CHEMISTRY COMMUNITY

Halle Villalobos 3E

Hi! Entropy is the measure of disorder in a system. A positive entropy corresponds to spontaneity, whereas a negtaive value is linked to a nonspontaneous reaction. I hope this helps!

Halle Villalobos 3E

Hi! If Q<K, the reaction will proceed towards the right to produce more products. If Q>K, the reverse reaction will be favored and more reactants will be produced. I hope this helps!

Halle Villalobos 3E

Hi! I always stick to kelvin since most constants use K. It doesn't hurt to spend the few extra seconds to convert!

Halle Villalobos 3E

Hi! Delta G is given in kJ.

Halle Villalobos 3E

Hi! I highly recommend going through each of the problems Lavelle posts on the outlines. They are extremely helpful in aiding your understanding of the content and he also uses some on the actual exam! His step-up sessions are also a great way to get extra help when it comes to solving these problem...

Halle Villalobos 3E

Hi! The change in Gibbs free energy is given by ∆G = ∆G˚ + RTln(Q). The values of ∆G˚ and T are given, so the only other value you need is Q. Calculate Q by dividing the partial pressure of the product raised to its coefficient by the partial pressures of the reactants raised to their coefficients. ...

Halle Villalobos 3E

Hi! It seems like you are executing the problem correctly. The proper R value to use for this problem is 8.314. I hope this helps!

Halle Villalobos 3E

Hi! We do not need to find heat capacity for this problem. You can calculate \Delta H_{vap} by dividing the amount of energy supplied to the sample from the heater by the number of moles of substance that were vaporized. Recall that 1W = 1J/s, and then convert the minutes provided into seconds and J...

Halle Villalobos 3E

Hi! $K = e^{(\Delta G_r/-RT)}$ . It seems like you did the calculation correctly, but I think the issue is with your notation. I hope this helps!

Halle Villalobos 3E

Hi! You can calculate \Delta H_{vap} by dividing the amount of energy supplied to the sample from the heater by the number of moles of substance that were vaporized. Recall that 1W = 1J/s, and then convert the minutes provided into seconds and J to kJ. Find the change in mass and divide that by the ...

Halle Villalobos 3E

Hi! To find the heat capacity of the calorimeter, use the equation C = \frac{q}{\Delta T} . To calculate q, convert the mass of the sample of compound A to moles using its molar mass, and then multiply the moles by the heat of combustion for compound A. Normally, when heat is produced or released, t...

Halle Villalobos 3E

Hi! Since SO2 consists of nonlinear molecules, the constant-pressure molar specific heat $C_P = 4R$ . I hope this helps!

Halle Villalobos 3E

Hi! Start by calculating the heat capacity of the calorimeter with C_{cal} = \frac{q_{cal}}{\Delta T} . Next, use the heat capacity of the calorimeter to calculate the heat released by the neutralization reaction. The heat lost by the reaction is given by the equation q_{reaction} = -q_{cal} = -C_{c...

Halle Villalobos 3E

Hi! For part a: When heat is produced or released, the sign of the heat of combustion is negative. However, since the sign is only used to denote that heat has been released, the absolute value of the heat of combustion is used in calculations. For part b: since heat is being released, the sign of q...

Halle Villalobos 3E

Hi! The change in internal energy of a system is $\Delta E = w+q$ . Remember to use the conversion 4.184 kJ/kcal for the first value so that your units match up. I hope this helps!

Halle Villalobos 3E

Hi! When K < 10^-4, the change in X will be so small that we can make it negligible. While the book considers X negligible when K < 10^-3, Lavelle notes that K < 10-4 is a better rule to work from. I hope this helps!

Halle Villalobos 3E

Hi! Lavelle emailed us the other day saying he has extended the deadline to Sunday of Week 5.

Halle Villalobos 3E

Adding an inert gas has no effect on the concentration and therefore no effect on the reaction.

Halle Villalobos 3E

Hi! There is a download link for respondus on our class's CCLE page. When taking the midterm or final, you will be prompted to launch the lockdown browser, which should automatically happen after clicking the button. Remember to make sure it is up to date before taking the next exams. I hope this he...

Halle Villalobos 3E

Hi! Kc is used to denote the equilibrium constant when using concentrations. It will depend on the context of the problem for which one you would use. I hope this helps!

Halle Villalobos 3E

Hi! pKa = -log(Ka). Lavelle has mentioned before in lecture that p is representative of taking the negative log of something. For example, we know that pOH is the -log[OH-]. I hope this helps!

Halle Villalobos 3E

Hi! I believe the error comes in when you calculate your value for Kb. You should have gotten 2.5*10^-7, which is small enough that you can use an approximation when solving for x. When taking the -log(x) remember that it is representative of the pOH since you are working with Kb. I hope this helps!

Halle Villalobos 3E

Hi! I would suggest double-checking your work before you plug in your values into the quadratic formula. To make the process more efficient, I would suggest uploading the program QUADFORM onto your calculator, if you haven't already. That way, you can directly input values that were previously calcu...

Halle Villalobos 3E

Hi! When values of Ka or Kb are < 10^-4, they are considered small enough that we can make the approximation of x equaling 0. I hope this helps!

Halle Villalobos 3E

Hi! You would have to multiply the enthalpy by -2 to use the reverse reaction and to change the coefficients. This will change the enthalpy to 208 kJ for the equation: 2MCl3(aq) -> 2MCl3(s). I hope this helps!

Halle Villalobos 3E

Hi! Lavelle noted in a lecture that if the K value is less than 10^-4, the x can be negligible. Another way of checking is to see if the x value is less than 5% of the initial concentration given. I hope this helps!

Halle Villalobos 3E

Hi! When pressure doubles due to a decrease in volume, the concentrations double and K remains constant. This is because concentration = # moles (n)/ volume (V) and n is constant. Therefore, concentration and volume are inversely proportional. In the case of pressure increasing due to the addition o...

Halle Villalobos 3E

Hi! It is the same procedure as when you would use concentrations. Remember to make sure all your units are in bar/atm when using partial pressure. I hope this helps!

Halle Villalobos 3E

The ICE box helps us find the equilibrium concentrations of a given reaction. The I stands for the initial concentration, C stands for the change in concentration, and E stands for the equilibrium concentration. Lavelle works out an example problem in the Chemical Equilibrium Part 3 Audio-Visual Top...

Halle Villalobos 3E

Hi! Yes, Ka*Kb will always be equal to Kw, which we know = $[H_3O^+][OH^-]$ = 1.0 * 10^{-14}. I hope this helps!

Halle Villalobos 3E

Hi! Lavelle noted in his audio-visual topic video that at the molecular level, reactants are still forming products and products are still forming reactants. The rates of the forward and reverse reactions are the same. I hope this helps!

Halle Villalobos 3E

Hi! To my understanding, no. The change in concentration is due to an increase/decrease in volume. And since K is a ratio, it remains constant no matter how much you alter the volume of the system. I hope this helps!

Halle Villalobos 3E

Hi! You would choose the value that does not exceed the initial concentrations given. Also, ignore any negative values of x. I hope this helps!

Halle Villalobos 3E

Hi! It really depends on what units are given to you and the context of the problem. But generally speaking, you would use Kp when working with gases because they are the only ones that have partial pressures. If you need to convert, use the ideal gas law (PV = nRT) to determine the concentration wh...

Halle Villalobos 3E

Hi! A reaction is exothermic when delta H is negative and is endothermic when delta H is positive. Based on this, you can figure out which direction either a cooling or heating will shift the K. I hope this helps!

Halle Villalobos 3E

Hi! It is important to remember these basic formulas:
pH + pOH = 14
pH = -log[H+]
[H+] = 10^(-pH)
pOH = -log[OH-]
[OH-] = 10^(-pOH)

Halle Villalobos 3E

Hi! The conjugate acid of HSO4- would be H2SO4 since you would be adding an H+.

Halle Villalobos 3E

Hi! A single bond has one sigma bond. A double bond has one sigma and one pi bond. A triple bond has one sigma and two pi bonds.

Halle Villalobos 3E

Hi! You would use "-ate" at the end of a metal name if it is an anion complex.

Halle Villalobos 3E

Hi! They are essentially the same thing. It's just personal preference in the nomenclature one uses.

Halle Villalobos 3E

Hi! I use exact numbers during calculations and wait till my final answer to round.

Halle Villalobos 3E

Hi! This website explains the differences between the two and provides examples:
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/14%3A_Acids_and_Bases/14.07%3A_Strong_and_Weak_Acids_and_Bases
I hope this helps!

Halle Villalobos 3E

Hi! They are the same thing. The lower the pH, the higher concentration of hydrogen ions in the solution.

Halle Villalobos 3E

Hi! I believe there is a link to a google drive test bank folder in the class GroupMe that has past exams. Good luck with finals!

Halle Villalobos 3E

Hi! The two amine groups in a and c are arranged so that they are not able to coordinate simultaneously to the same metal center. It is possible for them to form a bridging ligand, but that is not classified as chelating. Therefore, the only molecule that can function as a chelating ligand is b. I h...

Halle Villalobos 3E

Hi! This YouTube video by The Organic Chemistry Tutor may help explain VSEPR Theory to you: https://www.youtube.com/watch?v=DBrq31w8vC4
I hope this helps!

Halle Villalobos 3E

Hi! The carbon atom has single bonds to four other atoms via four equivalent sp^3-hybridized orbitals. The oxygen atom has single bonds to two other atoms and two lone pairs of electrons. Therefore, oxygen is also sp^3-hybridized. I hope this helps!

Halle Villalobos 3E

Hi! Resonance structures are multiple Lewish structures that collectively represent a molecule. Given that all three of the compounds have different atom connectivities, none of them are resonance structures of another. I hope this helps!

Halle Villalobos 3E

Hi! Delocalized pi bonds occur when pi orbitals extend over more than two atoms. In terms of Lewis structures, this occurs with resonance structures involving double and triple bonds. I hope this helps!

Halle Villalobos 3E

Hi! The weekly discussion grades refer to the 5 weekly Chemistry Community posts. By midnight tonight, you should have a total of 40. I hope this helps!

Halle Villalobos 3E

Hi! You should have at least 35 posts by Sunday midnight. We are required to post a minimum of 5 each week , so the number of posts over 5 do not carry over to the next week's requirement. You can check how many posts you have by clicking the "Quick links" icon on the top left and then goi...

Halle Villalobos 3E

Hi! A boiling point is a temperature at which a liquid turns to a gas state (water to water vapor). A melting point is when something goes from a solid state to a liquid state (ice to liquid water). I hope this helps!

Halle Villalobos 3E

Hi! I found this video on youtube that might help clarify things: https://www.youtube.com/watch?v=2VRTUuCnoxQ I hope this answered your question!

Halle Villalobos 3E

Hi! The answer is yes. A molecule is non-polar if the charge distribution is symmetric and polar if the charge distribution is asymmetric. After drawing your VSEPR model, you can analyze the molecule. If it has symmetry around the central atom, the bond dipole moments will "cancel out" and...

Halle Villalobos 3E

Hi! I found this video on youtube that might help: https://www.youtube.com/watch?v=rlx5dgKxWmo I also think we will begin to memorize the angles as we practice using the structures over time. I hope this helps!

Halle Villalobos 3E

Hi! According to the textbook, "although higher values of l (corresponding to g-, h-, . . . orbitals) are possible, the lower values of l (0, 1, 2, and 3) are the only ones that chemists need in practice." I hope this helps!

Halle Villalobos 3E

Hi! A Lewis structure in which the formal charges of the individual atoms are closest to zero typically represents the lowest energy arrangement of the atoms and electrons. I hope this helps!

Halle Villalobos 3E

Hi! I would assume yes. Midterm 2 covers everything in outline 3, which states that we should know the characteristics of Lewis acids and bases.

Halle Villalobos 3E

Hi! Formal charge exaggerates the covalent character of bonds by assuming that the electrons are shared equally, whereas the oxidation number exaggerates the ionic character of bonds. Formal charges depend on the particular Lewis structure you write, oxidation numbers do not. I hope this helps!

Halle Villalobos 3E

Hi! The electron configuration of Sc+ is[Ar] 3d^1 4s^1. This is because the 3d orbitals are compactly arranged around the nucleus. Introducing a second electron into a 3d orbital produces more repulsion than if the next electron went into the 4s orbital. Therefore, the configuration mentioned is the...

Halle Villalobos 3E

Hi! Similar to you, I have a hard time staying motivated and focused to do my work when I am in the comfort of my own room. I found the Pomodoro technique to be extremely helpful for me to prevent burnout and to stay on task. I also make sure all my devices are on do not disturb and don't check them...

Halle Villalobos 3E

Hi! I found this webpage that might help: https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Chemistry_for_Allied_Health_(Soult)/11%3A_Properties_of_Reactions/11.01%3A_Oxidation_Numbers It does note that there are many elements for which there are no specific rules, but there are...

Halle Villalobos 3E

Hi! I found this old post that may address your issue: viewtopic.php?t=31149 I hope this helps!

Halle Villalobos 3E

HI! To my understanding, it has to do with the covalent character within ionic bonds, and a greater polarizability means a greater covalent character. Since they have a large distortion within their electron shell (polarizability), they are less soluble in water. I hope this helps!

Halle Villalobos 3E

Hi! The octet rule refers to the tendency of atoms to prefer to have 8 electrons in the valence shell. When atoms have fewer than 8 electrons, they tend to react and form more stable compounds. There are a few exceptions that Lavelle went over in previous lectures. Make sure to remember those! I hop...

Halle Villalobos 3E

HI! When solving a problem regarding the photoelectic effect, we know that hv= \Phi - E_k . hv represents the energy of the photon. \Phi is the energy required to remove an electron (threshold energy). E_k is kinetic energy. Knowing any two of the three values in this equation will help you solve fo...

Halle Villalobos 3E

Hi! m_l is the magnetic quantum number that labels different orbitals of a subshell (gives the orientation of the angular momentum). It has the allowed values of l, l-1, ..., -l . There are 2l+1 different values of m_l for a given value of l and therefore 2l+1 orbitals in a subshell of quantum numbe...

Halle Villalobos 3E

Hi! For part b, you would use E=hv and plug in the given values to get E=(6.626*10^{34}J.s)(2.50*10^{-16}s^{-1})=1.66*10^{-17}J . For part c, the photon must have enough energy to eject the electron and cause it to move at 3.6*10^3 km.s^{-1} . To find the energy of the photon, add th...

Halle Villalobos 3E

Hi! I have not taken chemistry since my sophomore year so I am a little rusty as well. I have found Khan Academy and The Organic Chemistry Tutor videos on youtube to be extremely helpful in explaining concepts. I highly recommend taking a look at both sources. I hope this helps!

Halle Villalobos 3E

Hi! I found this video on youtube that might answer your question. https://www.youtube.com/watch?v=YIpc4RNhuK4 I hope this feeds your curiosity!

Halle Villalobos 3E

Hi! We will be allowed a periodic table and formula sheet while taking the midterm. The formula sheet includes the mass of electrons, protons, and neutrons. As for masses of certain elements, remember that 1kg = 10^3 g as we should always convert to kg in our calculations. I hope this helps!

Halle Villalobos 3E

Hi! Lavelle recently emailed us saying how he fixed the order of the Sapling homework to match what will be on the midterm. We should be familiar with the first 19 questions and the rest will be covered in week 4. I hope this helps!

Halle Villalobos 3E

Hi! To find the mass of a He atom you would perform the following equation: $\textup{1 He atom}*\frac{\textup{1 mol He}}{6.022*10^{23}\textup{ He atoms}}*\frac{\textup{4.00 g}}{1\textup{ mol He}}*\frac{\textup{1 kg}}{10^3\textup{g}}\approx 6.64*10^{-27}\textup{kg He}$ . I hope this helps!

Halle Villalobos 3E

Hi! First we rearrange Heisenberg's indeterminacy equation to $\Delta x \geq \frac{h}{4pi*m \Delta v}$ . Plugging in our known values gives us $\Delta x \geq \frac {h}{(4pi)(1.68*10^{-27})(10)}$ which results in a minimum uncertainty in position of 3.1 nm. I hope this helps!

Halle Villalobos 3E

HI! You would use Heisenberg's indeterminacy equation of $\Delta p \Delta x \geq \frac{h}{4pi}$ . Knowing that $\Delta p = m \Delta v$ , we can reaarange the equation to $\Delta x \geq \frac{h}{4pi * m\Delta v}$ . From there it is just plug and chug. I hope this helps!

Halle Villalobos 3E

You would use the De Broglie equation where wavelength = Planck's constant/(mv). The answer to the second question is yes because all matter has wavelike properties but only noticed for objects (like electrons) with a small mass and high velocity.

Halle Villalobos 3E

I am assuming you would multiply the uncertainty by Avogadro's constant since its unit is (mol^-1).

Halle Villalobos 3E

For part A, use Ek =(1/2)mv^2. The mass for an electron is 9.11*10^(-31) kg.
For part B, you would divide the work function by Avogadro's constant since the units for work function are (kJ/mol)
For part C, you would divide the sum of the kinetic energy and workfunction by Planck's constant

Halle Villalobos 3E

The extent of change of the electrical field is the slope of the wave. Since the electrical field corresponds to the amplitude and frequency is being decreased, the slope of the wave decreases.

Halle Villalobos 3E

Yes! The photon's energy is equal to that of the energy that excited the photon.

Halle Villalobos 3E

Hi! I think this previous thread would help clarify your question: viewtopic.php?t=1138#p2312
I hope this helps!

Halle Villalobos 3E

Hi! I would say that the units you use do not really matter unless the problem asks for a specific one. Usually, I see problems utilize mL when the value in L is a very small decimal.

Halle Villalobos 3E

My TA told us that Lavelle really encourages us to show our work. I'm not sure if we would necessarily get decked off points for not doing it, but I think it's a good habit to get into, especially when you go back to check your work when you get a question wrong.

Halle Villalobos 3E

Hi! I found this web page (https://sciencenotes.org/use-4-signific ... os-number/) that states using the 6.022*10^23 version is better for doing chemical calculations since you will be working with atomic masses, which are generally 4 sig figs. I hope this helps!

Halle Villalobos 3E

Hi! You also should've recently received an email from your TA with a download link to the program Respondus which is a locked browser we will use for quizzes and midterms.

Search found 85 matches