CHEMISTRY COMMUNITY

Brooke Yasuda 2J

A catalyst weakens the bonds of the species, ensuring that less energy is required to obtain the necessary activation barrier energy. This is how the reaction rates of the forward and reverse reactions are increased.

Brooke Yasuda 2J

A reducing agent has a tendency to reduce another species (which means it itself is being oxidized) and an oxidizing agent has a tendency to oxidize another species (which means it itself is being reduced). The way you determine if something is an oxidizing or reducing agent is by looking at the ele...

Brooke Yasuda 2J

This may be the longer way, but it works. To find the units of the rate constant, you just want to remember that your rate needs to have units of (mol/L/s). So when you have your expression which is something like rate = k*[x]^n[y]^m, you can determine that k multiplied by the rest of the expression...

Brooke Yasuda 2J

Yes, also remember that the value of delta G can help to determine the direction of the reaction that is favored. For example, a more negative delta G means that it is spontaneous and favoring the forward reaction. As the reaction progresses and moves towards equilibrium, the concentrations of the r...

Brooke Yasuda 2J

One way you could do this problem is by using the equation for a second order half life reaction and calculating the K value. With this you can then use the integrated rate law for a second order reaction to find the time needed to reach the concentration that is being asked.

Brooke Yasuda 2J

Yes, so the value of n would be 2 in this case. As you can see, one half reaction is 2Ce4+ --> 2 Ce 3+. On the left side of the equation, 2 * 4 = 8 and the right side, 2*3 = 6. So, you have a net charge difference of 2. On the other hand, we have 3 I- --> I 3-. In this case, we have a -3 on the left...

Brooke Yasuda 2J

Yes, the anode is the product and the cathode is the reactant. This makes sense because in the anode the concentrations of reactant are decreasing as it forms product in the cathode.

Brooke Yasuda 2J

I believe you forgot that the n value is equal to 2, not 1. So you would have t divide .05916 by 2 to get the correct answer

Brooke Yasuda 2J

The cell potential of the cathode and anode half reactions are usually found in a table unless you have closely related elements in which case you would calculate the cell potential. But otherwise, they are usually given and you would just have to determine which half reaction is oxidation and reduc...

Brooke Yasuda 2J

The n value that is referenced in this equation is the number of electrons that is transferred in the redox reaction. So yes, you should write and balance the half reactions and then plug in the value of n into the equation to find the Gibbs Free Energy.

Brooke Yasuda 2J

Yes and when Ecell is equal to zero, Gibbs free energy is also zero.

Brooke Yasuda 2J

I think what happened is that the equation is Ecell = Eanode - Ecathode. So you have -.698 = Eanode - (.34 V). So I think you just forgot that the species that is oxidized is subtracted from the species that is reduced.

Brooke Yasuda 2J

You have to first balance the half reactions so that there is conservation of charge. The molar coefficient of electrons is going to be your n value.

Brooke Yasuda 2J

When balancing a redox reaction under acidic conditions, you use H2O and Hydoxide ions (H+) to balance the number of Oxygens and Hydrogens on both sides. And then you finish the redox by balancing electrons for the reaction. On the other hand, for basic conditions, you use H2O and Hydronium ions (OH...

Brooke Yasuda 2J

You add H2O molecules in order to balance the number of oxygens that are on both sides of the reaction. After you balance the number of Oxygen molecules you balances the number of Hydrogens with H+. After doing this, you finally balance charge by adding electrons to the respective sides of the react...

Brooke Yasuda 2J

Maximum work occurs right before current flows because immediately when current begins flowing, the electric potential begins to decrease until it reaches zero. So, when asked a question about maximum potential, it is right before the circuit is closed or right before any reaction begins.

Brooke Yasuda 2J

A battery is dead when the system is at equilibrium. This occurs when there is a buildup of charge in the cathode and the anode, so there will no longer be any flow of charge. This can be avoided by having something like a porous disk to maintain a neutral state in the anode and cathode.

Brooke Yasuda 2J

For this you have to look at the Ph of the solution and the PKa. Because the Ph is more acidic than the PKa, there will be no dissociation. Dissociation will only occur if the Ph is less acidic than the PKa because then the salt could dissociate and make the solution more acidic to reach the Pka val...

Brooke Yasuda 2J

Yes, so the importance of this property comes in when you are balancing your redox half reactions. In redox, you want to make sure that the amount of electrons lost is equal to the amount of electrons that is gained. Just like other problems from the last section, you use molar coefficients to balan...

Brooke Yasuda 2J

Yes, and it's important to note that when there is that buildup of charge and the battery is "dead", the system is at equilibrium so the gibbs is zero and there is no energy for useful work

Brooke Yasuda 2J

Gibbs is a state property because all of its components, enthalpy, temperature, and entropy, are also state properties.

Brooke Yasuda 2J

an adiabatic process is one in which there is no change in q during the reaction, so no heat is exchanged.

Brooke Yasuda 2J

And for irreversible you can also look at the q of the surroundings because this could affect the total entropy change.

Brooke Yasuda 2J

The total entropy change also equals zero in an isothermal reaction. In this case, the q of the surroundings is equal and opposite in sign to the q of the system. So, the entropy change total is zero

Brooke Yasuda 2J

I think we only need to know the constants for a monoatomic gas b and if a question has a diatomic gas, the required constants will be given to us in the problem.

Brooke Yasuda 2J

We can approximate here and assume that it is a reversible reaction because the surroundings are so large that any change in heat will have little to no effect on the the surroundings.

Brooke Yasuda 2J

I would look at the equation sheet so that you know what equations will be provided. Other than that, you should probably know all of the equations from lecture and the equations used in homework problems

Brooke Yasuda 2J

a bomb calorimeter is tightly sealed and insulated which means that heat cannot be transferred with the surroundings. Also, the volume of the system cannot change so there can be no work done. This means that it is an isolated system.

Brooke Yasuda 2J

In an irreversible expansion, the external pressure is not changing. This means that there will be sudden expansion against an external pressure. In contrast, a reversible expansion is occurring when the system is in equilibrium. A very small incremental change in the external pressure causes a slow...

Brooke Yasuda 2J

I think that for most problems, if the information is not give you just have to know the standard conditions and assume these are the values in the problem.

Brooke Yasuda 2J

if it says anything about the system being in equilibrium or being reversible, you know that the external pressure can change

Brooke Yasuda 2J

When there is no expansion or compression, which is work, we can say that Delta U is equal to the q value.

Brooke Yasuda 2J

A bomb calorimeter is a calorimeter specifically for a reaction in which the volume remains constant. Because volume is constant, the pressure continues to build up and so this is a special calorimeter for a constant volume reaction.

Brooke Yasuda 2J

remember it is not technically breaking bonds, as in covalent bonds between Hydrogen and Oxygen molecules. It is breaking intermolecular forces between water molecules. Therefore, energy is needed to overcome and break the intermolecular forces which requires heat.

Brooke Yasuda 2J

Since the volume is constant, the pressure continues to build up as the reaction continues. Therefore, the special calorimeter for this type of reaction is a bomb calorimeter.

Brooke Yasuda 2J

Yeah, and in calorimetry problems we can say that all of the heat lost by the reaction is gained by the calorimeter, and vice versa. So the heat lost by the surroundings is gained/absorbed by the reaction

Brooke Yasuda 2J

So when AlCl3 dissociates in water, Cl- is the conjugate base of a strong acid so will not react at all with water. However, Al will coordinate with six water molecules. After it coordinates with water, it can actually give off protons, or acidic hydrogens, making it a weak acid. So you would have t...

Brooke Yasuda 2J

When you look at which value is going to be used, you have to look at whether or not when this value is subtracted from a reactant or product, it will be negative or positive. So assume you start off with some value of reactant and no product. As the system moves towards equilibrium, product is form...

Brooke Yasuda 2J

It's because of the ICE table that is set up. The equation is H2 + Cl2 --> 2HCl. So, when you do an ICE table, you represent the change as some variable like x. So the H2 and Cl2 concentrations will decrease both by x, and the HCl concentration will increase by 2x. It is 2x because there is a 1:2 ra...

Brooke Yasuda 2J

The equation would be NH3 + H2O --> NH4 + OH. It's given in the problem that 1E-3 Molar Hydroxide is present at equilibrium. Therefore, this must be the final equilibrium concentration of NH4 and OH because it is a 1:1 ratio. Percent deprotonation is the molar value of NH4+ formed at equilibrium div...

Brooke Yasuda 2J

From this language you can infer that the concentration given is the initial concentration, and 2.4% of this initial concentration deprotonates.

Brooke Yasuda 2J

X must be smaller than .22 because the initial concentration of HCL is .22 M. If the x is larger than .22, then the final equilibrium concentration of HCl would be negative and negative concentrations are not possible.

Brooke Yasuda 2J

Yes. If the partial pressure of CO2 increases, then Q is going to be greater than K. As a result, the reverse reaction is going to be favored, and the partial pressure of H2 decreases.

Brooke Yasuda 2J

Yes, you look at temperature that is given in the problem. And this also helps to solidify the concept that equilibrium constants are temperature dependent. So at different temperatures, the equilibrium constants are different.

Brooke Yasuda 2J

So the moles of the reactants are given and the volume is also given. With this, you can calculate the concentration of N2 and O2. Then, you set up an equilibrium ICE table. When you do the "change" part of ICE, remember to look at the molar ratios of the reactants and products. Then, calc...

Brooke Yasuda 2J

Yes, when the partial pressure of NH3 decreases, this will cause the Q value to be smaller than K. As a result, the reaction will favor the reactants and so the O2 concentration would increase

Brooke Yasuda 2J

When you have a reaction involving ionic compounds in a solution you want to rewrite the equation as the net ionic equation. This means, not writing anything that repeats on both sides of the equation, and would cancel out. This includes spectator ions, which are ions that come from an ionic compoun...

Brooke Yasuda 2J

so liquids that act as solvents can be involved in a reaction, such as dissolving an ionic compound. However, because the liquid such as water is the solvent, there are so many more water molecules per molecule of the ionic compound, that there is water on both sides of the equilibrium constant equa...

Brooke Yasuda 2J

When a question asks about which is more stable, it's like looking at lewis structures and determining which has stronger bonds and is thus, less likely to react. In this case, because there is more leftover unreacted Cl2 than F2, this shows that Cl2 is less likely to react into Cl and is, therefore...

Brooke Yasuda 2J

A way you could look at it is either that in the second experiment there is a greater number of moles of O2 within the same volume so the partial pressure of O2 must be larger as well. Or you could look at it in the perspective of equilibrium constants. For the system to be at equilibrium, the Kp mu...

Brooke Yasuda 2J

Yes, this will ensure that the equilibrium constant will remain the same given that the reaction and temperature of the reaction do not change.

Brooke Yasuda 2J

Most of the time it will be an electron and it will tell you the radius or diameter and you have to solve for the uncertainty in the velocity. When you are doing Heisenberg problems with electrons remember mass can't have uncertainty because it has a known value, and so you will most likely be solvi...

Brooke Yasuda 2J

It is still in alphabetic order because you look just at the ligand names and you have aqua and oxalato. Aqua is first alphabetically, so is written first.

Brooke Yasuda 2J

Acetic acid has the CH3 group, which actually makes the anion less stable because instead of delocalizing the electrons like we want, it contributes to the high electron density around the Oxygen in the anion. Formic acid does not have this CH3 group and instead just has the Oxygen single bond which...

Brooke Yasuda 2J

He said in class that we can use the new naming system or what is used in the textbook so long as the answer is correct.

Brooke Yasuda 2J

If an electron is in a higher energy orbital, then it is also in an orbital that is farther away from the nucleus. As a result, the radius of the atom is larger.

Brooke Yasuda 2J

He said that we can use either one, just make sure the answer is correct.

Brooke Yasuda 2J

When an electron is an infinite distance from the nucleus, we say that the energy is zero. So, when the electron comes closer to the nucleus, the Energy is negative. An electron from a higher to lower principal quantum number will be coming closer to the nucleus, so will become more negative. This w...

Brooke Yasuda 2J

Yes, there will be coordination compounds on the final. For problems I would just go over the questions from the homework and make sure you understand the questions from the chapter guidelines.

Brooke Yasuda 2J

The reaction for NH3 is NH3(aq) + H2O(aq) --> NH4+(aq) + OH-(aq)

Brooke Yasuda 2J

en is ethylenediamene and edta is ethylenediamenetetraacetic acid. You can find these in the table in the textbook also

Brooke Yasuda 2J

when they ask about which is more tightly bound look at the bond length. And to do this, look at the size of the atoms, so S and O. Because O is smaller than S, its bond length is shorter which means that it is stronger and more tightly bound.

Brooke Yasuda 2J

Every molecule has the potential to form dipole induced dipole induced interactions. Dispersion forces, or dipole induced interactions, are when electrons in the electron cloud randomly, by chance, can shift to one side creating a temporary dipole moment. The partial positives and partial negatives ...

Brooke Yasuda 2J

In choice III, while it is true that there are favorable attractions resulting from the chlorine atoms being close to the Hydrogen atoms. However, in this orientation there are also Chlorine atoms that are facing each other, and this will cause repulsions that are not favorable. In contrast, choice ...

Brooke Yasuda 2J

When you're doing a boiling point problem first make sure that you are looking at intermolecular forces, not intramolecular forces, because this has to do with phase changes. When I look at boiling point problems I first look to see what types of intermolecular forces are involved and I look for the...

Brooke Yasuda 2J

The number of electrons has to do with London Dispersion forces. The more electrons a molecule has, the more polarizable it is. And as we saw with the equation Potential Energy is proportional to - (polarizability of molecule 1) multiplied by (polarizability of molecule 2) divided by the distance to...

Brooke Yasuda 2J

CHI3 because Iodine is a larger atom which means that it has more electrons, and therefore, has stronger induced dipole induced dipole interactions.

Brooke Yasuda 2J

VSEPR is a rule that describes how atoms in a molecule are likely to move as far away from each other as possible to reduce repulsions. Therefore, when you are drawing your geometric shape of a molecule, count how many bonds there are and you have to memorize how the number of lone pairs on the mole...

Brooke Yasuda 2J

There are always induced dipole induced dipole intermolecular forces between all molecules and these are temporary dipoles. Dipole dipole moments and other dipoles occur when molecules are polar, so they are permanent.

Brooke Yasuda 2J

Between all molecules there exists induced dipole induced dipole intermolecular forces, even if the molecule is nonpolar. In this case, O3 has a bent geometric shape, which means that it is a polar molecule, and therefore, has dipole-dipole interactions

Brooke Yasuda 2J

In this case, both molecules have both induced dipole induced dipole and dipole dipole intermolecular forces, but because pentane is a much smaller molecule, it has less electrons and a smaller surface area. Because of this, pentane will have weaker induced dipole induced dipole forces, and therefor...

Brooke Yasuda 2J

A dipole moment is a charge separation that is caused by differences in electronegativity differences. Therefore, if the two atoms that are bonded together are the same element, they have the exact same electronegativity and will be non polar. Otherwise, if there is a difference in electronegativity...

Brooke Yasuda 2J

So this molecule has three resonance structures because there is a double bond and two single bonds between the carbons and oxygens. As a result, because we know that in resonance, the actual structure is a blend/hybrid of all of the resonance structures, all of the bond lengths are the same. As a r...

Brooke Yasuda 2J

Polarizing power is related to positive nuclear charge of a cation. Generally, more polarizing power means that it has a smaller radius and therefore is holding onto its electrons very tightly. These atoms do not have a lot of electron shielding, so has a very strong positive charge.

Brooke Yasuda 2J

Most of the time when you have a molecule that can form an expanded octet, such as Xenon, you know that it has eight valance electrons. In bonding, you're trying to reduce formal charge, so you can form double bonds between Xenon and other atoms to reduce the formal charge as much as possible.

Brooke Yasuda 2J

Yes, when there is an odd number of valance electrons, you want to follow the same bonding conventions as usual and put negative charges on the most electronegative atoms

Brooke Yasuda 2J

Remember that formal charge equation is #valance electrons - ((bonding electrons/2) + electrons in lone pairs)

Brooke Yasuda 2J

if you are trying to find the mass of one atom of potassium to substitute into de broglie's equation, you would have 1 atom of K, multiply by the reciprocal of avogadro's number to get moles. From moles convert to grams, from grams convert to kilograms. From there substitute into de broglie's equati...

Brooke Yasuda 2J

threshold energy is also called the work function. Remember in the photoelectric effect experiment, the incoming energy of electromagnetic radiation needed to reach a certain threshold energy, or threshold frequency, to be able to eject an electron. You can calculate this with the equation that stat...

Brooke Yasuda 2J

We won't need to know exact numbers, just trends. Electronegativity increases up and to the right of the periodic table.

Brooke Yasuda 2J

High electronegativity means that it has a high attraction to electrons. Therefore, atoms that have high electronegativity are likely to form bonds by accepting electrons from other atoms, to fill its valance shell

Brooke Yasuda 2J

yeah we are supposed to for sure know the wavelength of visible light. Also, it's easier and recommended that you memorize the wavelengths instead of frequency because the numbers are much easier to memorize!

Brooke Yasuda 2J

yes, when an atom is in a noble gas configuration, its valance shell is completely filled which means that it is stable. This is why the ionization energy is going to be extremely high. When the valance shell is filled, the atom is stable, and is not going to want to gain or lose any extra electrons.

Brooke Yasuda 2J

just remember that for ground state electrons, the electrons are in the lowest energy levels possible. So the electrons are not in an excited state.

Brooke Yasuda 2J

Even thought they are in the same shell, n = 2, they are in different subshells. The 2s subshell provides slight electron shielding to the 2p subshell which means that the effective nuclear charge is not felt by the 2p electrons as strongly.

Brooke Yasuda 2J

Yeah, look at the periodic table. But also, you can just know certain facts like when n = 1, there is only the s subshell. For n = 2, there is only the s and p subshell. for n = 3, there are s, p, and d subshells, and so on. And when writing electron configurations, you always write in the order of ...

Brooke Yasuda 2J

constructive interference is when two waves interacting with one another are in phase with one another. So, when the waves interact the amplitude of the waves add together. Destructive interference, on the other hand, is when the waves are out of phases, so when they interact with one another, the w...

Brooke Yasuda 2J

Yes, there is also the f orbital which is at the very bottom of the periodic table, but for the purposes of our class, we don't need to memorize anything about the f block.

Brooke Yasuda 2J

I think in class Dr. Lavelle said that we won't be asked to draw the orbitals on tests, but in general, you can guess the shapes based on the axes that they are on. So just remember which way is the x, y, and z axis.

Brooke Yasuda 2J

For the grades I think that it just depends on who your TA is. Every TA updates the grades at different times in comparison to each other.

Brooke Yasuda 2J

Atomic spectroscopy is the analysis of electromagnetic radiation that is emitted or absorbed and corresponds to the spectral lines that we see. These lines as we talked about in class, are unique to every element and are a result of the particle characteristics of electrons. This is because spectral...

Brooke Yasuda 2J

Yes, so for how to approach or attack these problem, you should start off by looking at what is given and what you are looking for. For example, in part a you are given the molarity of your concentrated solution (.50 M). You are also told that the final product you are looking for is 1.00 L of .50 M...

Brooke Yasuda 2J

When you look at light as a wave, the intensity of light refers to the amplitude of the wave. In the world, you will see the increase in the amplitude of a wave as an increase in the brightness of the light source.

Brooke Yasuda 2J

For this test, just make sure that you have your ID number memorized!

Brooke Yasuda 2J

The work function is the minimum amount of energy a photon needs to have in order to eject an electron. In other words, it is the threshold energy. You can solve this by the equation E(photon) - work function = E(excess/KE).

Brooke Yasuda 2J

I do not think that you will be marked off if you don't rearrange the equation. What matters is that you show your work on how you get the answer and that the actual answer is correct.

Brooke Yasuda 2J

a pm is 1E-12 m. And yes, the answer should usually be in meters but it doesn't really matter. I would say it depends on what the question is asking for.

Brooke Yasuda 2J

First you would convert all of the masses of compounds into moles. From there, you would look at the molar ratios of the potassium and sodium ions to their prospective compounds, and calculate the moles of these ions within the solution.

Brooke Yasuda 2J

When you begin with a dilution or molarity problem, first, just remember that molarity is the total moles of some solute (mol) divided by the total volume of a solvent (L). So with all of the given information, if you have something in grams I would begin with converting it to moles, and then carefu...

Brooke Yasuda 2J

Also, when balancing equations with Oxygen in the problem, Oxygen does not appear as simply one Oxygen because it is a diatomic molecule. This means that Oxygen forms molecules of two atoms bonded together to make O2. Therefore, if there is Oxygen involved in the reaction like in the combustion of h...

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