CHEMISTRY COMMUNITY

vpena_1I

I would be prepared for either. Multiple choice would make the most sense for the computer to easily grade, but this can also be done with fill in the blank. I don't think we would be able to get partial credit with the latter, however.

vpena_1I

By definition, a closed system does not allow for the exchange of matter. A closed water bottle is an example of a closed system, its volume will not change over the course of the reaction.

vpena_1I

What it is doing is decreasing the concentration of the solution. Diluting the anode will increase the cell potential, and diluting the cathode will decrease the cell potential.

vpena_1I

I believe it will always decrease because reactions tend to move towards equilibrium, not against it.

vpena_1I

Work is positive when it is done on a system (at least in our course). Think the video might've been using a physics equation.

vpena_1I

It's usually easier to do bigger concentration over smaller concentration, but yeah, either way will get you to the same answer.

vpena_1I

All gas constants are in Kelvin, so unless you really wanted to convert these to Celsius, its easier to convert Celsius to kelvin from the get-go.

vpena_1I

A rate constant is specific for the reaction at a given temperature. Change the temperature, change the rate constant.

vpena_1I

I know that the rate law of an elementary unimolecular reaction is first-order and that of a
bimolecular elementary reaction is second-order. Don't know what else you can apply molecularity to.

vpena_1I

I'm guessing if you want certain measurements to be accurate, you'd wanna use a model in which you can control all variables. I don't think doing the kinds of experiments we talk about are very helpful in the situations you're describing.

vpena_1I

Bigger concentration over smaller concentration.

vpena_1I

If I'm understanding your question correctly, you would flip the oxidation one, so that your electrons are on the right side of the equation, and on your left for the reduction half reaction.

vpena_1I

you flip whichever one would give you the most positive delta E*cell

vpena_1I

An activated complex is the structure that results in the maximum energy point along the reaction path

vpena_1I

Probably because it's cheaper than diamond, and not soft like graphite.

vpena_1I

It was on the chapter outline, so I would assume so. Electrolysis is the chemical decomposition produced by passing an electric current through a liquid or solution containing ions.

vpena_1I

How does one begin this problem? For the first-order reaction A + 3B--->C, when [A]0= 0.015 M, the concentration of B increases to 0.018 M in 3.0 min. (a) What is the rate constant for the reaction expressed as the rate of loss of A? How do you incorporate [B] into the integrated rate law, and why d...

vpena_1I

7B.3(c) Determine the rate constant for each of the following first- order reactions, in each case expressed for the rate of loss of A: c) 2A--->B + C, given that [A]0 = 0.153 mol /L and that after 115 seconds, the concentration of B rises to 0.034 mol/L What differential/integrated rate law(s) woul...

vpena_1I

You wanna compare reactions in which the concentration of the species you are trying to solve with respect to is the only one being changed. That way, you know whatever change there is in the initial rate is purely the result of that species. If one of your "other species" is zero order ho...

vpena_1I

For 7A.15, the solutions manual says to look at experiments 2 and 4 to solve for the order with respect to A. I thought this could only be done if in the experiments, the concentrations of all other species are constant while A is the only one being changed.

vpena_1I

For homework 6N.5(a), I'm good up until having
0.06V= -(0.0129 V) ln[H+]^2

How do I use this to find the pH? Im not sure how the solutions manual went from this step to
0.06V= -(0.0257 V) ln[H+]
= -0.0257V x (2.303 log[H+]
pH=1.0

vpena_1I

I know that the anode is on the left and cathode is on the right. My question is, that for species that are in the same phase, occurring in either cathode or anode, and separated by a comma, is there a particular order to them in the cell diagram? Eg Pb4+ +Sn2+ ---> Pb2+ +Sn4+ C(gr)| Sn4+(aq), Sn2+(...

vpena_1I

For species separated by a comma, is there a particular order to the reactants/products? I remember Dr. Lavelle telling us in lecture that they should be in the form of Reactants(aq), Products(aq) , but I've seen them written in the solutions manual as Products(aq), Reactants(aq) . Does it depend on...

vpena_1I

Is this the correct way we should order our cell diagrams? --> solid|gas|aqueous || aqueous|gas|solid
This is the first I've heard of it being that way, the book wasn't too clear about the exact order.

vpena_1I

also, why is H+(aq) included in the cell diagram while H2O(l) is not. Both are in the given balanced reaction.

vpena_1I

For 6M.5(a), the cell diagram is
Hg(l)|Hg2 2+(aq)||No3-(aq), H+(aq)|NO(g)|Pt(s)

What is the electrode for the anode? I thought it would also need a platinum electrode since none of the species are a solid.

vpena_1I

Is that the only specific thing?

vpena_1I

In 6L.3(e), the following cell diagram is given, Pt(s)| Sn4+(aq),Sn2+(aq) ||Cl-(aq)|Hg2Cl2(s)|Hg(l) the addition of half-reactions gives: Sn2+(aq) +Hg2Cl2(s)---> Sn4+(aq) +2Hg(l) +2Cl-(aq) I thought Dr. Lavelle said in lecture that species separated by a comma would be in the form R,P. The balanced ...

vpena_1I

Is there a way of telling what species are reactants/products when given just the cell diagram? I can tell once I see the standard potentials, but is there a way to know without searching them up? For example, 6L.3(b) gives: C(gr)| H2 (g)| H+ (aq)|| Cl- (aq)| Cl2 (g)|Pt(s) I thought the species in b...

vpena_1I

In ∆G*=-nFE*, is n the number of moles of the species that is being reduced?

vpena_1I

Why was a platinum electrode used for both cells in 6L.5(b)? The solutions manual says: "Pt(s)| I-(aq) | I2(s) || Ce^4+(aq), Ce^3+(aq) | Pt(s) An inert electrode such as Pt is necessary when both oxidized and reduced species are in the same solution" When is says oxidized and reduced speci...

vpena_1I

In example 6K.2 (Balancing a redox equation in basic solution), it's done in this way though, and all other solutions in the solutions manual have done it like this too. For example, "6K5(c) Cr^3+ --->CrO4^2- balance O: Cr^3+ +4H2O --->CrO4^2- balance H: Cr^3+ +4H2O +8OH- --->CrO4^2- +8H2O I do...

vpena_1I

How do we know when to put a line between species and when to put a comma? Theres always a line between the species and the conducting electrode, but what goes between the two other species?

vpena_1I

How do you know if a galvanic cell is working reversibly?

vpena_1I

I'm confused as to how the solutions manual began this problem. P4(s) ---> H2PO2-(aq) +PH3(aq) takes place in a basic solution. I started with: P4 ---> H2PO2- balanced the P: P4 ---> 4H2PO2- balanced the O: 8H2O +P4 ---> 4H2PO2- balanced the H: 16OH- +8H2O +P4 ---> 4H2PO2- +16H2O balanced the charge...

vpena_1I

Someone in another thread wrote that there was a typo in the textbook. I believe the question should say
Cl2 ---> HClO +2Cl-

vpena_1I

In our balanced equation, does it matter where our H20 and H+ go?
For example, I wrote:
4Cl2+ 5H2O +S2O3^2- --->2SO4^2- +10H+ +8Cl-
but the answer in the book was:
4Cl2 +S2O3^2 +5H2O - --->8Cl- +2SO4^2- +10H+

Does this matter at all? I don't wanna lose points on a test.

vpena_1I

I read somewhere online that one of the rules for finding the oxidation number of an element is that H is always +1.
If that's the case, how is H2 neutral? Or is the "rule" just an oversimplification?

vpena_1I

So is ΔGrxn used when were trying to find the change in free energy at a different temp and/or when the reaction is not at equilibrium?

vpena_1I

What is the difference between ΔGrxn and ΔG*rxns?

vpena_1I

I think when you are referring to calculating the standard change in gibbs free energy, that is G not, the delta H and delta S in this case are the delta H not and delta G not at standard temperature and pressure, 25 degrees C and 1 atm. Thus, the delta H and delta S remain constant while the tempe...

vpena_1I

For the change in molar Gibbs free energy of NH3(l) --> NH3(g), how is it that you can plug in any temperature for ΔG=ΔH-TΔS when the ΔH and ΔS are the change in enthalpy/entropy of vaporization? I thought you would need to use the specific temperature at which NH3 vaporizes to be able to use ΔHvap ...

vpena_1I

Why are most enthalpies of formation in the appendix negative?
For example, ΔHf (H2O, g)= -241.82 KJ/mol.
I was expecting it to be positive considering it's 1 molecule being made from 2 different molecules, H2 and O2. I might be confusing it with exothermic/endothermic reactions. Thanks!

vpena_1I

That is correct, though it is not very common. The initial molarity of the reactants would have to be pretty small for that to happen.

vpena_1I

Enthalpy is the internal energy added to the product of the pressure and volume of the system. It is a state function because it only depends on the initial and final conditions, and not on the path taken to establish these conditions.

vpena_1I

I think you meant it the other way around. Internal energy is a state function, while work and heat are not. The reason being, at least the way I think of it, is that an object can store internal energy. An object cannot, however, store work or heat. Work and heat is the process of internal energy b...

vpena_1I

Calculate the pH of (b) 0.055 m AlCl3(aq)

I know that AlCl3 will decrease pH and I need to do an ICE table, but the solutions manual writes the equilibrium expression as
Al(H2O)6^3+(aq) + H2O(l)<-->H3O+(aq) + Al(H2O)5OH^2(aq)

What happened to the Cl3, and how did they get this expression?

vpena_1I

It's possible to have a negative pH for a very acidic solution.

vpena_1I

Yes, temperature is the only factor that influences K. Q is influenced by pressure (when volume is decreased, NOT when adding inert gas) or adding/removing reactants/products.

vpena_1I

Equilibrium problems involving gasses can use either Kp or Kc, it just depends what values you're given. BUT this does not mean they are interchangeable. If you are given molarity values, you must use Kc as the Kp value would be different.

vpena_1I

An inert gas is a noble gas. They do not react with many substances, so they only affect pressure, not concentration or K.

vpena_1I

For problems that give you the molarity of a strong acid or base, you do not need an ICE table because you can assume the acid/base dissociated completely. If you are given only the molarity of a weak acid/base, you will need an ice table because you do not know the concentrations of the other solut...

vpena_1I

For a problem like 6D.15, the Kb of the conjugate base of NH4Cl is given, but how would you get Ka from this and then pH?

vpena_1I

Osmosis is the net movement of solvent molecules through a selectively permeable membrane into a region of higher solute concentration. This results in the concentration on both sides of the membrane to be equal. Osmotic pressure, then, is how much pressure it would take to prevent osmosis from occu...

vpena_1I

It is the other way around. A positive delta H corresponds to an endothermic reaction in which the products formed have greater energy than the reactants. Energy can be thought of as a reactant in this case, which explains why an increase in heat will result in greater product formation.

vpena_1I

Yes, given moles of reactants and products, you will find molarity of each and use it to calculate Kc of the equation, not Kp. Only answer in Kp if the reactants and products are given in terms of bar or atm.

vpena_1I

Dissociation of a diatomic molecule, X2(g)<-->2 X(g) occurs at 500 K. The equilibrium state of the reaction is shown in 1 and the equilibrium state in the same container after a change has occurred is shown in 2. Which of the following changes will produce the composition shown? (a) Increasing the ...

vpena_1I

My guess is that there are 2 typos, the first being that you are supposed to use table 5G.2. Here, it says the equilibrium constant of the reaction N2O4(g)<-->2NO2(g) at 298 K is 6.1x10^23. The second one being that the equilibrium constant of 2NO2(g)<-->N2O4(g) at the same temp should just be the i...

vpena_1I

taking the -log of a smaller [OH-] gives you a higher pOH, making the pH smaller.

vpena_1I

pH would be negative if an acid's Ka is greater than 1. This occurs is super acidic acids.

vpena_1I

Most of them only have 2, because the highest energy level for a transition metal is the s orbital. If it is not 2, I'm pretty sure they will give us the oxidation number in the form of Roman numerals.

vpena_1I

Look at the periodic table. If the letters are completely different than the way the element is spelled, its probably latin. Though, I believe for the test, we only really need to know Iron(Ferrate), copper(Cuprate), and lead (plumbate). Some other ones are silver (Argenate), gold (Aurate), and tin ...

vpena_1I

Pi bonds are weaker, yes. They cannot rotate, or else they would break, but I don't know exactly what would force that rotation. Either way, this is the reason why pi bonds can't form hybrid orbitals.

vpena_1I

In a coordination complex? If you are writing the formula for a coordination complex, the ligands do not have to be in a particular order, as long as the central metal cation is first, and whatever is in the coordination sphere is in brackets. If you're going from the formula to writing the whole na...

vpena_1I

It will not form a double bond; doing so would change the formal charge of whatever the central atom is. Instead, the oxygen will carry that negative charge, which it is happy to do so considering its electronegativity.

vpena_1I

A Lewis acid is anything that accepts a pair of electrons, a bronsted acid also accepts electron pairs, but specifically at an acidic hydrogen.

vpena_1I

I'm not really sure if I'm wording this right, but the hybridization of 2 bounded atoms can be, for example, σ(C2sp2, C2sp2).

How would this look between a lone pair of electrons and an atom?
How would this type of question be asked?

vpena_1I

If they are the same atom directly opposite of each other, they will cancel, as in the case of CH4, but if just one of those hydrogens is different, as in CH3Cl, they won't cancel completely and will result in a dipole.

vpena_1I

Lone-Bonding pairs have higher repulsion strength than bonding-bonding pair, so the presence of a lone pair of electrons will push bounded atoms closer together, decreasing their angle.

vpena_1I

It can have up to 4. The lone pairs on the oxygen can attach to 2 hydrogens and the 2 hydrogens within the molecule can each bind to a lone pair.

vpena_1I

A Bronsted base is a proton acceptor, a Lewis base is an electron pair donor. All bronsted bases are Lewis bases, but not all Lewis bases are bronsted bases. The reason for this being that a Lewis base can donate its electrons to a number of cations, not just to H+. You can figure out what's what by...

vpena_1I

A Bronsted acid is an H+ donor (proton donor), and a bronsted base is an H+ acceptor (proton acceptor). In order for a bronsted base to accept an H+ ion, it uses its lone pair of electrons to bind to it. You can then see the connection between a bronsted base and a Lewis base (<- electron pair donor...

vpena_1I

The Lewis acid in a coordinate covalent bond is the transition metal cation. I don't know if it has a name any more specific than that.

vpena_1I

The Pauli Exclusion principle states that no 2 electrons occupying the same orbital can have the same spin. This is due to the fact that electrons have the same, negative charge and repel one another.

vpena_1I

The first part! Though the electron does not have to be ejected, it just moves to a higher orbit by absorbing energy. The lines on the absorption spectrum are unique to each element as each element can only absorb certain energies or wavelengths.

vpena_1I

Interesting question.
Theoretically, yes, the numbers can be plugged into the De Broglie formula to prove so, but experimentally, I think it'd be very difficult to measure just how slow the object would have to move.

vpena_1I

I'm not sure if I'm interpreting your question correctly, but an electron in a hydrogen atom only occupies the 1s orbital. After 2 electrons, multi-electron atoms will fill 2s, 2p, and so on. These orbitals hold electrons of higher energy than what the electrons of hydrogen atoms occupy.

vpena_1I

Yes, considering CH3(CH2)CH3 is written separately and not C3H8, you know that these are 3 different molecules that are covalently bonded. The only way for that to be possible is in a rod shape.

vpena_1I

A triple bond between the C and O gives them both a full octet with the 10 available electrons. Because each has a lone pair, you end up with a positive and negative charge. This, however, cancels out. Any other configuration would get you a formal charge even further away from 0.

vpena_1I

I believe you'd be able to calculate energy with E=1/2mv^2, and then use E/ = - h R/n^2. Solving for n will give you the principle quantum number.

vpena_1I

Usually just looking at the a/symmetry is enough to tell whether a molecular is polar or not.

vpena_1I

A dipole results from the uneven distribution of electrons within a molecule, so covalently bonded atoms with different electronegativities will usually have a dipole moment. They can then interact with other molecules with dipoles, or induce a dipole that didn't originally have one. It's just a mat...

vpena_1I

An instantaneous dipole occurs to the random movement of electrons in an atom/molecule. The uneven distribution leads to an instantaneous dipole, and it can't be predicted.

vpena_1I

We expect double bonds to be shorter than single bonds within the same molecule, but through experimentation, we can see that the observed bond length, or the actual bond length, is a blend of the different resonance structures possible for that molecule. The real structure is known as the resonance...

vpena_1I

Yes, as long as they have the same overall charge as the molecule/ion.

vpena_1I

When looking at transition metals, you count the electrons in the s orbital and d orbital as its valence electrons.

vpena_1I

I don't think there are any periodic trends for determining LDFs since they occur when atoms/molecules interact with other atoms/molecules. LDFs are, however, influenced by electronegativity in that atoms in a molecule can be more electronegative than others and cause a dipole moment.

vpena_1I

You start off by calculating the charges of each element if they were to reach noble gas configuration. For example, magnesium is [Ne] 3s2, so it will most likely lose 2 electrons and have a charge of +2. Sulfur gains 2 electrons and acquires a charge of -2. You then combine these elements with such...

vpena_1I

Using E=hv/入, you find that the energy of the red ruby laser is 2.86x10^-19 J, and the energy of the GaN laser is 4.90x10^-19 J. The work function can be converted to J by using 1 eV=1.602x10^-19 J. 2.93 eV(1.602x10^-19)= 4.69x10^-19J. This means a photon would need at least 4.69x10^-19 J of energy ...

vpena_1I

Ionic bonds are the transfer of electrons between metals and nonmetals, whereas covalent bonds are the sharing of electrons between nonmetals.

vpena_1I

In general, you take electrons from the s subshell in order to create a half-full or full d subshell, as this configuration is more stable. Palladium is a good example of that, but I'm not sure why platinum doesn't follow the same rule.

vpena_1I

Vanadium's electron configuration would be written [Ar]3d3, 4s2. Remember 3d has lower energy than 4s, so it is written first. You would only take electrons from the s orbital if it's to complete a half-full or full d subshell, as in the instance of Cr:[Ar]3d5,4s1.

vpena_1I

Mn has electron configuration [Ar]3d5,4s2. It will typically only lose its outermost 2 electrons to form Mn2+: [Ar]3d5, which is a pretty stable configuration. Remember that half-full (d5) and full (d10) sub-shells have lower energy, and are therefore more stable. This is the same reason why Iron wi...

vpena_1I

The number of orbitals an electron can occupy is dependent on what sub-shell you're talking about. The s subshell has 1 orbital, p has 3, d has 5, and f has 7. Hund's rule states that electrons will occupy different orbitals with a parallel spin until all orbitals are occupied. They will then pair u...

vpena_1I

Yes, but as the mass of an object increases, the wavelength becomes practically undetectable.

vpena_1I

In this case, the electrons are in different shells, as seen by the principle quantum number, n. As n increases, so does energy. If you were looking at electrons in the same shell, then s would have less energy than d. s<p<d<f.

vpena_1I

Are you talking about cases like chromium? I know half-full (d5) and full (d10) d sub shells have lower energy than if the electrons were spread out in different, incomplete subshells. It's the symmetry that results in a more stable electron configuration.

vpena_1I

I'm pretty sure you're correct. The spin magnetic quantum number describes the momentum of the electrons. This has direction (+ or -) and magnitude (1/2). Based on this definition, the electrons will spin around an axes in opposite directions, though, I've heard the same that it's not really that si...

vpena_1I

a) False. The more electrons an atom contains, the more shells they will occupy. Those at lower energy levels will "shield" outer electrons from the electrostatic attraction of the positive nucleus. These outer electrons will experience an effective nuclear charge. b) True. They are the cl...

vpena_1I

In what case would we use the formula En=-hR/n^2?
We went through a few examples using ΔE=Ef-Ei =, but I don't know what a problem requiring the first equation would look like.

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