy is required to break every bond in the reactants and add how much energy is released when every bond in the products is formed. However, a shortcut is to...

andrewcj 2C

It is important to remember that because of the Ideal Gas Law, gases in equilibrium have concentrations proportional to their partial pressures. This is why both may be used in equilibrium constant calculations, although not with each other.

andrewcj 2C

In short, the Law of Mass Action just states that for a set temperature, the equilibrium constant K does not change.

andrewcj 2C

What exactly do you mean by aqueous solution? From what I understand, we include ions in solution in equilibrium constants. However, pure liquids and solids are excluded because they have an activity of 1, and equilibrium constants are technically calculated with activity, not concentration.

andrewcj 2C

I'm not sure if this rule applies in this class, but my TA gave out a worksheet today in which we were supposed to approximate that subtracting X from a concentration doesn't change the concentration. The worksheet asks for you to check if the concentration of the product, or X, was less than 5% of ...

andrewcj 2C

If you see reverse and forwards reactions as arbitrary, I think it makes ICE tables easier. There is not a universal way to tell which way of a reaction is forward, it just depends on how its written. If ICE tables make sense to you for forwards reactions, just imagine the reverse reaction as the fo...

andrewcj 2C

When adding chemical equations, the reactants stay reactants and the products stay products. Therefore, in the combined equation, all reactants would be in the denominator and would be multiplied together. A similar logic can be applied to the products for the numerator. This is equivalent to multip...

andrewcj 2C

If there were, would that be a problem? Assuming the gas constant R was given in the correct units and the temperature was specified, I would think there would be enough information to solve a problem even if both partial pressures and concentrations were part of the information provided.

andrewcj 2C

Yes, the reaction quotient Q can indicate the direction of the reaction by comparing it to K. If Q < K, then the products side is favored and the forward reaction will proceed, and vice versa if Q > K.

andrewcj 2C

I think that increasing the partial pressure of a gas that doesn't participate in a reaction such as He has a similar effect as increasing the pressure of the system, assuming the volume of the reaction vessel is kept constant. If the partial pressure of He is increased without any of the other part...

andrewcj 2C

Thank you!

andrewcj 2C

In the post exam, there is a question 19b, but no question 19a. However, the question references the previous part, and doesn't provide sufficient information to solve it in its current form. I also tried using figures from question 18, but the answer I got does not match any of the available choice...

andrewcj 2C

It is always used in equilibriums to show a dynamic constant state of simultaneous reaction.

andrewcj 2C

CH 4 has weaker London forces than CCl 4 because it is smaller and has fewer electrons, which means the intermolecular forces in CH 4 are weaker than those in CCl 4 since both molecules are nonpolar. This means it takes less energy to break the attractions between molecules in a sample of CH 4 , res...

andrewcj 2C

i think for 14A, you only need to know how to calculate those values. In 14B, you can use those values in calculations and potentially get other values such as pH from Ka and pKa, but for now you don't need to worry about that.

andrewcj 2C

As I understand it,if a complex has the central metal atom as a part of a ring, it is a chelate.

andrewcj 2C

Why is HF a weaker acid than HI? When an acid dissociates, the bond must be broken between the H + and the conjugate base. If the bond is strong, which it is in the case of HF, this makes it harder and less of the acid dissociates, resulting in less H + present in the resulting solution and a highe...

andrewcj 2C

More generally, equilibrium arrows are used whenever there is an equilibrium present. This can be in reactions involving acids and bases, or just any reaction in general that can have an equilibrium such as the equilibrium between NO₂ and N₂O₄.

andrewcj 2C

The conjugate acid is [C₅H₅NH]⁺.

andrewcj 2C

It's worth noting that while H₂SO₄ is a strong acid, once it gives off its first proton, it becomes a weak acid. The HSO₄^- ion acts as a weak acid, as while it can give off its second proton, it usually doesn't.

andrewcj 2C

I feel like this is obvious, but it hasn't been mentioned so I'll say it. You can also compare pH (lower pH means stronger acid).

andrewcj 2C

Bond strengths is the most-correct reason that HCl is a stronger acid compared to HF. If we only consider electronegativity, we only focus on how strongly F can pull H's electron, which doesn't necessarily imply how easy it is to pull the proton off of HF.

andrewcj 2C

I don't think it matters, as if you don't, common sense tells you that the subscript is 1, and if you do, you definitely wouldn't get marked down for it.

andrewcj 2C

knowing bond angles of vsepr structures and how lone pairs can affect those bond angles is probably a good idea

andrewcj 2C

do diatomic molecules have a bond angle?

andrewcj 2C

it can bond with any N, O, or F atom with lone pairs available.

andrewcj 2C

The fact that transplatin can't bind to DNA means that it doesn't stop replication, which is the purpose of chemotherapy drugs.

andrewcj 2C

No. It's important to remember that in molecules with both single and multiple bonds, there only exist hybrid bonds of equal energy, and the localized double bonds depicted in lewis structures only exist as a limitation of lewis structures. This is why resonance structures exist: because the multipl...

andrewcj 2C

It's important to remember that the shaded/dashed triangles still represent bonds, just they help show a 3D shape in 2D.

andrewcj 2C

Some shapes come from others, so it might not be necessary to explicitly memorize every shape. For example, bent comes from a tetrahedral shape with 2 lone pairs, but you don't have have to necessarily memorize bent. Instead, just think about a tetrahedral shape and remove 2 atoms.

andrewcj 2C

Generally, an A represents the central atom, an X represents an atom bonded to it, and an e represents a lone pair. For example, the formula of a tetrahedral shape would be AX₄, and the formula for a bent shape would be AX₂e₂.

andrewcj 2C

A good way to think about this is to compare a known polar molecule to a known nonpolar molecule. Take H 2 O and CH 4 for example. We know water is polar, and we can see that the net dipole moments do not cancel, and the molecule is left with a positive dipole by the hydrogen atoms and a negative di...

andrewcj 2C

I learned that you should try to minimize the absolute charge on the atoms in the molecule/ion. Therefore, if an ion has an overall +1 charge, it should be achieved by 0 charges and a single +1 charge, instead of say a -3 and a +4 charge.

andrewcj 2C

Recall the rules for drawing Lewis structures. The central atom is the one that is the least electronegative, which means it has the least electron pulling power. Therefore, it would probably be best for it to not have a negative charge.

andrewcj 2C

The expanded octet comes from access to d orbitals, which can be used to create additional bonds to the 4 allowed by the s and p orbitals. In general, if an atom in a molecule can achieve a lower formal charge and make the molecule have a more logical formal charge by making more than 4 bonds (there...

andrewcj 2C

A friend gave me the mnemonic "Lewis has e" which reminds me that Lewis acids and bases focus on electron exchange. After this, I just think about H+, which is acidic, and how it can accept electrons. Therefore, Lewis acids are electron acceptors.

andrewcj 2C

I believe the final is cumulative, but it is generally a good idea to remember the concepts covered so far, as they may come up in future chem classes.

andrewcj 2C

I suppose the trial and error gets easier with experience, as it is possible to identify patterns such as certain functional groups or complex ions in a structure.

andrewcj 2C

If one atom is more electronegative than the other, the electrons will spend more time on average around that atom, and the character of the bond would become more ionic.

andrewcj 2C

Be wary of resonance structures any time you see a multiple bond, as you just have to think about if you could put the multiple bond somewhere else.

andrewcj 2C

They do not always violate the octet rule. A simple counterexample would be HCl, which follows the rule.

andrewcj 2C

A trick I use is to just memorize that fluorine is the most electronegative element, and note its position on the periodic table.

andrewcj 2C

Debora Fernandez Clemente_ 4H wrote:would the bond length vary when it is a double or triple bond?

Yes! Multiple bonds will be shorter than single bonds.

andrewcj 2C

While my high school AP Chem teacher emphasized that this is a very unscientific way to think about it, I think it makes sense to think about electron affinity as how badly a neutral atom "wants" an electron. So for elements such as fluorine, which is so close to achieving a stable noble g...

andrewcj 2C

Does anyone else find the wording a little ambiguous? Because I think you can make the argument that fluorine is extremely reactive too, due to how close it is to a noble gas electron configuration, but fluorine is definitely not an s-block metal. In this case, fluorine readily takes electrons from ...

andrewcj 2C

Another way to look at this (I think) is to look at bond order. If a molecule has resonance, its bond order will be higher than a molecule with only single bonds. In other words, its "average" bond strength will be in between that of a single bond and a double bond, and since multiple bond...

andrewcj 2C

I'm having a little trouble with problem 1E.5. It reads as follows: Which of the following statements are true for many-electron atoms? If false, explain why. (a) The effective nuclear charge Z eff is independent of the number of electrons present in an atom. (b) Electrons in an s-orbital are more e...

andrewcj 2C

I believe there's a table on Lavelle's course website with those masses.

andrewcj 2C

Reading involves visible light.
Getting a dental x-ray involves x-rays.
Using a microwave results in microwave radiation.
Tanning involves UV rays.

andrewcj 2C

In the formula E = hv, E is measured in Joules and v is measured in Hz, or s^-1. By dimensional analysis, we see that by multiplying kg x m^2 / s by 1/s, we get a units of kg x m^2/s^2. Since Joules is a measure of energy/work, it is equivalent to force x distance. Breaking this down further, force ...

andrewcj 2C

Part of the experiment demonstrating the photoelectric effect revealed that more electrons can be ejected from metal by shining light on it if the wavelength of the light was increased. If light was viewed as a wave, we would expect increasing intensity to correlate to more electrons being ejected. ...

andrewcj 2C

Wavelength refers to the length of a wave, which is measured in meters. Just in case you needed the conversions, 1 nm = 10^-9 m. I also think that these conversions will be provided on tests.

andrewcj 2C

Dimensional analysis will reveal that the units cancel out, and you end up with an answer in grams when you multiply the 2 quantities.

andrewcj 2C

I thought it was by wavelength, not frequency, although I guess because they're proportional it doesn't matter.

andrewcj 2C

I think in the future, we may be asked what color a specific wavelength of light given off is (i.e if an electron emits a photon with a wavelength of 400, what color is the light?)

andrewcj 2C

One additional thing for my reply, make sure you convert nm to m.

andrewcj 2C

I think it goes like this:
( $\Delta$ p)( $\Delta$ x) = (1/2)(h/2 $\pi$ )
( $\Delta$ p) = m( $\Delta$ v)
From here you can plug in (0.01)*(0.05 nm) for $\Delta$ x and solve.

andrewcj 2C

Can someone explain 1.3 to me? The question is as follows: Which of the following happens when the frequency of electromagnetic radiation decreases? Explain your reasoning. (a) The speed of the radiation decreases. (b) The wavelength of the radiation decreases. (c) The extent of the change in the el...

andrewcj 2C

My best guess would be that rounding caused the masses of the individual elements to be a little smaller than they really are, which creates a small discrepancy in the summed mass and the total mass. In real life, our instruments are not always perfect, and this might be trying to show that.

andrewcj 2C

I really think 5293 has 4 sig figs, not 5. I think if it was 5293.0, it would have 5 sig figs.

andrewcj 2C

Suppose the activation energy is too high for the reaction to proceed without a catalyst. Does that make a difference?

Search found 102 matches