CHEMISTRY COMMUNITY

Shubham Rai 2C

We can substitute the [N2O2] with [NO]^2 because we can use the equilibrium concept that K = [N2O2]/[NO]^2 because the second step in the reaction is slow and creates a bottleneck. So you just have to solve for [N2O2], which is K[NO]^2.

Shubham Rai 2C

What you are doing is finding the leftover [A] because if you just do (2molA/1molB)(0.034molB/L), it gives you the change in [A] based on how much the [B] changed. That's why you have to subtract the change from the initial to get the [A]t.

Shubham Rai 2C

It doesn't matter what the ratio is but you should logically get a whole number as 1/4 doesn't make sense when dealing with rates and n value. In one of the cases you will have to flip the ratio after you do the calculations.

Shubham Rai 2C

The Kr shouldn't really change as when you multiply the rxn, both the reactants and products will be affected and should cancel the multiplication.

Shubham Rai 2C

The difference is probably due to what the [A] value is. The - is probably associated with reactants while the equation with the - is associated with products.

Shubham Rai 2C

In this reaction, the element Cr is oxidized while Mn is reduced. Thus, Cr3+ will be the reducing agent and MnO2 will be the oxidizing agent.

Shubham Rai 2C

This is a derived equation from the formula of delta G (standard) = -RTlnK where you solve for K for two different T value. You have to combine the two different T equations to find the equation ln(k2/k1)=(-delta H/R)(1/T2 - 1/T1). So technically it is on the equation sheet, but you can memorize it ...

Shubham Rai 2C

You would add Pt to the Ce4+ side of the diagram as it has no solids to conduct the charges. We use Pt when there are no solids in the half reaction.

Shubham Rai 2C

Platinum is added as it is an inert metal that is able to conduct the current but doesn't interact with the solutions itself. Platinum can be added to either, cathode and anode, or both. Its usually used when there is no solid to conduct the charge in the cell.

Shubham Rai 2C

After balancing the two half-reactions, the number of electrons will be the value of n for the equation.

Shubham Rai 2C

One of the half reactions has to be reversed in order to balance the number of electrons on both sides of the overall reaction.

Shubham Rai 2C

When referencing entropy, the temperature should be 0 degrees Kelvin as that's when a compound will not have any vibrational or rotational entropy.

Shubham Rai 2C

When calculating the standard cell potential, we must to look at appropriate half reactions and combine them to get the appropriate cell potential. Most likely if there is no half-reaction for a compound, then it probably can't be used to find the standard cell potential.

Shubham Rai 2C

You will have to draw an octahedral shape with two particles of X and 4 particles of Y. For cis, you have to look at the arrangements were the X particles can be close to each other. This would result in 12 different orientations while for tran you have to make sure the X particles are opposite to e...

Shubham Rai 2C

The key wording of the problem is "assume ideal behavior", this basically states that assume all the unknown values are that of an ideal gas. In this case that would be that the n value is 1 mole.

Shubham Rai 2C

1 mol should be used in the equations whenever the problem states that the gas is an Ideal Gas and thus has ideal gas properties. When dealing with ideal gas, you assume 1 mole if not explicitly stated.

Shubham Rai 2C

Yes, you are correct that the degeneracy of BF3 would be 1 as you can't distinguish between the positions the F atoms occupy. For BF2Cl the degeneracy value would be 3 as the Cl could occupy any of the three bonds in the triangular planar shape.

Shubham Rai 2C

Any form of heat transfer into the system should satisfy the need for nonspontaneous changes as long as the energy is sufficient enough to cause the change.

Shubham Rai 2C

I doubt they would mark you off for switching the signs from K to C but most equations use the value of K for a proper answer and if you use the C value you would get an incorrect answer. So it better to just stick to K.

Shubham Rai 2C

When the process is reversible and the pressure is held constant, there are changes in the temperature that causes the gas to expand or contract, resulting in the work done on or by the system.

Shubham Rai 2C

Enthalpy is equal to the internal energy plus the pressure times the volume. Thus, it can be calculated for any reactions. For reactions with a fixed volume, the change in v is 0, therefore, the enthalpy value equals the internal energy.

Shubham Rai 2C

Melting and vaporizing do have to do with breaking small intermolecular bonds, however, I think that the majority of the heat is transferred to the molecules themselves and is translated into kinetic energy. I infer this as the molecules in the liquid state move around at greater speeds than those a...

Shubham Rai 2C

This analogy was related to the concept of a reaction occurring spontaneously. I think that Dr. Lavelle just wanted to convey the idea that reactions will have a certain outcome, say the rock rolling down, if not influenced by the surrounding. The rock is not likely to go up the hill, similarly, a r...

Shubham Rai 2C

Conjugate acid is the molecule formed on the product side when a base reacts with water and gains hydrogen. Similarly, a conjugate base is the molecule formed when the acid loses hydrogen during a reaction. Thus, each reaction has a conjugate acid/base pair.

Shubham Rai 2C

Since it is a strong acid it doesn't have a Ka value as it dissociates completely in the reaction.

Shubham Rai 2C

The simplest way to look at is that for an exothermic reaction, Kc decreases with the increase in temperature, while for an endothermic reaction the Kc increases with increase in temperature. The amount that the Kc value changes by is probably dependent on the reaction.

Shubham Rai 2C

I agree Q is used when the reaction is not at equilibrium. The Q value then will allow us to predict the direction the reaction will proceed when compared to Kc.

Shubham Rai 2C

Usually, it depends on the problem. You can ignore the X when the K value is smaller than 10^-3. However, for the numerator, it would be best to keep the x as usually, that's the x value that is being solved for.

Shubham Rai 2C

I agree, typically use Kc for aqueous solutions while Kp for gases. You can also convert Kp to Kc using the equation Kp = Kc [RT]^ng, where ng is the sum of the constants of products and reactants.

Shubham Rai 2C

When the is a coefficient in front of a product, you must use it while constructing the equation for K when solving for x. The coefficient is used as a power value in the equation. Ex. 2NO would be [NO]^2 in a K equation.

Shubham Rai 2C

The temperature will be unique for each of the problems as reactions can take place at different temperatures and the Kc value is also affected by the temperature thus each reaction will have a unique value of K (Kelvin).

Shubham Rai 2C

Partial Pressure is basically the pressure exerted by one gas in a mixture of multiple in a specific volume. You don't have to use partial pressure for gases per se, but gases are often measured by their pressure instead of their concentration. Also, you can't measure pressure for other states thus ...

Shubham Rai 2C

I agree you should use molarity during the ICE table problems as you have to deal with the value of K, and K requires you to solve for the molarity. So it's more convenient to use the molarity.

Shubham Rai 2C

So far in this class, we only focus on the central atoms. That and because the atoms usually are not hybridized and thus we don't have to account for that.

Shubham Rai 2C

I don't think there is a specific method for determining the polarity of a single atom. When talking about polarity, it is a molecule with two or more atoms. To find the polarity of a molecule, you just have to look at the electronegativity and then treat them as vectors. If the vectors cancel then ...

Shubham Rai 2C

I am not completely sure if we need to know this or not. But as for how do we know which is the connecting atom, the atom after the k is the atom that connects to the metal. In the first case N bonds with Fe from NCS while in the second one S bonds with Fe.

Shubham Rai 2C

The radical should be included in the hybridization as it would still influence the bonds by repelling them away from itself.

Shubham Rai 2C

Double bonds still count as a single region of electron density. This is a crucial idea for the VSEPR model and the shapes of the molecules.

Shubham Rai 2C

I think Professor Lavelle said that in this chem the arrow should point to the negative charge.

Shubham Rai 2C

To me, it seems that the electrons would be the cause of the repulsion as the elections with similar spins would repel each other, causing repulsion between the atoms itself.

Shubham Rai 2C

I think Dr. Lavelle said that this happens as the molecule wants to be in a state where the lone pairs are not in 90 degrees from the least number of bonds. So if the lone pairs occupy the axial position, they will be 90 degrees to 3 bonds while if they don't then it would be only 2 bonds.

Shubham Rai 2C

a) The elemental composition would give us the empirical formula CH4O, which agrees with the molar mass. This would give us a Lewis structure of the compound methanol. The Lewis structure would have a single bond between C and O with three H atoms bonded to O with single bonds and the fourth H bonde...

Shubham Rai 2C

The number of sigma and pi bonds don't influence the shape or structure of a molecule. This is because when determining the molecular shape, instead of the number of bonds we look at the electron regions as a single charged region.

Shubham Rai 2C

The first bond to form is always the sigma bond. There can only be one sigma bond between two atoms and any bonds formed after are the pi bonds due to the way the electrons of each atom interact with each year. Thus, a double bond has one sigma bond and pi bond, while triple bonds are made of one si...

Shubham Rai 2C

Molecule shapes can be predicted based on the Lewis dot structures using the VSEPR theory. It states that electron pairs in the valence shell of an atom repel each other, resulting in the molecular geometry and shape.

Shubham Rai 2C

The negative values that we discussed represent the fact that -250KJ/mol of energy is needed to break a mole of the bonds. It is negative as the surrounding loses energy in the process of breaking the bonds.

Shubham Rai 2C

This question is quite interesting and really got me thinking. Personally, I don't think chemical structures can be expressed in 4-D. The reasoning for my thinking is because all these chemical structures exist in our world and thus must be bound by the 3-D model of our world. Even if there were 4-D...

Shubham Rai 2C

Based on what we have learned so far, to me it seems that we can never be certain of electrons location in the orbitals. However, when a significant amount of the electrons are supposedly at one side of the atom, they can create a dipole moment that then induces a dipole moment in the adjacent molec...

Shubham Rai 2C

That list compares the bond strength of the three different interactions that occur around the central atom involving electrons. The list just states that the repulsion between two pairs of lone electrons is greater than the repulsion between a lone pair and a bonded pair. Finally, the interaction b...

Shubham Rai 2C

Yes, when there are elections in the 4s and 4p shells, 3d shell doesn't contribute to the number of valence electrons due to the fact that valence electrons are calculated from the highest energy levels. The d orbitals only contribute to the valence electrons for any transition metals as the d orbit...

Shubham Rai 2C

To my understanding, the Delta negative region is referencing the part of a molecule where the difference in the electronegativity between the atoms of a molecule forms a negative charge. This is also the concept of the dipoles, where one part of the molecule become negative and the becomes positive...

Shubham Rai 2C

There are 5 valence electrons in [Ar] 3d10 4s2 4p3 (Arsenic), and 7 valence electrons in [Ar] 3d5 4s2 (Manganese). The transition metals can use the electrons in their d orbitals to bond while the other elements can't. Thus, Arsenic only has valence electrons in the s and p shells, while Manganese h...

Shubham Rai 2C

H and He are exceptions to the Octet Rule because for them a full shell is only 2 electrons (the 1s shell). Li and Be are metals, and according to the periodic trends they are likely to lose electrons and become cations (Li+, Be2+). When that happens, they would have an electron configuration of a &...

Shubham Rai 2C

You are supposed to always write the lowest energy levels based on their n number. Initially, it is easier to fill the 6s orbital but when the electrons begin to fill the 5d orbital, the overall energy of the 5d orbital is lower than the 6s orbital. Thus, we have to write the configuration from the ...

Shubham Rai 2C

I don't think it matters if its positive or negative, but we usually keep it positive as to not get confused while working with more than one electrons in the p, d, and f orbitals. This is because all the electrons must be of the same charge before they start pairing up. But there is probably a much...

Shubham Rai 2C

When the electron undergoes the transition from 1s-orbital to a 2p-orbital all of the choices are correct. This is because, when it jumps to a higher value of n (n=2) it gains energy as the n increased from 1 to 2. Also, the value of l increases as the p subshell represents l = 1 while the s subshel...

Shubham Rai 2C

There will be 7 values for l when n=7 but the highest value of l would be a 6. This is because l includes 0 as one of the values and to find the highest value of l we can use the equation (n-1).

Shubham Rai 2C

Yes, they are the same thing. As far as I know, they represent the same concept that no electron can be found within the region they represent. However, the plane might represent a plane shaped like a square, but the angular node could be other shapes such as the cone in the dz2 orbital.

Shubham Rai 2C

This is because the work function, or the energy required to cause an electron ejection is fixed. So, based on the equation E = hv, as the frequency increases, the energy of the photon increases. ANd because the work function is a fixed value, the remaining energy is transferred to the ejected elect...

Shubham Rai 2C

In this case yes!! Just use the equation E = 1/2 m v^2. Using the uncertainty in the velocity in the equation will give you the uncertainty in the kinetic energy of the electron.

Shubham Rai 2C

When the hydrogen gets excited, the electron doesn't completely leave the hydrogen. The electron just absorbs the energy and gains energy levels, meaning it is in an excited state but still influenced by the pull from the hydrogen atom's nucleus. When the electron goes back to the ground state(where...

Shubham Rai 2C

Yes, first convert all the values to their SI Unit equivalent. Then, find the molar mass of the compound in question using the periodic table. After getting the molar mass, fin the moles of the substance by dividing the amount present by the molar mass of that substance. Finally, to find the number ...

Shubham Rai 2C

The number of sig figs that you should use should be based on the measured value in the question, in this case, 25.2 kg. Thus, you should have 3 sig figs for your answer.

Shubham Rai 2C

That equation is derived from the equation c = n/V where c represents the molarity of a solution, n the moles of the substance being dissolved and V represents the volume of the solution.

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