CHEMISTRY COMMUNITY

Ivy Lu 1C

You can use either method of solving for E, but just make sure you stay consistent and don't combine the two. If you use cathode-anode, then don't flip the sign of the E for the anode.

Ivy Lu 1C

It still depends on what you're solving for. If you're not given pressure and solving for pressure, then you would use 8.206 x 10-2 L·atm·K-1·mol-1, because that would cancel out so that you get atmospheres for the units of pressure. If you're not given pressure and trying to solve for work, then yo...

Ivy Lu 1C

A metal will dissolve in the solution if the standard cell potential of the redox reaction is positive, meaning the standard Gibbs free energy is negative because that would mean the reaction is spontaneous.

Ivy Lu 1C

A carbonyl is a functional group with C double bonded to O. Aldehydes and ketones are both carbonyls. Aldehyde has the carbonyl group bonded with either one or two hydrogens. Ketone has the carbonyl group bonded with two carbons.

Ivy Lu 1C

Redox reactions in basic solutions don't need to have OH- in the final balanced equation if they cancel out. OH- would only show up in the final equation when the two sides of the equation don't require the same amount of OH- when balancing.

Ivy Lu 1C

That information should be given on a test because we are not allowed to use graphing calculators.

Ivy Lu 1C

They are not interchangeable. If you want to use log instead of ln then remember that ln(X) = 2.303log(X). If an equation has ln, then you should use ln. If an equation has log, then you should use log. This way it's easier to avoid error.

Ivy Lu 1C

Unless it's given [A]0 then [A] is just the concentration at the given time, not the initial concentration.

Ivy Lu 1C

When determining the rate laws of a reaction with multiple reactants, we use the reactant of the smallest concentration because that reactant will determine the rate of the overall reactions. Since the other reactants are in excess, they don't really affect the rate of the reaction. This way, we are...

Ivy Lu 1C

A reaction rate is independent when an increase of concentration doesn't affect of the rate of the reaction. This type of reaction would be a zero order. When given a graph, it is when the graph of concentration vs. time is linear.

Ivy Lu 1C

A 0 order reaction would have a linear graph of concentration vs time. The slope of the line equals -k, and the y-intercept equals the initial concentration.

Ivy Lu 1C

You would have to look at the experimental data to find the order or reactions because the chemical equation doesn't tell you anything about the order. You would have to look at a table of data and solve for the order or look at different graphs and determine what the order is.

Ivy Lu 1C

It should be safe to assume that that it's in standard state, if no additional information about concentration or partial pressure is given.

Ivy Lu 1C

When you take the natural log (ln) of both sides, you get rid of e because ln is equal to log_e. Once you do this, you just solve for deltaG.

Ivy Lu 1C

You want the standard potential of the reaction to be positive because that would make deltaG be negative (deltaG=-nFE) and k be greater than 1.

Ivy Lu 1C

The n in the equation is the moles of electrons transferred. To find the electrons transferred, you'll have to write the two half-reactions and balance them.

Ivy Lu 1C

The oxidizing agent is the substance that's getting reduced. The reducing agent is the substance that's getting oxidized.

Ivy Lu 1C

The charges should equal each other on each side. If they don't equal each other, then the redox reaction is not balanced.

Ivy Lu 1C

A salt bridge connects two separate beakers while a porous disc is a separation between two solutions in the same beaker. They both serve the same purpose of allowing ions to flow between the two solutions.

Ivy Lu 1C

A negative delta G is spontaneous because there is a lost in free energy, and going from high free energy to low free energy is favorable. A positive delta G is nonspontaneous because there is a gain in free energy. For that to happen, energy would have to be added, making the reaction nonspontaneous.

Ivy Lu 1C

||represents a salt bridge. | separates the elements of different phases. A comma separates differences in oxidation states.

Ivy Lu 1C

The heat capacities, 3/2*R=Cv and 5/2*R=Cp, relate to when there is an ideal gas at constant volume/pressure. For this problem, the water is in its liquid state, so the heat capacity of an ideal gas would not apply.

Ivy Lu 1C

∆G° is the standard free energy of the reaction. This means that the reactants and products are at their standard state: 1 atm for gases and 1 M of aqueous solutions, typically at 25°C. ∆G is just the free energy of the reaction that doesn't have to be under the standard state conditions.

Ivy Lu 1C

The 6 is raised to avogadro's number because the question is asking for the molar entropy, so that's the entropy per mole of SO2F2.

Ivy Lu 1C

The question says to assume ideal gas behavior, but no moles of gas are given, so it's safe assume that there's only one mole of N2 gas to make the calculations easier.

Ivy Lu 1C

Change in enthalpy, delta H, is equal to heat loss/gained under constant pressure. If there's constant pressure than in the equation delta S=q/T, q can be replaced by delta H. Under these condition, change in entropy would be directly proportional with change in entropy.

Ivy Lu 1C

If you look on Lavelle's website, there is a link to solution manual errors. For question 9.13, R should be multiplied by 5/2, so the equation would be S=nCln(T2/T1). The work in the solution manual is just wrong for that part.

Ivy Lu 1C

The 101.325 J.L-1.atm-1 is just the conversion from L.atm to Joules. On the test, this number will be given to you on the constant and equations sheet.

Ivy Lu 1C

If the system is heated up, then the work is positive because work has to be done on the system for it to take in heat. If the system loses heat, then the work is negative because the system did work to lose heat.

Ivy Lu 1C

Part b does more work because a reversible process is the maximum output for the process because it's done slowly, so less energy is lost. In a irreversible process, more heat is lost because it's done quickly.

Ivy Lu 1C

There is a plateau even though heat is being added because all the heat being added will go to breaking the bonds first, before the temperature will start changing. While the bonds are breaking, the temperature isn't changing because the energy added will only go to breaking the bonds.

Ivy Lu 1C

Heat capacity is an extensive property because it does depend on the amount of substance. However, specific heat capacity is per gram of substance and molar heat capacity is per mole of substance, so they are intensive properties.

Ivy Lu 1C

Heat, q, refers the amount of work the system can do. Enthalpy, H, refers to the amount of heat absorbed or released by the system. Heat can equal enthalpy if under constant pressure.

Ivy Lu 1C

Since the pH is related to the amount of concentration of hydronium ions, the pH of an acid in a solution could be greater than 7. If the concentration of acid is very little than the solution would have a higher pH because there wouldn't be much acid in the solution.

Ivy Lu 1C

To see if a ligand if a bidentate, tridentate, or hexadentate, you look to see how many lone pairs there are one different atoms. The best way to see this is by drawing a lewis structure. For example, a bidentate would have two lone pairs, each on different atoms.

Ivy Lu 1C

It wouldn't be wrong to use concentration, but normally you would use partial pressure if all the reactants and products are gas. If you were given concentrations, then you could just use the concentrations.

Ivy Lu 1C

P4 is non polar. Since all of the atoms in the bond are the same, there is no polarity in the bonds between the atoms, so the entire molecule is non polar.

Ivy Lu 1C

A polydentate ligand has more than one donor atom, so it can occupy more than one binding site at a time.

Ivy Lu 1C

The reason OH2 is used is to how that the oxygen is being bonded to the central atom not the hydrogen. It would be written OH2 when attached to the right of the central atom and written H2O when attached to the left of the central atom.

Ivy Lu 1C

If two molecules have the same VSEPR formula, then they have the same shape. From the formula, we can only estimate what the bond angle is, though they may vary a bit for different molecules. We can't find out the true bond angle with the VSEPR formula, that has to be found experimentally.

Ivy Lu 1C

Polar molecules have a molecular shape where the dipole moments don't cancel each other. Nonpolar molecules have molecular shapes there the dipole moments do cancel each other out. It also depends on what the atoms are because dipole moments are different, they might not always cancel out.

Ivy Lu 1C

A biradical is a molecule with two unpaired electrons and a lone pair is when two valence electrons that aren't part of bonding for an atom. For a biradical, the two unpaired electrons on usually on different atoms.

Ivy Lu 1C

Resonance structures are delocalized, which means the bond pair is distributed to multiples pairs of atoms and not just one pair of atoms. Because of this, the energy is lower which also makes the molecule more stable.

Ivy Lu 1C

This only refers to electrons in individual atoms. For each atom, no two electrons can have all four quantum numbers be the same due to the Pauli exclusion principle. Because of this, each orbital has only two electrons, and they have opposite spin.

Ivy Lu 1C

Zn1+ has that configuration because when you're writing the electron configuration for an cation you take electrons from the outermost orbital. In this case, you're taking the electron from the 4s orbital.

Ivy Lu 1C

In the cases where 4s comes before 3d, it just has to do with the charge of the atom. When n increase, the subshells are closer together and the charge of the atom can cause 4s to be filled first like with potassium and calcium.

Ivy Lu 1C

For part b and c they ask about the values of m l which is the orbital. Those values are from -l,-l+1,..., l. The orbital is the different possible shapes for each subshell. Subshells refer to l. l=0 is the s-orbital, l=1 is the p orbital, and l=3 is the d orbital. That's why for part d, n=4 so l=3,...

Ivy Lu 1C

In atomic spectroscopy, the electron doesn't completely leave the atom because it is still being pulled towards the positively charged nucleus of the atom. Also with the photoelectric effect, the detector that detects the speed of the ejected electron is slightly positively charged so it'd draw the ...

Ivy Lu 1C

First, you would convert the 5.00 mg to atoms. Once you have that, you just multiply by the Joules per atoms that you got from part a. The atoms of sodium should cancel leaving you with just Joules.

Ivy Lu 1C

Photons don't have mass so you wouldn't really be able to use the equation p=mv. Sorry, the first time I read that as proton.

Ivy Lu 1C

When you use 5.77g P4 again in the second question, it's just to convert it to P4O6 because you want to find how much P4O6 came out of the first equation to use in the second equation. You would use the mole ratio so that 1 mole P4 equals 1 mole of P4O6, and then solve for how much oxygen you need, ...

Ivy Lu 1C

You need to solve for the masses of carbon, hydrogen, and nitrogen there is based on the masses of CO2, H2O, and N2 given, and then add up the masses of C, H, and N. You should see that the sum of the masses for C, H, and N isn't equal to the mass of caffeine given in the question, so that means som...

Ivy Lu 1C

There are two equations to the problem. For the first equation, if you use all the P4 and calculate the amount of O2 needed, you get 4.47g O2. Since you originally had 5.77g O2, you still have 1.30g O2 left. For the second equation, you use 5.77g P4 again but in the context of the second equation to...

Ivy Lu 1C

I don't still don't really get how to do this problem. Can someone explain how to do it more?

Search found 54 matches