CHEMISTRY COMMUNITY

Hiba Alnajjar_2C

Chem_Mod wrote:Test 2 answer key

Will the answer key for Tuesday's test be posted as well? Thank you!

Hiba Alnajjar_2C

You could probably send pictures of your homework to your TA so that you could get credit for it.

Hiba Alnajjar_2C

As a follow-up question, will review sessions take place through services like Zoom as a result of the transition to online instruction? Thank you!

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The rate law for the formation of NOBr is equal to the rate law of the slow elementary step, as the fast step does not have much of an effect on the overall rate compared to the slow step. Therefore, your rate would be equal to k[NO][Br2]. The slow step is the step that takes the longest to occur, w...

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When trying to determine the order reaction, you should compare two experiments where the other reactant concentrations are held constant and the reactant you are analyzing is changing. For example, given the reaction A + B-->C, to find the order with respect to A, you would want to compare two expe...

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You would get a linear plot when you graph 1/[A] vs. time. This makes a bit more sense when you look at the integrated rate law for a second order reaction: 1/[A]=kt+1/[A]o

(In this plot, k would be the slope).

Hiba Alnajjar_2C

You should look at which elements' oxidation numbers change. If the oxidation number increases, then the species is being oxidized. For example, if it goes from 2- to 0, it is being oxidized. If the oxidation number decreases, then the species is being reduced (since it's becoming more negative). If...

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O3 is being reduced, but the half reaction should not be going from O3 --> O2. Rather, it should be going from O3 to BrO3-. Here, the oxidation number of oxygen goes from 0 to -2, which is why this the reduction half reaction (as electrons are being gained).

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The rate-determining step is equivalent to the slow step. Therefore, by comparing each of the elementary steps' rate laws with the overall reaction rate law, you can determine which step is the rate-determining step: it's whatever step gives you a rate law identical to the overall reaction rate law....

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Yes. If it has a high standard reduction potential, then it means reduction is favored. The species being reduced is the oxidizing agent, so something that is strongly favored to be reduced would be a strong oxidizing agent.

Hiba Alnajjar_2C

Focus 5G.3, 5G.4, 5J.3 should correspond to the second page of the thermodynamics outline.

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Hi Tiffany,
I got the same answer as you! I agree with your point about n being equal to 1, particularly since the oxidation number only changes by 1.

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I believe that the oxidization of a particular species allows the other species to be reduced, which is why the species being oxidized is referred to as the reducing agent. When that species is oxidized, it loses electrons that the other species then gains, allowing it to thus reduce the other speci...

Hiba Alnajjar_2C

Did you check Appendix 2B? The standard reduction potentials can be found there. :)

Hiba Alnajjar_2C

Standard potentials are given as reductions, not oxidations. Therefore, in the reduction of Fe2+, Fe2+ would be the reactant and Fe would be the product. In the given reaction, Fe2+ is being oxidized, which is why it's flipped and Fe2+ is the product rather than the reactant. But the standard potent...

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The species being reduced is the oxidizing agent. Similarly, the species being oxidized is the reducing agent. Essentially, the species being oxidized loses electrons which the species being reduced then gains, making it the "reducing agent" as it allows for the other species to be reduced.

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Another important aspect of electrolysis brought up in lecture was the difference between a galvanic cell and an electrolytic cell. Because the overall redox reaction in an electrolytic cell is not favorable, electrolysis requires the use of an external power supply. In this example, electrical curr...

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Your logic is correct. Using the equation delta G = delta H - (T)(delta S), it is apparent that the value of delta G will be close to that of delta H when you have a value of delta S close to or equal to 0. Therefore, the answer with the smallest change in entropy, as you mentioned, would be the co...

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Your logic is correct. Using the equation delta G = delta H - (T)(delta S), it is apparent that the value of delta G will be close to that of delta H when you have a value of delta S close to or equal to 0. Therefore, the answer with the smallest change in entropy, as you mentioned, would be the cor...

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The change in temperature would be the same for Celsius and Kelvin (if you convert to Kelvin and calculate the difference, you'll notice it's the same as the difference in Celsius). However, if it's just temperature (such as in delta G = -RTlnK), I believe that you have to convert to whatever units ...

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It should cover the content we learned after the midterm. The main topics include Gibbs Free Energy and electrochemistry. Professor Lavelle will likely mention what exactly will be covered either in lecture or on the website.

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Selena Yu 1H wrote:For 5G.15, for the standard GFE I got -12.35 kJ and for my final answer, I actually got -2.7 kJ/mol instead of -27 kJ/mol

I also got -2.7 kJ/mol, did you happen to figure out what went wrong? Thank you!!

Hiba Alnajjar_2C

Could someone explain when to use Cp or Cv? Is Cp used when the pressure is constant, while Cv is used when the volume is constant? Thank you!

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In this problem, delta S and delta U are equal to 0 when referring to the change in entropy/internal energy over the course of the entire process, which the question is asking for. Entropy and internal energy are state functions, meaning we can look at the initial and final states directly without ...

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In this problem, delta S and delta U are equal to 0 when referring to the change in entropy/internal energy over the course of the entire process, which the question is asking for. Entropy and internal energy are state functions, meaning we can look at the initial and final states directly without h...

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Irreversible expansion refers to work done against a constant external pressure. To calculate the work, you would use the equation w = -P(delta V). In an irreversible expansion, there is a big difference in internal and external pressure, which is why the expansion is much quicker than a reversible ...

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ATP hydrolysis is listed under the second page of the Thermodynamics Outline, which Dr. Lavelle announced would not be on the upcoming midterm. Only content from the first 3 outlines and the first page of the Thermodynamics Outline will be covered on the exam.

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I believe that the formula sheet we received for Test 1 had all of the relevant formulas we needed, up to the thermodynamic equations we are currently studying. The bottom half of the online formula sheet that was not included in Test 1 deals with electrochemistry and kinetics, which aren't covered ...

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I believe that the standard entropies of substances will be given to us, if needed. It doesn't seem like we need to memorize them.

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Yes, it depends on the units that the given heat capacity is in. For example, if you're given the molar heat capacity, you should be using moles. On the other hand, if you're given the specific heat capacity, you should be using grams. Looking at the units is your best bet in determining whether you...

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When a process is isothermal, delta U is equal to zero. Therefore, q = -w (since your equation would become 0 = q + w), just like you said. Essentially, as work is done by the system, heat flows back into it, making delta U zero. q and w would have to be equal and opposite for the conditions of an i...

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Either should be acceptable. The problem will most likely specify whether or not your answer should be in J or kJ

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Given that the process is adiabatic, we know that q=0. According to the 1st Law of Thermodynamics, delta U=q+w. Since q=0, we can conclude that delta U will be equal to w in an adiabatic process.

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When expansion takes place, work is being done and is therefore negative. When compression takes place, work is positive.

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Heat capacity refers to the amount of energy needed to raise the temperature of a substance by 1 C (or K). Specific heat capacity is the amount of energy needed to raise the temperature of one gram of a substance by 1 C (or K). A crucial difference is that heat capacity is an extensive property whil...

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I'm not sure if you'd have to memorize it but it's fairly simple and easy to remember. To convert from Celsius to Kelvin, add 273. Likewise, to convert from Kelvin to Celsius, subtract by 273.

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Because enthalpy is a state function, it is not path-dependent, meaning only the final and initial states matter in calculations. As a result, values of enthalpy are additive. Hess's Law is used to add up several reactions' enthalpies to find the total reaction enthalpy, which we would not be able t...

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Delta U is equal to q + w. When no energy is transferred as heat, one can say that delta U is equal to w (meaning the change in internal energy is equal to the work). As described above, the volume would be constant in this situation.

Hiba Alnajjar_2C

8.314 J*K^-1*mol^-1 is the gas constant, R. Example 4C.1 does a good job of explaining the use of this.

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No, we do not need to memorize any Ka or Kb values. You should be able to interpret them and/or calculate them though.

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On Outline 2, the last bullet point (see below) states that we should be able to calculate the pH in particular examples. Do we have to memorize chemical formulas for these examples for the upcoming exam or can we expect to be given the reaction/chemical formula? Thank you!! • Calculate the pH in th...

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In an endothermic reaction, heat can be viewed as a reactant. Just as increasing the amount of reactants drives product formation, increasing heat in an endothermic reaction will also drive product formation.

Hiba Alnajjar_2C

Since the last example Dr. Lavelle discussed during Friday's lecture pertained to calculating the pH of a buffer solution, I believe that it could potentially be on the exam, as it was part of the content covered in lecture.

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Changes in concentration do not result in a different K value. When we discuss the reaction shifting, this means that the concentration of either the reactants/products will change in order to minimize the effect of a change in concentration. The concentrations of the reactants and products will be ...

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If you multiply a chemical equation by a factor, the value for K would be the original value to the square root of the factor. For example, if you multiply a chemical equation by a factor of 3, your new K value would be the original K to the power of 3.

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If it doesn't specify Kc or Kp, either should be fine. However, you should be able to tell what the question wants based on what's given. For example, if it gives you concentration values, you would probably have to find Kc.

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Will titrations (particularly weak acid/strong base or strong acid/weak base) be tested on next week's exam? Or will we not be covering them this week? Thank you!

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I had the same issue as you with this problem, as typically you do have to balance the reaction before writing the equilibrium expression to ensure that your stoichiometric coefficients are correct. However, I checked the answer key and its stoichiometric coefficients match the equation as given (4:...

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A very low value of K indicates that there are more reactants than products at equilibrium. As described above, in the autoprotolysis reaction, very rarely does H2O dissociate to form H3O+ and OH-. As a result, you have a greater concentration of H2O than H3O+ and OH-, and therefore a greater concen...

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805097738 wrote:do you omit pure solids and liquids while solving for K too?

Yes, that is correct.

Hiba Alnajjar_2C

Yes; when solving for Q, you write the expression in the same exact way you would for K, omitting pure liquids and solids. The central difference between Q and K is that K represents the value at equilibrium, while Q is the ratio of products to reactants at any time during the reaction. But you woul...

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When writing the expression for K, include gases and aqueous substances and leave out solids and pure liquids. Liquids are not included in the equilibrium constant because there is no considerable change in concentration, as the number of molecules involved in the reaction is much smaller than the n...

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The Gibbs free energy is equal to -RT*ln(K) where K is the equilibrium constant.

Hiba Alnajjar_2C

I completely agree with the previous responses about online resources! I just wanted to add that the textbook also covers ideal gases in Focus 3A-D, which I also found to be particularly useful!

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I believe that the answer to this question can either be in terms of partial pressures or concentrations since it doesn't specify whether it wants Kc or Kp. I believe that partial pressures are more commonly used for gases as opposed to concentrations, but both options are correct.

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Could somebody clarify the difference between molecular shape and molecular geometry? In high school, I had learned it as electron geometry & molecular geometry so I would appreciate clarification about the difference. Does molecular shape treat bonds and lone pairs as the same, whereas molecula...

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During discussion, my TA pointed out that drawing the Lewis structure of the molecule can help you determine where an H could logically be attached. For this example, you would draw out C6H5O-'s structure and identify that the O atom is the best place to attach another H.

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Could someone expand upon potential options for the reduction of acid rain & explain the reasoning behind those solutions? Much appreciated!

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Yes! Water is amphoteric since it can act as both an acid and a base.

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Based on the outline posted on Dr. Lavelle's website, it does not appear that we need to know the names of acids and bases for Chem 14A. It appears that the problems will be dealing with formulas rather than names like bromous acid.

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Given that you're looking for the longest wavelength (and therefore the least amount of incoming energy required), you can assume that the Kinetic Energy is equal to 0. Then, you could rearrange the formula so that the work function is equal to h*c divided by the wavelength. You are essentially usin...

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I think that Dr. Lavelle was mainly doing so in order to help us understand how these reactions work, so I am not entirely sure we'll be tested on it. However, if you were, you would simply have to draw out the Lewis Structures like we've been doing all quarter so it's really nothing too new :) It w...

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The order matters for the actual name, not the formula. As long as the ligands are after the transition metal, the order is arbitrary. You only have to pay attention to organizing them alphabetically when writing out their name.

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I don't believe that ironate would be correct. From what I have understood based on the chart and textbook, ferrate seems to be the best choice for those examples.

Hiba Alnajjar_2C

Though it would be best to ask your TA, I believe that memorizing the chart sent by Dr. Lavelle is the best thing you could do. Most of the examples in the textbook and in class do not appear to extend beyond what is included in that chart, so I think you'd be fine just learning that!

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It is possible for a polydentate ligand to not be a chelating ligand. A good example of this is in Textbook Question 9C.7, where certain isomers are unable to form a ring based on their structure.

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In addition to VSEPR, it'll also cover intermolecular forces & sigma/pi bonds.

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Due to repulsion from lone pairs, it is sometimes more favorable for a lone pair to be in the equatorial position rather than axial (or vice-versa). This is an especially prevalent concept when looking at trigonal bipyramidal and octahedral structures. It is more favorable to put a lone pair in the ...

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With trigonal bipyramidal structures, lone pairs are preferred to be in the equatorial position rather than axial (above/below), since there is less repulsion if you put the lone pair in the equatorial position. Therefore, you would have one atom and two lone pairs bound in the equatorial plane, whi...

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SiF4 would have a higher boiling point. Because F is larger than H and contains more electrons, it is more polarizable. Because it is more polarizable, its electrons are more easily distorted, meaning its London Dispersion forces are stronger than that of SiH4. Because of these stronger intermolecul...

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Yes, it would have a bond angle of 180 degrees. And for AXE2, it should also be 180 degrees. The molecular geometry indicates the actual observed shape since it accounts for repulsion from lone pairs, while the electron geometry just indicates the number of bonding domains. Therefore, you should acc...

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Yes, since hydration is an example of ion-dipole interactions. If there is a larger extent of hydration, this means that the cation is strongly interacting with H2O's Oxygen's partial negative charge.

Hiba Alnajjar_2C

The test should cover everything we've done after the midterm, so the concepts after Focus 2D, including intermolecular forces and VSEPR. From what I understood during yesterday's lecture, we won't be required to actually draw VSEPR models but, rather, we'll have to identify the molecular shape &...

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It's everything we've done after the midterm. Since the midterm went up to Focus 2D, it would make sense for the test to cover 3F and 2E. However, asking your TA would be a good idea.

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Not sure when they'll be posted, but Dr. Lavelle announced that we would be getting them back during Wednesday lecture, so maybe after Wednesday?

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I believe that Nitrogen, Oxygen, and Fluorine are hydrogen bond acceptors due to their high electronegativities. Because Hydrogen would have a partial positive charge, a highly electronegative atom would form a stronger hydrogen bond than an electropositive one.

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The series of short lines represents the hydrogen bond between the two H2O molecules (making it an intermolecular force). This is done to differentiate from the intramolecular covalent bonds, which are solid lines.

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I believe that we will be tested on all the concepts we've learned throughout the quarter. All the concepts, to some extent, build upon one another so it might be helpful to continue reviewing.

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First, you have to make sure the information that you've been given is in the right units. Wavelength should be in m and the work function should be in J/electron. Therefore, you have to make these conversions from nm to m and kJ/mol to J/electron before you can start. Using the wavelength, use de B...

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Yes! Essentially, Dr. Lavelle discussed how it is necessary to correct the ionic model to account for the fact that electrons are shared, to some extent, between a cation and anion due to these distortions. With regards to solubility, you are correct in stating that the more covalent character a com...

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It might be helpful to check out the constants sheet on Dr. Lavelle's website so that you can determine which formulas are given to you and which ones you need to memorize. As for knowing when to utilize certain formulas, I find the best way of learning that to be doing practice problems. That way y...

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Because these three compounds are ionic rather than covalent, it would be inaccurate to connect them like we have typically been doing. Ionic compounds involve a transfer of electrons, rather than a sharing of electrons (which the lines in Lewis structures traditionally represent).

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I believe that they should be separated around the phosphide structure, as this is an ionic compound. In contrast to covalent compounds, no electrons are being shared between the ions. Rather, electron(s) are being transferred, meaning you shouldn't draw a connection in your structure.

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When removing electrons to form an ion, you remove electrons from the outermost shell. In this case, that's the 4th shell (n=4), which is occupied by 2 electrons in the 4s orbital and 1 electron in the 4p orbital. As a result, you remove these three electrons before removing any from 3d.

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Given that there's going to be a midterm review session focusing on the Fundamentals, I assume that the Fundamentals will be on the midterm. However, I would confirm with either Dr. Lavelle or one of the TAs.

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The valence electrons are those found in the outermost shell , not sub shell. As a result, for Sb, you would include the 2 electrons from the 5s sub shell as well as the 3 electrons in the 5p sub shell, rather than just those in the 5p sub shell. A helpful trick would be to identify the highest valu...

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Yes, h is Planck's constant (6.626 x 10^-34 J*s)

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For the Week 5 homework, should we do questions from The Quantum World unit or Chemical Bonds? Or would either work? Thank you!

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LL14A19

Hope this helps!

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Yes, isoelectronic atoms are those that have the same number of electrons! It might be important to note that isoelectronic atoms may also have similar chemical behavior and may react in a similar way because they have the same number of valence electrons (and electrons in general).

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The "6" refers to the value of the principal quantum number, n, essentially telling you that the electron is in the 6th shell. Since n=6, we can determine that the possible values of l are 0,1,2,3,4,5 (as it goes from 0 to n-1). As you pointed out, the specific sub shell we're focusing on ...

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An atom is considered to be a cation when it's lost an electron(s) from its neutral state and is thus positive (a common method used to remember this is that "cats" are paws-itive). On the other hand, if an atom gains an electron(s), it becomes more negative and is considered to be an anio...

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I believe that the quantity of energy found in E=hv is just the smallest unit of energy for that particular frequency. It might help to think of the equation as E=(number of photons)*hv and, so, if your number of photons is just 1, then you're left with E=hv. You could then take that energy and mult...

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For parts b and c, you essentially do dimensional analysis, just as we did in the high school review unit. Because your answer to part a gives you the amount of energy emitted per atom, you can use dimensional analysis to convert from mg to g to mol to atoms. After finding the number of atoms of Na,...

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The mass of an electron is equal to 9.11 x 10^-31 kg. To find the wavelength of an electron, you would use the de Broglie equation- the wavelength of an electron is not a constant since it depends on the energy of the electron, which is different for electrons occupying different energy levels.

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Yes! That is the correct unit for energy in the E=hv equation. It gives you the amount of Joules of energy per photon.

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Arranging these types of electromagnetic radiation in order of increasing frequency is synonymous with arranging them based on increasing energy (since energy and frequency are directly proportional; the higher the frequency, the higher the energy) or decreasing wavelength (the higher the frequency,...

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You can definitely solve this problem without the Rydberg formula. I solved this problem using E=hv and E=-hR/n^2. If you convert the given wavelength to frequency [c=(wavelength)(frequency)], you can then use E=hv to find the change in energy. This change in energy is equal to the energy of the fin...

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Have you checked out the outlines on the Chem 14A website? I find those pretty helpful for determining which sections of the book to read depending on the unit. You could also ask Dr. Lavelle or your TA for suggestions on what to read. Hope this helps!

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You can find the minimum frequency required for electrons to be ejected using the value for the minimum amount of energy required for electrons to be ejected. It ultimately depends on which information you're given, but if given the minimum amount of energy, you can use the equation E=hv to solve fo...

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