CHEMISTRY COMMUNITY

Rachana Jayaraman 1H

I think they forgot to include it.
For a) delta H is equal to q since it is constant pressure.
For b) I'm not sure if you can calculate it directly since it is not constant pressure.

Rachana Jayaraman 1H

A is the frequency factor and represents the number of collisions with correct orientation over time. It has to be given to you since it depends on the reaction and is used mainly for the Arrhenius equation: K = Ae^(-Ea/RT) or lnK = -(Ea)/(RT) + lnA.

Rachana Jayaraman 1H

If the forward reaction has a greater activation energy than the reverse reaction, it is endothermic. This is because the potential energy of the products is higher, therefor it is easier for the products to reach the energy barrier than it is for the reactants. If the forward reaction has a lower a...

Rachana Jayaraman 1H

It is a concentration cell, so the half reactions are: H2(g) --> 2H+(aq) + 2e- (anode) 2H+(aq) + 2e- --> H2(g) (cathode) The overall reaction will be: 2H+ (cathode) --> 2H+ (anode) The standard potential of the cell is 0 since the half reactions are just flipped. Q = [H+, pH 4]^2/[H+, pH 3]^2 which ...

Rachana Jayaraman 1H

It's a mistake. The errors in the solution manual pdf for the 6th edition shows the correct answer.

Rachana Jayaraman 1H

The difference in the sign of the slopes is due to the value plotted on the y-axis. For first order reactions, you are plotting ln[R] over time. Since the [R] decreases over the course of the reaction, ln[R] at a specific time t will be a smaller value than ln[R]o since [R]t < [R]o. If you rearrange...

Rachana Jayaraman 1H

I think so, because having ions in an aqueous state implies that the reaction involves water.

Rachana Jayaraman 1H

When the overall equation is given to you, it has already been determined which reaction occurs in the anode and which reaction occurs in the cathode. In the case of MnO4- + 8H+ + 5Ce^3+ --> 5Ce^4+ + Mn^2+ + 4H2O, Mn goes from an oxidation state of +7 to +2 so it is reduced, and Ce goes from an oxid...

Rachana Jayaraman 1H

The best way to find the number of electrons is to write the individual half reactions for the overall reaction and make sure that they are balanced. For 2Ce(4+)(aq) + 3I(-)(aq) --> 2Ce(3+)(aq) + I(3)(-)(aq), look at the reduction of Ce(4+). Ce (4+) + e- --> Ce(3+), and you know from the balanced eq...

Rachana Jayaraman 1H

Cl2 is both the oxidizing and reducing agent in this case. The oxidation number of Cl in HClO is +1 and the oxidation number for Cl- is -1. Therefore the oxidation reaction is Cl2 to HClO in an acidic solution: Cl2 + 2H2O ---> 2HClO + 2 e- + 2H+ The 2 e- is because each Cl atom loses 1 e- to go from...

Rachana Jayaraman 1H

Do you mean reacts with 5H2O? If you balance the sulfurs, you have: S2O3(2-) ---> 2SO4(2-) In this case, you have 8 oxygens on the product side and 3 on the reactants side. Therefore you add 5H20 to the reactants to balance the number of oxygen atoms. The equation should be: S2O3(2-) + 5 H2O ---> 2S...

Rachana Jayaraman 1H

I believe that this question has a typo, and it should be Cl- not Cl2 on the products side. Therefore the reduction half reaction is: Cl2 + 2e- -----> 2Cl- Usually when a reactant is both the oxidizing and reducing agent you write both half reactions using that reactant. I think in most cases in one...

Rachana Jayaraman 1H

I think that usually for each side (anode and cathode) it is usually written as reactant|product or reactant,product if they are both aqueous. The Pt(s) or other inert electrode is on the outer edges. It should look like this:

anode_reactant|anode_product||cathode_reactant|cathode_product

Rachana Jayaraman 1H

Usually it is constant pressure, unless it's a two step question where you are changing volume and temperature. In the second case, you do constant volume because you do not want it to change again.

Rachana Jayaraman 1H

Dr. Lavelle said it would be 8 questions, but I am not sure on the breakdown of the test.

Rachana Jayaraman 1H

Isothermic reactions have reversible expansion, which do the most work. Some of the potential energy is lost during irreversible expansion since the pressure is unchanging, whereas reversible expansion is the maximum possible pressure at each stage. The best way to understand it is to think of work ...

Rachana Jayaraman 1H

The integrals were used to show how the equations for work were derived. The actual equations you need to know are w = -PdeltaV and w = -nrtln(V2/V1). The first equation is used under constant pressure and the second is used for isothermal expansion (constant temperature). When volume is constant, w...

Rachana Jayaraman 1H

For temperature changes in a particular state (solid, liquid,gas), you use q = mcdeltaT. For phase changes specifically, you do deltaH*n where n is the number of moles. The delta H corresponds to the change that is taking place such as fusion, vaporization. You should remember for opposite processes...

Rachana Jayaraman 1H

No, the equation for the heat of a calorimeter is q = CdeltaT. The heat capacity is always going to be in terms of J(kJ)/C.

Rachana Jayaraman 1H

I think that more than the type of calorimeter you use, you need to understand how to do the calculations. Usually it will give you the amount of heat absorbed by the calorimeter, so you need to recognize that the system is losing heat. You take the negative of the heat absorbed by the calorimeter, ...

Rachana Jayaraman 1H

The equilibrium will be unaffected by adding inert gas at constant volume. This is because the reactions and products at equilibrium so their respective pressures will be the same. You can think of this in terms of PV = nRT. Since volume, moles, and temperature are constant, pressure is constant. If...

Rachana Jayaraman 1H

I think that the one mole of the substance correlates to its enthalpy of formation since the units is kJ/mol. Usually the reactions are written such that the coefficient of the substance is 1, but if you don't have exactly one mole, you multiply the enthalpy of formation by the number of moles. Also...

Rachana Jayaraman 1H

When Ka is small, that means that the acid is weak and that the equilibrium favors reactants over products. This is because a very small percentage of the acid actually deprotonates. Consider the Ka expression: [H3O+][A-]/[HA]. If Ka is small (<<10^-3), then the [HA] is significantly larger than [A-...

Rachana Jayaraman 1H

Ka * Kb = Kw = 1.0 *10^-14 since the K expressions cancel to form [H30+][OH-].
Another way to write it is pKa +pKb = 14.00, where pKa = -logKa and pKb = -logKb.

When Ka decreases, Kb increases and vice versa. This explains the conjugate seesaw.

Rachana Jayaraman 1H

The best way is to identify the action that will counteract the change in the system. Adding reactants will result in the production of products because that will result in some of the reactants being used up, and the ratio will be restored. The same applies to adding products; reactants will form s...

Rachana Jayaraman 1H

The book gives a table of Ka and Kb values for common weak acids and bases. In this case, you have to separate NH4Cl into its ion, NH4+ and Cl-. Cl- is a neutral ion since it is the conjugate base of HCl, which is a strong acid. However, NH4+ is the conjugate acid of NH3, which is a weak base, so it...

Rachana Jayaraman 1H

The strong acids are HCl, HI, HBr, HNO3, H2SO4, HClO4, and HClO3.
The strong bases are Group 1 & 2 hydroxides such as NaOH, LiOH, KOH, Ca(OH)2, etc.
Usually you will be able to tell if an acid/base is weak if it has a Ka/Kb value.

Rachana Jayaraman 1H

The inert gas doesn't react with the reactants and products that are at equilibrium. As a result, the concentration of the products and reactants is still the same and the K value is constant.

Rachana Jayaraman 1H

For a), the partial pressure of NH3 should increase. Since P = [conc]RT an increase in partial pressure signifies an increase in the concentration of NO, so the reaction will shift towards the reactants. As a result, more NH3 will be produced and the partial pressure of NH3 will increase. Another wa...

Rachana Jayaraman 1H

I think brackets are usually used to denote concentration, such as [J] is the concentration of solution/gas J, whereas parentheses are used to show that you are multiplying values. In the ideal gas law, Dr. Lavelle used brackets when rewriting the formula as P = [conc.]RT, but for calculations it sh...

Rachana Jayaraman 1H

Yes, those rules are important to know. The first two will help you answer if the reaction is product or reactant favored, and the last three will help you answer/determine which direction the reaction will proceed. It would be hard to answer these sort of questions without the rules.

Rachana Jayaraman 1H

Polydentate ligands typically have more than one atom which has a lone pair on it to form a coordinate covalent bond. However, you do need to consider the molecular shape. If the ligand is linear, even if it has two atoms with lone pairs, it cannot be polydentate since only one of the atoms can bind...

Rachana Jayaraman 1H

It would be best to memorize which acids/bases are strong and which ones are weak. This will be important for identifying the pH of salt solutions, and in general it will tell you whether the reaction goes to completion or is at equilibrium. Although we haven't done equilibrium yet, you should be ab...

Rachana Jayaraman 1H

Resonance stabilizes the conjugate base (anion) because of the delocalized electrons, which spread the negative charge over more atoms. This makes the acid stronger because it makes it more likely for the acid to lose a proton, and the conjugate base is less likely to accept a proton since the negat...

Rachana Jayaraman 1H

It would be helpful to know common ones like carbonate, sulfate, chlorate, nitrate, etc. I would memorize table 17.4 in the 6th edition and 9C.1 in the 7th edition.

Rachana Jayaraman 1H

The Roman numerals represent the oxidation number of the central metal atom/ion. The oxidation number of the metal atom and the oxidation numbers of the other ligands should add up to the overall charge on the coordination compound/ion.

Rachana Jayaraman 1H

I think it actually would be minimal repulsion because if there are two equatorial lone pairs,they would interact with only two atoms at less than 90 degrees (the two axial atoms) and one atom at <120 degrees. If they were the two axial lone pairs, they would interact with all three equatorial atoms...

Rachana Jayaraman 1H

For part c) when you draw the Lewis structure, you see that B is the central atom and it is bonded to 4 H atoms. This indicates that it has 4 regions of electron density, so it must be sp3 hybridized since there will be 4 hybridized orbitals, one for each bond ( 1 s orbital + 3 p orbitals = 4sp hybr...

Rachana Jayaraman 1H

The electron arrangement is trigonal bipyramidal, so it can form two axial bonds and three equatorial bonds. The equatorial bonds form an equilateral triangle, or trigonal planar, shape. So for AX2E3, which is the formula for I3-, the three electron pairs are where the equatorial bonds would form. T...

Rachana Jayaraman 1H

The electron arrangement for a bent shape molecule is usually trigonal planar, so it has 1 lone pair and 2 bonds. The bond angle for trigonal planar is 120, but the lone pair electrons create greater repulsion than they would if they were a bonding pair. The added repulsion pushes the two bonds down...

Rachana Jayaraman 1H

Be contributes 2 electrons and each chlorine contributes 7 electrons for a total of 16 electrons. Each chlorine has 3 lone pairs and there are 2 bonds, which account for all the electrons. Be, since it is in Group 2 and period 2, is an exception to the octet rule since it only needs a complete 2s su...

Rachana Jayaraman 1H

The S atom has a lone pair, which creates added repulsion since lone-bonding pair repulsion is greater than bond-bond pair repulsion. This repulsion pushes the bonds further from the lone pair and closer together, thus decreasing the bond angle.

Rachana Jayaraman 1H

O2 does not have dipole moments because there is no difference in electronegativity between the 2 O atoms. That's why O2 is nonpolar.

Rachana Jayaraman 1H

Dipole moments are partial positive or negative charges on an atom or in a molecule due to uneven sharing of electrons. The arrow points from the positive to the negative dipole because the electrons are being pulled towards the more electronegative atom or molecule. Polarizability and dipole moment...

Rachana Jayaraman 1H

Polarizability depends on the number of electrons and the size of the molecule. The more electrons there are, or the larger the molecule, the more polarizable it is. This is because the electrons are further from the nucleus, and the electron clouds can be distorted more easily. I don't think you ca...

Rachana Jayaraman 1H

I don't think you need to use it, but you do need to understand the parameters and what it represents.

Rachana Jayaraman 1H

If it is hard to visualize it, the best way to find the number of unpaired electrons is to draw out the orbitals of the valence shell to see the configuration. Or else you can figure it out based on the group that the element falls under. For the s-block, elements in group 1 will have one unpaired e...

Rachana Jayaraman 1H

The number of valence electrons in the d-block is usually the group number plus 2 (for the 2 electrons in the preceding s orbital) in the ground state. For example, scandium's configuration is [Ar]3d^14s^2, so it has 3 valence electrons. This may only work for the first 2 rows though because further...

Rachana Jayaraman 1H

For identifying a mystery element, first count the number of electrons in the molecule by adding the number of bonding electrons and the number of electrons in lone pairs. Then sum up the number of valence electrons for each given atom in its free state, like you would in the first step of writing t...

Rachana Jayaraman 1H

While it was organized before the orbits were discovered, the 4s orbital is filled before the 3d orbital, so elements that have electrons in the 3d orbitals have more protons, thus a higher atomic number, than elements with electrons in only the 4s orbital.

Rachana Jayaraman 1H

Another way to remember it is by the group the element is in.
Elements in group 1-2 are in ns orbital
Elements in groups 3-12 are in the (n-1)d orbital
Elements in groups 13-18 are in the np orbital.
n is the period number of the element.

Rachana Jayaraman 1H

The general rule is that cations are smaller than the noble gas/atom which are smaller than the anions. This is because in cations have more protons in their nucleus, resulting in a stronger attraction that pulls the electrons closer. Anions have fewer protons in the nucleus, so the attraction is we...

Rachana Jayaraman 1H

There is an exception to electron configurations in which if there are 4 or 9 electrons in the d orbital, then instead of having 2 electrons in the following s orbital, you add an electron to the d orbital to make it 5 or 10 electrons. This is because a half filled or complete d orbital has a lower ...

Rachana Jayaraman 1H

Nodal planes are planes in which the probability of finding an electron is zero. These planes usually occur along the axes where the nodes almost intersect. As for the cone-shaped nodal plane, he was discussing the 3dz^2 orbital. There are two cone-shaped planes, thus 2 nodal planes for the orbital....

Rachana Jayaraman 1H

As for the relation between the quantum numbers, l = 0,1,... n-1; This indicates that there are always n subshells for a shell with a quantum number n. ml = -l, ...0,... l; So if l = 3 (f-orbital), the ml values are -3,-2,-1, 0,1,2,3. The relationship between l and ml can also be simplified as ml = ...

Rachana Jayaraman 1H

On the actual test you would round 0.25 to 0.2 because of sig figs, but I think that the answer is 0.25 on this question is to prevent confusion since people may not be considering sig figs while taking the assessment.

Rachana Jayaraman 1H

Your steps are correct. I would recommend double checking your calculations at each step to make sure you did not enter any numbers incorrectly. You could also try doing all the calculations at once like in the solutions manual: (3500 kg Al)(1000 g Al/1 kg Al)(1 mol Al/26.98 g Al)(5 mol Al2O3/10 mol...

Rachana Jayaraman 1H

You should first find the moles of each of the compounds and then use molar ratios to find the appropriate amount of K+ for part a and S2- ions for part b from each compound. You can then add the total moles of the ions and divide by the volume of the solution, or in this case 500. mL (0.500 L)

Rachana Jayaraman 1H

Sig figs refer to the entire number. The rules are: 1. leading zeros are never significant (ex: 0.00456 = 3 sig figs) 2. trailing zeros are significant if decimal point is specified (ex: 120. = 3 sig figs) 3. Imbedded zeros are always significant (ex: 1034 = 4 sig figs) So in the case of 1.02, it is...

Rachana Jayaraman 1H

It kind of matters what the ion is bonded to when determining the concentration of the ion. You need to first find the concentrations of the individual compounds and then use the appropriate molar ratio to determine the ion's concentration. Once you have found the concentration of the ion from each ...

Rachana Jayaraman 1H

Yes, but he said the most useful ones for the class will be:
k kilo 10^3
c centi 10^-2
m milli 10^-3
and
A (Angstrom), which is 10^-10 and is used for bond lengths, but is not a SI unit.

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