CHEMISTRY COMMUNITY

Andrea Zheng 1H

Try searching "Joyce". She uploads worksheets!

Andrea Zheng 1H

They got this by using the equation ln(k) = -Ea/(R*T). Because you are finding the difference in two different k values, you would use ln(k')-ln(k), which is ln(k'/k). On the other side of the equation, you would have Ea (which is constant) and R (which is constant) and T (which changes). This means...

Andrea Zheng 1H

I used the method of directly plugging in the values of the Ea into the formula of A*e -Ea/(R*T) to get the ratio of e^x for each of the reactions. I divided the second e^x number (Ea=125) by the first e^x number (Ea=75) and got the factor the k increases by, as A remains constant in both re...

Andrea Zheng 1H

If you look at experiment 1 and experiment 4, you see that the concentration of A and B remained the same, while the concentration of C goes from 700 mmol/L to 400 mmol/L. Because concentration of C is the only thing changing, any changes in the initial rate would stem from changes in C, IF the rate...

Andrea Zheng 1H

You would use ln[A] for first order reactions. For these reactions, rate=k[A]=-1/a * d[A]/dt. If you set a=1 and then you separate the variables, you get k*dt=-d[A]/[A] or (-1/[A])*d[A]. If you integrate both sides, you would get kt=-ln[A], or -kt=ln[A], as the integral of (-1/[A])*d[A] is ln[A]. Th...

Andrea Zheng 1H

If you are given the units of K, you can determine the order of the reaction by looking at those units, as they should cancel out with the reactant's units to get \frac{mol}{L*sec} , which is the unit for the rate of the reaction. You know that the [reactant] is \frac{mol}{L} . For example: Rate con...

Andrea Zheng 1H

You can use either the equation for decomposition or composition, as it will yield the same answer. As an example, part (a): If you used the equation of composition, P + 5/2Cl2 -> PCl5, you would find that the entropy of the reaction was negative. This means that at higher temperatures, the reverse ...

Andrea Zheng 1H

E^ocell is the standard state cell potential, meaning the condition of [1M] for solutes or 1 bar for gases is met. Ecell is the cell potential at non standard state conditions. You can use the Nernst equation to find Ecell using E^ocell.

Andrea Zheng 1H

A reducing agent is the substance that has the ability to reduce other substances, by losing electrons and being oxidized.The reduced species is the thing that is reduced (gain electrons) An oxidizing agent is the substance that has the ability to oxidize other substances, by gaining electrons and b...

Andrea Zheng 1H

Professor Lavelle used a single horizontal line to represent the interface between phases in contact with each other (including a porous disc/wall) and a double horizontal line to represent the salt bridge.

EDIT: a comma is used when the oxidized and reduced species are in the same phase.

Andrea Zheng 1H

Yes, 1V= 1J/Coulomb, so 1 Volt is 1 Joule of Work per Coulomb of charge transferred

Andrea Zheng 1H

The Gibbs Free Energy is the energy associated with a reaction that can be used to do work. This is a function of the enthalpy of the reaction minus the temperature multiplied by the entropy. It can be used to see the spontaneity of a reaction, with a negative Gibbs Free energy representing a sponta...

Andrea Zheng 1H

Kp for any reaction is = [products]/[reactants] but you wouldn't include solids or liquids in this eq constant. Because the reactant is a solid, the only thing that affects Kp is the products. Therefore, Kp = [products] = 0.13. This means the [CO2(g)] = 0.13 atm.

Andrea Zheng 1H

Because work is not a state property, you can't only look at the initial and final state. You would have to add up all the work done in each step of the pathway to get the total work done.

Andrea Zheng 1H

For a constant pressure system, you would be able to use the equation W=-P*deltaV, as the P would be constant and thus can be multiplied by the change in volume to get the work done.

Andrea Zheng 1H

Because this problem is giving you the molar concentrations (given mmol and L), you would use the column K_c, as the book says that the K_c column is the eq constant for the molar concentration of gases.

Andrea Zheng 1H

Change in internal energy would be 0 for an ideal gas in the case when there is constant temperature (isothermal). Because change in temperature is 0, then change in internal energy is 0, as the two variables are related through the equation: deltaU = 3/2*n*R*deltaT.

Andrea Zheng 1H

If you are referencing problem 9.3 in 6th Edition textbook, for part (a), you would use the equation deltaS=deltaQ/T . You would get deltaQ = +65J and T = 25 + 273 = 298 K. This gives you a deltaS of 65/298 = 0.22J/K You would use the same process for part (b), except with 373 K instead of 298. For ...

Andrea Zheng 1H

Degeneracy is the number of ways of achieving a given energy state. For example, the orbitals in the 2p sublevel are degenerate, as they have the same energy. Degeneracy can be calculated through the equation degeneracy = X^N, where X = possible microstates and N= # of particles.

Andrea Zheng 1H

I think the solution manual is just giving a visual way of explaining what you did in your head. It is just showing how the combustion reactions cancel out to form the equation given in the exercise and why you would flip the sign for Hc of C2H6. I don't think you would necessarily need to show all ...

Andrea Zheng 1H

Hc is the standard enthalpy change of combustion. This means that the given values are the change in enthalpy when each substance reacts with O2, and thus the change in enthalpy when the substances are the reactants in the equation. Because C2H6 is a product in the equation given in the exercise, yo...

Andrea Zheng 1H

Cv = (3/2)R and Cp = (5/2)R would be true for any monoatomic ideal gas, so you would use these equations in those cases.
If a molecule is linear, it wouldn't be a monoatomic ideal gas, so you would use the other equations for constant pressure and constant volume.

Andrea Zheng 1H

The molar entropy would also increase as a substance goes from solid to liquid to gas.

Andrea Zheng 1H

Yes, you would be solving for K_p.

Andrea Zheng 1H

Water in liquid form would not affect the Q or K of a reaction. Because water(l) doesn't affect Q, if you added water to a system at equilibrium, the Q would equal K, meaning the system remains at equilibrium. This is only if water is in liquid form. If it is a gas then you would use that in calcula...

Andrea Zheng 1H

First, you would find the initial concentration of each substance. For H2, it is 0.4 mol/3L = 0.133M. For I2, it is 1.6mol/3L=0.533M. For HI, it is 0M. Using the balanced chemical reaction, change in H2 is -x, change in I2 is -x, and change in HI is +2x. This means the eq composition of each are: H2...

Andrea Zheng 1H

I believe it should just be +2x, not c+2x, because I used eq composition of SO3 to be 2x to solve the problem and it said I got the correct answer. The c may have come from problem 12, where you were asked to find what the value of c would be (where c=+2x)

Andrea Zheng 1H

a) To solve this problem, you would use the reaction Cl2(g) <-> 2Cl(g). From table 2, you see that the K for this reaction is 1.2*10^-7. ICE TABLE: Because it asks for the equilibrium composition, you can use the ICE table (initial, change, equilibrium). To set this up, you would need to find the in...

Andrea Zheng 1H

Given only K, you would have to make an ICE table and then use the final equilibrium line from your table in the expression for K (K=[eq line of products]/[eq line of reactant 1][eq line of reactant 2]). Because you are given the K value, you would set the expression equal to the K value of 0.56. Yo...

Andrea Zheng 1H

yes, you should use the same process as (a) but use the value of F2 <-> 2F of 1.2*10^-4. All the other calculations should remain constant. The ICE table would be 0.001-x for F2 and 2x for 2F. This would make the Kc value (2x)^2/(0.0010-x)=1.2*10^-4. Solving for x using the quadratic formula, you wo...

Andrea Zheng 1H

First, it says it is in a 3L vessel, so you would divide all the given values by 3 to get the moles per 1 liter, as that is the molarity used in the calculation. Looking at the equation, the As is solid, so you would not need to account for it in the Qc. Solving the problem, you would divide the two...

Andrea Zheng 1H

This is how you would name it under the IUPAC Name Convention. Lavelle uses a different convention, but he says that he accepts both.

There is a sheet on Lavelle's website that has ligands and their names, including the IUPAC Name Convention.

Andrea Zheng 1H

Yes, you would use aqua. Lavelle has posted a "Naming Coordination Compounds" sheet on his website that has some common ligands and their names.

Andrea Zheng 1H

For an acid, the conjugate base would just be the structure of the molecule after the acid donates a proton (when it is deprotonated). For a base, the conjugate acid is when the base accepts the proton (when it is protonated). There is a more detailed description and examples in the textbook on pg 4...

Andrea Zheng 1H

I posted this on another question, I hope this helps! Bronsted looks at protons, while Lewis looks at electron pairs. A Bronsted acid is something that is a proton donor, while a Bronsted base is a proton acceptor. A Lewis acid is an electron acceptor, while a lewis base is an electron donor. see th...

Andrea Zheng 1H

Bronsted looks at protons, while Lewis looks at electron pairs. A Bronsted acid is something that is a proton donor, while a Bronsted base is a proton acceptor. A Lewis acid is an electron acceptor, while a lewis base is an electron donor. see this link for examples: http://www.personal.psu.edu/user...

Andrea Zheng 1H

Hybridization is based on the number of electron densities around the atom, so a good rule to follow is that the number of regions of electron density equals the number of hybridization orbitals.

Andrea Zheng 1H

Molecules that undergo hybridization will still follow the rules we learned before about sigma and pi bonds. For a single bond (ex: C 2sp3 - H 1s in CH4), there would be a single bond between the atoms and thus a sigma bond between the atoms. For a double bond (ex: C double bond C in C2H4), there wo...

Andrea Zheng 1H

In lecture, we were told that it is present in molecules with N,O,or F. We were also told that for a H bond to occur, there needs to be an H atom covalently bound to an electronegative atom and close to another electronegative atom that has an available lone pair. It needs a lone pair because the lo...

Andrea Zheng 1H

I believe we don't need to memorize the exact electronegativity of elements, but that we should know general trends, as this is useful in figuring out types of bonding or polar/nonpolar molecules.

Andrea Zheng 1H

a) Xenon is larger and has more electrons. This means it will have increased attractive forces, making the melting point higher. b) There will be a higher vapor pressure when the attractive forces are not as strong. Because water has hydrogen bonding, the diethyl ether will have the higher vapor pre...

Andrea Zheng 1H

https://www.khanacademy.org/science/org ... igma-bonds

Andrea Zheng 1H

sigma bonds have orbitals end to end and allow the bound atoms to rotate. pi bonds have orbitals that overlap side to side and don't allow bound atoms to rotate. a single bond is a sigma bond and a double bond is one sigma one pi bond

Andrea Zheng 1H

Elements in period 3 have an empty d-orbital, so it does not have to obey the octet rule. Hydrogen, helium, beryllium, and lithium are exceptions to the octet rule because they form duplets. Radicals are also exceptions to said rule.

Andrea Zheng 1H

A compound can be radical if it has an unpaired electron.

Andrea Zheng 1H

Normally bond lengths are given based on experimental data.

Andrea Zheng 1H

The oxidation number is determined by the electrons that would be added/lost to form a noble-gas configuration. Because Cl has 7 e-, it would add one electron to get to the electron configuration of Argon. Because it adds one electron, it would have an oxidation number of -1.

Andrea Zheng 1H

4f14 is the full orbital of elements from lanthanum to Ytterbium (14 elements). Lutetium is the first element in the 5d orbital.

Andrea Zheng 1H

You got the correct wavelength, however, you just switched n1 and n2 in the equation. The equation should be: wavelength = Rydberg constant *(( $\frac{1}{{{n_{1}}^2}}$ ) - ( $\frac{1}{{n_{2}}^2}$ )). If you use this, you should get the correct answer of $n_{2}$ = 3.

Andrea Zheng 1H

this question says the wavelength is equal to 102.6 nm. This wavelength corresponds to the Lyman series, a set of lines in the UV region of the spectrum with n1 = 1. This is why the answer uses the assumption that n1 = 1. If the wavelength corresponded to the Balmer series, then n1 = 2.

Andrea Zheng 1H

Ge: [Ar] 3d10 4s2 4p2, so it would be 4p orbital
Mn: [Ar] 3d5 4s2, so 4s
Ba: [Xe] 6s2, so 6s
Au: [Xe] 4f14 5d10 6s1, so 6s

Andrea Zheng 1H

Electron affinity is the energy released when an electron is added to a gas-phase atom. If the electron affinity is positive, then energy is released when the electron is added to the atom. This would occur if the electron has a lower energy when it occupies the atom's orbital. If the electron affin...

Andrea Zheng 1H

a) use the Rydberg equation found on page 7 of 6th edition textbook. You already know Rydberg's constant, and you are given n2=4 and n1=2. You just plug these values in and you can get the wavelength. This should be 486 nm. b) The Balmer series consists of lines with n1 = 2 and the Lyman series cons...

Andrea Zheng 1H

The work function is the energy required to remove an electron from a metal. In class, it was also referred to as the threshold. In order for an electron to be removed from the metal, the E=h*freq has to be greater than or equal to the work function (threshold energy). (see textbook pages 12-13 for ...

Andrea Zheng 1H

In part (b), they are asking for the energy that is needed to remove the electron. This means that you would use the equation E=h*frequency. This is why the solution uses the value (2.5x10^16 1/s), as this is the value that is given for frequency. Because you already know h (Planck's constant), you ...

Andrea Zheng 1H

You would want to use the De Broglie equation for this question (wavelength = h/m*v). You would need to look up the mass of the electron and then you are given Planck's constant and the velocity. From there, you would get the wavelength of the ejected electron. Make sure to convert everything to SI ...

Andrea Zheng 1H

for (a), you would use the equations c=(wavelength)*(freq) and E=h*(freq) to derive the equation E=h*c/(wavelength). Because you are given h, c, and wavelength, you can calculate the energy given off by an excited sodium atom. Using this energy calculation, you can answer parts (b) and (c) by conver...

Andrea Zheng 1H

You would use the equations E=h*(freq) and c=(wavelength)*(freq). From these equations, you derive the equation wavelength=(h*c)/(E). Because you are given E, h, and c, you can then solve for wavelength.

Andrea Zheng 1H

There is a mystery metal in this problem, signified by the letter M in M(OH)2. They give you the molar mass of M(OH)2 so that you can get the molar mass of the mystery metal from there. The sulfide of this metal would be the compound that results from the mystery metal bonding with the element sulfi...

Andrea Zheng 1H

There's Khan Academy which has videos and other resources for a lot of different chemistry topics: https://www.khanacademy.org/science/chemistry

Andrea Zheng 1H

In lecture, Lavelle used the mass composition to provide the mass (g) of an element given a sample of 100 g (ex: given a mass composition of C = 45%, the mass of C would be 45g). Because the problem directly gives you the mass (g), you don't need to calculate the mass composition, as you can already...

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