CHEMISTRY COMMUNITY

Manya Bali 4E

Yes I believe so, but be careful in how you order the elements. I don't know your exact questions but if it asked for reducing power, then that means that how well a substance is oxidized. Oxidizing power would be how well a substance is reduced. Depending on how you flip the equations, you would ge...

Manya Bali 4E

Let's say you have a reaction A + B --> C + D The equation for the rate of the reaction is as follows: rate = k[A]^n [B]^m The powers of [A] and [B] (n and m in the equation) are not affected by the coefficients of the chemical equation. If we had experimental data that showed that doubling [A] (the...

Manya Bali 4E

I think it would be similar to the HW. For example (from the 6th edition), question 15.53 asks which of the proposed reaction mechanisms fit the rate law. I attached a picture of the problem in case you don't have the 6th edition.

Manya Bali 4E

I think the wording does matter. For me it was easier to think of it this way. Reducing power means that the substances are being oxidized. Thus increasing reducing power means Eo of oxidation goes from lowest to highest. In this way, the statement "going from their neutral to second oxidation ...

Manya Bali 4E

Ok, in this case I understand since that's all that is given. However, when would you be able to differentiate between using Q with only concentrations or partial pressures and using Q with both? From previous chapters, we had to use the ideal gas law PV = nRT to convert from one to the other in ord...

Manya Bali 4E

E^o is the standard reduction potential. It is the reduction potential calculated at a 1 M solution or at 1 atm and at 25 degrees Celsius. E is the cell potential, and it can change based on concentrations of product/reactant, temperature, and other nonstandard conditions. The equation that relates ...

Manya Bali 4E

In number 37c in Ch. 14 from the 6th edition, the solution manual uses a formula for the reaction quotient that is, Q = P(H2)*P(CL2)/[H+]^2[Cl-]^2. I thought that Q (and K too) had to be only in terms of partial pressure or concentrations. Why is this formula acceptable? For reference, the problem a...

Manya Bali 4E

A battery does work so the amount of energy that leaves the system is -w. Cell potential, E = -w/Charge Rearranging the above equation, we get: Work done, w(max) = -charge x E (since E is the max potential) Charge can be rewritten as the moles of electrons transferred times Faraday's Constant = nF S...

Manya Bali 4E

Hello. I have the 6th edition, but I believe that this is the same problem (in basic solution). O3 -> O2 Add H2O on left side and 2 OH- on right side so that there are equal numbers of Os and Hs Add 2 electrons to left side so charges are -2 on both sides Half reaction: O3 + H2O + 2e- -> O2 + 2OH- I...

Manya Bali 4E

You should add water in the cell diagram if it is the substance being oxidized or reduced. Otherwise, you usually do not not need to put water in the cell diagram because if any of the other substances are aqueous, then water is assumed to be present.

Manya Bali 4E

Standard enthalpy and standard entropy (unless otherwise noted) are calculated under the standard temperature condition of 25 degrees celsius or 298K. So if you're using these values and the problem does not specify a temperature, you can assume it is 298K.

Manya Bali 4E

When the change in Gibbs Free Energy is negative, that means that the substance has lost energy or gone from a higher energy state to a lower and presumably more stable energy state. It makes sense that this process would be spontaneous as the substance would be releasing energy. If the change in Gi...

Manya Bali 4E

They are interchangeable. The difference in name comes from scientists who worked on similar ideas for an early battery around the same time frame.

Manya Bali 4E

Since entropy is a state function, the path that you take to create a change in entropy doesn't matter. This means, it doesn't affect the result if you calculate a change in entropy based on a reversible or irreversible pathway. However, work is NOT a state function, which is why for calculating wor...

Manya Bali 4E

For an isothermal or reversible reaction, the problem typically has to specify it since these aren't natural conditions in real life or the lab. For constant pressure, if there's nothing in the problem specifying it, you can usually safely assume that the reaction is taking place under constant pres...

Manya Bali 4E

If there's nothing specified, you usually use Cp since you can assume that under normal conditions, the pressure is constant. However, if you're calculating entropy with a temperature and volume change, you break up the calculations into 2 parts. One is the entropy change when the temperature is con...

Manya Bali 4E

How are average kinetic energy and U (internal energy) related? I know we have the equation U = nRT(3/2) for an ideal gas, and then in 8.103 (6th edition), the question asks for average kinetic energy. The solution manual gives the equation, average kinetic energy = RT(3/2) so I'm a little confused ...

Manya Bali 4E

The only way for a system to gain internal energy is for the surroundings to do work on it or for the system to absorb heat. It loses energy when it does work on its surroundings or releases heat. Thus, we have the delta U (internal energy) = q (heat) + w (work). In an isothermal, reversible system ...

Manya Bali 4E

It takes .5 hours for the water at 0 degrees celsius to go to 5 degrees celsius. It takes 10.5 hours for the ice to go from being ice to being water at 5 degrees celsius. The ice therefore must take some amount of time to become water at 0 degrees celsius and some other amount of time to go from bei...

Manya Bali 4E

I don't know which problem you're referring to, but you can figure out which constant you need to use based on the units of the numbers in your equation. Typically, you want all the units except joules to cancel out, but it depends on the problem. I don't think it should change the answer whether yo...

Manya Bali 4E

So you have to manipulate the reactions given, either by reversing them (in which case, the delta H of the reaction changes sign), or by multiplying them by a constant (in which case, delta H is also multiplied by the constant), such that when the reactions are added together, you get the final reac...

Manya Bali 4E

Heat capacity (kj/degrees C) is the heat required to raise the temperature of an object by 1 degrees Celsius. Heat capacity is extensive because it depends on the amount of substance you have. The amount of heat needed to raise a swimming pool's worth of water by 1 degrees Celsius overall will be mu...

Manya Bali 4E

There are two other ways. One is that the enthalpy of the reaction equals the sum of the enthalpies of formations of the products minus the sum of the enthalpies of formation for the products. The other way is to use Hess's method and add together the enthalpies of different reactions if, together, ...

Manya Bali 4E

Most of the time, if you don't see the K value for the acid or base you need, it's because the table lists a K value for the substance's respective conjugate acid/base. For example, let's say you need to find the Kb value for NaCH3CO2 (aq). In water, this substance would separate into its ions, Na+ ...

Manya Bali 4E

For all reactions, there is an extremely small concentration of H+ protons given off by water due to its autoprotolysis (10^-7 M). Since this number is so small, we don't add the value on to the H+ concentrations that come from weak acids because it wouldn't make a difference in the pH of the water ...

Manya Bali 4E

I believe you have to use a graphing calculator or a computer software to answer that problem. For tests, I believe that you will never have to solve more than a quadratic equation. If the variable seems more complicated (cubic, quartic, etc.), then check to see if the equilibrium constant is small ...

Manya Bali 4E

What numbers are you inputting? Don't forget to add the appropriate power for concentrations of substances that have coefficients in the chemical reaction. For me, I have Kc = [CH4][H2O]/[CO][H2]^3 = (.0478)(.0478)/(.152)(.157)^3 = 3.88. Let me know if you have any questions about where these number...

Manya Bali 4E

You could go about this two ways. One is to create a Q value, [CO2][H2]/[CO][H2O]. If the CO concentration is increased that means that the denominator for Q will be bigger than whatever this ratio is for K. Thus, overall Q will be less than K. To balance out this change and go back toward equilibri...

Manya Bali 4E

Yes. Percentage ionization is the amount of the weak acid that deprotonated (gave away a H+ to form H3O+).

Manya Bali 4E

I believe that if you square root what you gave, you get 1.449... x 10^-7. It's a weird case where I suppose you could round up or down based on how you decide. I wouldn't worry about it too much though since rounding will probably be more obvious in other problems.

Manya Bali 4E

Yes you can do that! I'm not quite sure when an amount is small enough for it to be essentially insignificant overall, but a good way to check is to do the problem backwards. In other words, once you find the equilibrium partial pressure values, if you calculate the Kp value and it is the same as wh...

Manya Bali 4E

P: Pressure: varies, can be atm, bar, and some others - it'll depend on the problem
V: Volume: Liters
n: moles
R: Gas constant, can change depending on what unit P is in
T: Temperature in Kelvin

Manya Bali 4E

On the 14B website, there's a link to a Constants and Equations PDF that essentially has everything we would need to know for the class. From there, you can see that the R value (Gas constant) is 8.314 J·K^-1·mol^-1 or 8.206 x 10^2 L·atm·K^-1·mol^-1 or 8.314 x 10^2 L·bar·K^-1·mol^-1 or 62.364 L·Torr...

Manya Bali 4E

11.1 (a) False, In class we learned that equilibrium means that the forward and reverse reactions are taking place at the same rate so it doesn't stop. The reactants are becoming products, and then the products are becoming reactants but because they're happening at the same rate, there is no overal...

Manya Bali 4E

Atoms that are able to withdraw electron density means that they are able to pull the electrons in the molecule toward them. I think it's easiest to explain with the example we did in class. In the anion ClO-, the lewis structure shows a single bond between Cl and O. Thus the formal charge on the O ...

Manya Bali 4E

I believe so. Metal (specifically transition metal) atoms are able to form the complexes in coordination compounds with chelating agents and others because of their extra d orbitals which allow them to form more bonds. I don't think it would be very common, if at all, for other ions to do this.

Manya Bali 4E

There's a list of ions under Naming Coordination Compounds on the 14A website. I don't necessarily think you should memorize them all but just being familiar with them is good. We've discussed most of them in class at length already (CN-, SO4^2-, etc.)

Manya Bali 4E

Yes so for VSEPR, I do not believe resonance structures matter very much as you're only looking at regions of electron density instead of types of bonds (as you said, double or triple bonds don't affect if a molecule is linear or trigonal planar, they are unrelated.) Regardless, I think that resonan...

Manya Bali 4E

Specifically, chelating compounds tend to have a ligand that is essentially a string of atoms attached to each other through sigma bonds. Sigma bonds, unlike pi bonds, are flexible so individual atoms in the string can rotate/move around. This characteristic allows the string of atom to bend so that...

Manya Bali 4E

No, coordination compounds are not limited to transition metals in the 4th period, but in this class and in general chemistry, I believe those are among the most common transition metals that form coordination compounds.

Manya Bali 4E

I believe that atoms such as Boron are still hybridized during bonding but since they can only hold fewer than an octet, their hybridized orbitals have to correspond accordingly. For example, boron in BF^3 would have one s orbital and 2 p orbitals that form 3 sp^2 hybridized orbitals. After all, wit...

Manya Bali 4E

From class, we know that the number of hybrid orbitals equals the regions of electron density. (a) One s orbital and 3 p orbitals hybridize to create 4 sp^3 orbitals. This means that there are 4 regions of electron density around the central atom so there must be a tetrahedral arrangement. (b) One s...

Manya Bali 4E

Single, double, and triple bonds all count as one region of electron density because they're all interacting with the same thing. The electrons in these bonds are being shared, given, or donated together so they're not acting against each other in terms of electron repulsion. Any type of bond can ha...

Manya Bali 4E

In 4.1 (6th edition) part a, the drawing depicts a central atom with 2 other atoms bonded to it. If there were no lone pairs of electrons, then the only regions of electron density would be those associated with the 2 bonding atoms. To be the most stable structure and reduce electron repulsion, we m...

Manya Bali 4E

ICl3 has 3 bonding pairs of electrons (I - Cl), and then the central atom, I, also has two lone pairs. As discusses in class, the electron repulsion of lone pairs and lone pairs is higher than that of a lone pair and a bonding pair which is higher than that of two bonding pairs. Because of these lon...

Manya Bali 4E

I think the difference is small but notable. To my understanding, polarity refers to the unequal sharing of electrons or the existence/strength of a dipole in a molecule. On the other hand, polarizability explains how easily the electrons (electrons cloud) can be distorted or how easily a dipole can...

Manya Bali 4E

What you say is true. The electronegativity of large atoms is less because with more electrons at higher energy levels, more of the core electrons are shielding those valence electrons and other electrons from the pull of the nucleus. On a simpler level though, bigger atoms have a bigger radius so t...

Manya Bali 4E

I also do not believe we will ever need to memorize different bond lengths. However, it's important to understand the trends behind bond lengths so that you can make predictions based on the information given. For example, triple bonds are shorter than single bonds. The bigger the atom, the larger t...

Manya Bali 4E

What's already been mentioned is true, but you shouldn't get so caught up with making the formal charge 0 that you forget about the octet rule and its exceptions. For example, carbon would never have more than an octet. In that cases, it would be better to have some formal charge that is not 0 (but ...

Manya Bali 4E

Yes, the structure where the formal charge(s) are closest to 0 is dominant. However, when there are resonance structures (for example, NO3-), that means that for the Lewis structures, we draw different configurations of electrons (single bond, double bond, etc) among the same configuration of atoms....

Manya Bali 4E

I believe it means up till section 3.11 (The Unusual Structures of Some Group 13 Compounds) which is under Exceptions to the Octet Rule.

Manya Bali 4E

Yes, periodic trends will be on the midterm! Some potential applications for these trends in questions include diagrams in which the atoms involved in a reaction are shown with differing sizes and we have to infer which atoms are which based on our knowledge of atomic radii. Also, when drawing Lewis...

Manya Bali 4E

Is degeneracy associated with a number? From what I understood, degenerate orbitals have the same energy level and become filled evenly. For example, when there are 3 2p orbitals in an atom, those orbitals all have the same energy level. s orbitals at different n levels, however, should have differe...

Manya Bali 4E

What's already been said is true, but I also wanted to add that it's possible for B (boron) to complete its octet if another atom provides two electrons in what's called a coordinate covalent bond. To reiterate the example from today's lecture, BF3 + F^- --> BF4^-. In the product, B forms an octet d...

Manya Bali 4E

Look at 2 elements in the periodic table that are diagonal to each other (touching corners), and see if there are any relationships between them. For example, like we learned in class, the size of atoms increases as we go down the periodic table (due to the number of electron orbitals/shielding) and...

Manya Bali 4E

So I believe atomic radius is just referring to the radius of a single atom at a time. Covalent radii refers to the radius of the two atoms involved in the covalent bond and can be found, as you say, essentially by taking the average (or dividing in half the distance between the two atoms' nuclei) i...

Manya Bali 4E

Question 2.33 from the 6th Edition asks: "Which of the following increase when an electron in a lithium atom undergoes a transition from the 1s-orbital to a 2p-orbital? (a) Energy of the electron (b) Value of n (c) Value of l (d) Radius of atom. " I understand that the specific electron wi...

Manya Bali 4E

These equations should both find the energy for a photon (J). However, they use different variables to find E such as p (momentum), c (speed of light), h (Planck's constant), and v (wavelength). Depending on what you're given, you can use either one. This is a bit more than you asked, but you can al...

Manya Bali 4E

When I think about a change in the electrical field, I think it refers to a change in amplitude. Waves can get bigger or smaller, and as that happens, their electric field also gets bigger or smaller. If you imagine the drawing of a wave, the waves get kind of flattened when the frequency decreases ...

Manya Bali 4E

In addition to what we discussed in class, there's a really nice summary in the textbook that cleared up this concept for me. Hopefully, it's helpful for you too! I've posted it below. Electromagnetic radiation (light) has wave-particle duality. "In the wave model, the intensity of the radiatio...

Manya Bali 4E

To answer your question: (a) You are correct. The speed stays the same (as according to the equation you posted above). (b) Yes. (c) So I believe when the question asks about a change in electrical field, it's referring to a change in amplitude. I think of it as though as the wave gets taller or sho...

Manya Bali 4E

Hello, So imagine the hydrogen atom, which has a nucleus and then one electron floating around the nucleus. This electron can be at many different energy levels depending on the state of the hydrogen atom/how "excited" the electron is. It may be close to the nucleus or far. If an electron ...

Manya Bali 4E

Yup! The photoelectric effect states that electrons are emitted from a metal when an incident light shines on the metal. However, the electron can't be emitted unless the energy of the photon is equal to or greater than the threshold energy/work of the metal. If the photon is able to emit the electr...

Manya Bali 4E

Hi. This is a question I was initially confused with, and I saw lots of posts on this community about it. In fact, I think there are endorsed answers to this question in the announcement section, which is at the top before the list of topics. In case you can't find it, the Chem_Mod has posted about ...

Manya Bali 4E

You can find out what metal it is without any complicated concepts! The problem gives you the molecular formula M(OH)2. We need to find what element M is. We know that the molar mass of the entire compound is 74.10 g/mol. We don't know what the molar mass of M (the unknown metal) is but we do know t...

Manya Bali 4E

Just like 12 inches make up 1 foot, 6.022 x 10^23 (Avogadro’s number) atoms make up one mole of atoms. Molar mass is the number of grams in one mole of a substance. On the periodic table under carbon, you can see the atomic mass of carbon which is 12.01 atomic units. This relates to moles because ev...

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