CHEMISTRY COMMUNITY

Nora Sharp 1C

expansion work in a irreversible change = -P*deltaV, and when there is no change in V, deltaV = 0 so work = 0. This only holds true if we only consider expansion work.

expansion work in a reversible change = -nRT*(ln (Vfinal/Vinitial))
if vfinal = vinitial, then any expansion work also is zero.

Nora Sharp 1C

If a system goes back its original internal energy, internal energy should be zero overall. I don't think you would have to flip the sign of any internal energy steps in this case, because the sign of internal energy would already be considered in your calculations of deltaU to make deltaU overall b...

Nora Sharp 1C

I usually interpret >> as "much larger than" but I could be wrong.

Nora Sharp 1C

Reaction rate units concerning one substance are always Molarity/time.

But for k:
zero order is M/time
first order is 1/time
second order is 1/(time*M)

Nora Sharp 1C

You're correct, I think the answer key in the book does say that the rate for oxygen should be .44*3.

Nora Sharp 1C

I'm not really certain what you mean by internal energy being positive, but here's how I solved Q5: Because step one concerns an adiabatic process, q = 0, and consequently internal energy in step one = work. But the second step returns the internal energy to its initial state before step one happene...

Nora Sharp 1C

I believe the rate laws that are involved in 15.1-15.6 are supposed to calculate the instantaneous rate of reaction at certain quantities of reactant, assuming that no product has formed yet (because a product can undergo the reverse reaction and become the product, so the equation would have to be ...

Nora Sharp 1C

The rate of formation generally refers to one of the products (d[B]/dt, for example). The rate of consumption is the rate at which one of the reactants is being consumed at the beginning of the reaction. If you divided the rate of consumption and formation by the coefficients of the compounds the ra...

Nora Sharp 1C

Isobaric procedures occur at constant pressure and isochoric ones occur at constant volume. Usually when it's mentioned in a test question, it tells you what assumptions you should make and what kind of equations to use. Use Cp and Cv for each respectively when calculating heat capacity, and in isob...

Nora Sharp 1C

A catalyst should be unchanged after the reaction that it facilitates occurs and is not consumed by the reaction. Because of this, it should not be treated the same as an intermediate, because intermediates are produced and consumed in the overall process of the reaction. That being said, a catalyst...

Nora Sharp 1C

I believe the reaction order does depend on the number of coefficients involved the reaction, but it more specifically depends on the coefficients of the reactants involved in the slow step.

Nora Sharp 1C

Q and K are written with the same scheme that we learned last quarter. for reaction aA + bB ---> cC + dD, K = ([C]^c * [D]^d)/([A]^a * [B]^b) In a galvanic cell, use the equation for the full reaction (combine the half reactions) to get your products and reactants. The substances that are favored to...

Nora Sharp 1C

Are you referring to concentration cells? If so, there are some explanations for that here:

viewtopic.php?f=142&t=20974

Nora Sharp 1C

Q is similar to K (products/reactants) but the amounts of products and reactants used in calculating Q represent the reaction when it is not at equilibrium and is instead in the process of reaching equilibrium. It allows us to compare the current ratio of products and reactants at a certain point in...

Nora Sharp 1C

Usually, the question should tell you whether the reactions are occurring in a basic or acidic environment.

Nora Sharp 1C

I think the two calculations that consider entropy increase as a result of temperature change and entropy increase due to volume change should be done entirely separate from each other. Therefore, since we already took entropy change from volume into account in a prior calculation, we assume that th...

Nora Sharp 1C

I assumed that cyclopentane has a lower entropy than 1-pentene because cyclopentane has to conform to a specific shape due to its carbons bonding in a closed ring structure, while 1-pentene is a straight molecule with two free ends.

Nora Sharp 1C

If a reaction is non-spontaneous, its deltaG is positive and therefore the formation of products is not favored. This means that it's favored for the reactants to stay reactants instead of undergoing the reaction, and if there were a great number of products under certain conditions, the reaction wo...

Nora Sharp 1C

viewtopic.php?f=130&t=26453

Here's a thread with some more information on the question you asked.

Nora Sharp 1C

I believe deltaG can be used to represent the total amount of energy that is released or absorbed into or by a system during a chemical reaction. This energy released can be used to perform different functions, hence deltaG representing the amount of energy free to do work. For example, if we said A...

Nora Sharp 1C

A process such as a reversible expansion is not seen in real world scenarios due to two reasons. First, expanding a gas reversibly involves incrementally decreasing the pressure by infinitely small amounts to make the gas expand, which would take forever to do. Secondly, all heat in a system expandi...

Nora Sharp 1C

In the case of a reversible expansion, the system does the maximum possible amount of work. All heat is converted into work, which models a perfect engine, something that doesn't exist. A reversible expansion is 100% efficient and would not be found in real life, to the best of my knowledge. I suppo...

Nora Sharp 1C

At equilibrium, the system is no longer trending towards a spontaneous change, and by consequence it has reached its maximum entropy state. (Entropy increasing can drive processes like diffusion, at which the system wouldn't be at equilibrium.) If an isolated system at equilibrium were to decrease i...

Nora Sharp 1C

A gas can still expand without internal energy changing because deltaS, the maximum number of positional states (entropy) of a system, is going up. If we assume that the gas is expanding into a vacuum at constant temperature (deltaU is zero), its expansion is still favored because deltaS is positive...

Nora Sharp 1C

I guess you could discern intensive and extensive properties by imagining a block of the substance you're examining, and then cutting it in half. If the characteristic you're trying to figure out is intensive or extensive changes, then it's extensive. If I had a block of water and (hypothetically) I...

Nora Sharp 1C

We need to calculate the temperature rise for copper and water separately because they have different specific heats. The specific heat gives us the amount of energy needed to raise a substance based on mass and temperature, so that's what we'll use: Copper: 500g*(.38 J*C^-1*g^-1)*(100 degrees C - 2...

Nora Sharp 1C

Part of your question has actually been already answered here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=75&t=25593 The equation w = - (external pressure)(change in volume) actually only applies to one of the expansion work scenarios given by the book. This equation applies to what the...

Nora Sharp 1C

Given that the final temperature of the metal will be less than the initial, the calculated energy lost by the metal will be negative. And given that the final temperature of the water will be more than its initial, the calculated energy gained by the water will be positive. So before modifying the ...

Nora Sharp 1C

If a system loses energy in the form of heat, q should be negative. But if a system absorbs heat, q should be positive.

When water changes from a liquid to a gas, water molecules absorb heat to break the attractive intermolecular forces between them, so q in this reaction is positive.

Nora Sharp 1C

If you are in Chemistry 14B, we have to do 3 comments per week now, but Chem 14A students will do 2 comments a week (Sorry about this ambiguity, I can't really tell what class you're in) I'm guessing that the week counts as Monday - Sunday, but I will ask my TA tomorrow to be sure. In one of my disc...

Nora Sharp 1C

When a strong base (or acid) is dissolved in water, they dissociate completely. Therefore .10 M HCl means .10M H+. However, watch out for strong acids/bases that have more than one H+ or OH- ion for every molecule. .10M Ba(OH)2 is actually .20 M OH - because there are 2 molecules of OH- for every mo...

Nora Sharp 1C

You are correct that K for a reaction only changes when temperature changes. However, when we halve all the stoichiometric coefficients for the reaction, we are actually changing the reaction we are using in our calculations. 2SO2 + O2 ---> 2SO3 is different than SO2 + (1/2)O2 ---> SO3. Therefore, t...

Nora Sharp 1C

While it is true that this formation would result in formal charges of zero for every atom, the accepted answer is still the one listed on the midterm solutions. There are two tip offs that let you know this is the case: First, note the way the molecule is written. The molecule you bring up actually...

Nora Sharp 1C

Question 11.7 states that the initial pressure exerted by 11 molecules of the gas X2 is 0.10 bar. I assume that X2 decomposes according to the equation X2 --> 2X. At equilibrium there are 12 molecules of X and 5 of X2. The solution manual states that the equilibrium constant of this equation is equa...

Nora Sharp 1C

The criteria that I gave for an Ideal Gas (no intermolecular attractions and no kinetic energy is lost in the collisions between molecules) was not actually every requirement that a gas needs to be ideal. Another requirement of Ideal Gases is that the individual molecules of the gas do not have volu...

Nora Sharp 1C

The Ideal Gas Law is used to analyze gas samples under the assumption that gases are "ideal gases," which have two criteria: 1) no intermolecular attractions and 2) no energy is lost in the collisions between molecules. No gas is an ideal gas, and the Ideal Gas Law becomes less accurate as...

Nora Sharp 1C

That figuration would not have the lowest formal charge possible. The oxygen on the left now bonds with carbon twice, so its formal charge is now 6 - 5 = 1 e-. The oxygen on the right now also has the formal charge 6 - 5 = 1 e-, (assuming it has the lone pair and the one radical electron) while the ...

Nora Sharp 1C

The molecule is not linear because both lone electron pairs have more repulsion than electrons from atoms bonded to the central atom, and there are four regions of electron density around the atom. Keep in mind that the "bent" shape is not entirely flat; the regions of electron density are...

Nora Sharp 1C

The H atom should have a single bond with the oxygen, which in turn should have a single bond with the carbon. The carbon atom should have a double bond with the remaining oxygen. In this case, HOCO is a radical because all the valence electrons of the atoms add up to form an odd number. The remaini...

Nora Sharp 1C

This exception remains true for most elements in Group 15, but as you go down the group it becomes less pronounced. By the time you reach Bismuth the exception no longer holds true. It appears that when atoms in group 15 have a much larger number of protons and electrons (are at much higher energy l...

Nora Sharp 1C

You are correct that Br is smaller than Na. However the Br atom discussed in this question is actually an ion with an extra electron. Adding another electron to Br increases its radius by a large amount. This is because the electrons now experience a reduced effective nuclear charge and are not draw...

Nora Sharp 1C

There does seem to be many exceptions when following electron affinity down a period, but theres a trend that can be very loosely applied: electron affinity increases across a period and decreases down a group. There are explanations for some of the exceptions, though there are a lot of said excepti...

Nora Sharp 1C

I looked online for an answer and this is what I got: "However, some of the third-period elements (Si, P, S, and Cl) have been observed to bond to more than four other atoms, and thus need to involve more than the four pairs of electrons available in an s2p6 octet. This is possible because for ...

Nora Sharp 1C

I think the octet rule is usually used when drawing lewis structures. This is the idea that bonding atoms generally end up with their electrons in the electron configuration of a noble gas; as in, they all have 8 electrons in their valence shells. The octet rule does have exceptions, however. It doe...

Nora Sharp 1C

Weak acids create "strong" conjugate bases because their dissociation reaction (the reaction of it releasing an H+ ion) is not favored. If you take HNO2 (nitrous acid) as an example, the reaction for HNO2 to produce a conjugate base would be as follows: HNO2 = H+ + NO2- In this case, NO2- ...

Nora Sharp 1C

I can't say for sure but the test schedule says that Test 3 will cover "new material up to October 27," while the midterm, which is cumulative, says "all material up to November 3." If I had to go off this I'd say Test 3 isn't cumulative AND covers any material featured in the le...

Nora Sharp 1C

Nodes are areas in orbitals where there is a zero percent chance of finding an electron. Orbitals like the p orbital are drawn as dumbbell shaped because they represent spaces around the nucleus in which electrons are more likely to appear. (This is all represented by the wave function). Nodal plane...

Nora Sharp 1C

I'm not sure about that. We should either ask if we need to know them or memorize them to be safe.

Nora Sharp 1C

This is what I wrote to explain choice c on another person's question about 1.3: From the textbook: Light is a form of electromagnetic radiation, which consists of oscillating (time-varying) electric and magnetic fields that travel through empty space at about 3 * 10^8 m/s, or at just over 670 milli...

Nora Sharp 1C

From the textbook: Light is a form of electromagnetic radiation, which consists of oscillating (time-varying) electric and magnetic fields that travel through empty space at about 3 * 10^8 m/s, or at just over 670 million miles per hour. This speed is denoted c and called the speed of light. Visible...

Nora Sharp 1C

Here's my idea of a question you might get that would involve the series. If course I have no way of knowing exactly what will be asked on future exams but here's my idea: The Balmer Series and the Lyman Series both correspond to certain types of electromagnetic radiation. The Balmer series includes...

Nora Sharp 1C

Hi Essly! Since CaCO3 is the only reactant in the equation, it has to be the limiting reactant.

Nora Sharp 1C

I believe you're correct. Adding an oxygen atom to the right side of the equation would be adding a product that isn't produced by the reaction in the first place, so now the modified equation isn't even a representation of the original.

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