CHEMISTRY COMMUNITY

Asia Yamada 2B

The reason why a larger Ka value correlates to a stronger acid is because Ka=[products]/[reactants] and the larger the Ka value is, the more products there are. When there are a lot of products, that means that a lot of the acid (reactant) dissociated which means there will be more H+ ions, making t...

Asia Yamada 2B

Yes, an increase in pressure would cause the reaction to shift in the direction that reduces the number of gas-phase molecules, but since the number of gas-phase moles on the reactant side and product side is the same, the equilibrium doesn't change.

Asia Yamada 2B

A) Yes, if the partial pressure of CO2 is increased, the reaction will then shift towards the reactants which decreases the partial pressure of H2 as a product. B) Yes, if the partial pressure of CO is decreased, the reaction will shift towards the reactants which means the partial pressure of CO2 w...

Asia Yamada 2B

When you decrease the volume, the pressure increases, so the equilibrium shifts to reduce the number of gas-phase molecules. However, the equilibrium constant, K, is not affected by a change in pressure.

Asia Yamada 2B

This is referring to Le Chatelier’s Principle. When you cause a disturbance to a system, the system tries to reestablish equilibrium by either shifting towards the products or towards the reactants.

Asia Yamada 2B

K is the ratio of products to reactants at equilibrium. Q is the reaction quotient that is calculated at any point in time during the reaction. It is similar to K where it is the ratio of products to reactants. You can compare Q to K to determine which way the reaction will proceed to eventually est...

Asia Yamada 2B

Yes Q < K, but that means that the reaction would proceed to the right, forming products. For part C, since you are given the initial concentrations and the value of Kc, you can use the ICE table to determine the concentrations at equilibrium.

Asia Yamada 2B

You can refer to https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf for the error in the value of Kc.

Asia Yamada 2B

Aqueous solutions are included in the equilibrium constant equation. However, pure substances like liquids and solids are not.

Asia Yamada 2B

Yes, Q is the reaction quotient that is calculated at any point in time during the reaction. You can compare Q to K to determine which way the reaction will proceed to eventually establish equilibrium where Q=K.

Asia Yamada 2B

When the concentration of products or reactants is changed, the system shifts in the direction that will reestablish equilibrium. So, if you decrease the concentration of the products, the reaction will shift to the right until the original product:reactant ratio is obtained. Similarly, if you decre...

Asia Yamada 2B

If you compress a system (volume decreases) and pressure increases, the system shifts towards the side with less moles to reestablish equilibrium. If you expand a system (volume increases) and pressure decreases, the system shifts towards the side with more moles to reestablish equilibrium. However,...

Asia Yamada 2B

If a reaction is endothermic, the reaction will absorb heat and shift to the right which increases the concentration of the products and therefore increases the value of the equilibrium constant. On the other hand, if the reaction is exothermic, that means that the reaction gives off heat and shifts...

Asia Yamada 2B

If a reaction is endothermic, the reaction will absorb heat and shift to the right which increases the concentration of the products and therefore increases the value of the equilibrium constant. On the other hand, if the reaction is exothermic, that means that the reaction gives off heat and shifts...

Asia Yamada 2B

Yes, you would omit As because it’s a solid and the concentration of a pure substance like a solid will not affect the reaction. The equilibrium constant would then be [H2]^3 / [AsH3]^2.

Asia Yamada 2B

Electronegativity was experimentally calculated using both ionization energies and electron affinities, so you cannot determine those numbers yourself. I think that a table of electronegativity values should be provided to you, and from there, you can calculate the electronegativity difference betwe...

Asia Yamada 2B

Nonmentals are more electronegative than metals and this is a periodic trend. Electronegativity increases across a period and decreases down a group. It increases across a period because even though the number of electrons are increasing, they’re being added to the same shell, so it doesn’t affect s...

Asia Yamada 2B

Electronegativity is how strongly an atom attracts electrons from other atoms.

Asia Yamada 2B

Ionization energy is the minimum energy needed to remove an electron from an atom in the gas phase. Electron affinity is the energy released when an electron is added to a gas phase atom. Electronegativity is the electron pulling power of an atom.

Asia Yamada 2B

To look at it more concretely, the electronegativity of Carbon, Fluorine, and Hydrogen, are 2.5, 4.0, and 2.1, respectively. The electronegativity difference of CF4 would be 4 - 2.5 which is 1.5. The electronegativity difference of CH4 would be 2.5 - 2.1 which is 0.4. Since 1.5 is bigger than 0.4, t...

Asia Yamada 2B

You can determine if a species is acidic or basic by looking at whether it donates a proton or accepts a proton. If it donates a proton, then it’s acidic. If it accepts a proton, then it’s basic. Amphoteric species are ones that can react with both acids and bases depending on what other species the...

Asia Yamada 2B

Yes, I agree that it’s similar to 2F.5. You would go through the same process as you did before to find the hybridized orbitals.

Asia Yamada 2B

I think that electronegativity is related to the s-character of the hybridized orbitals. The greater the s-character of the hybridized orbitals, the greater the electronegativity because the electrons are held tighter to the nucleus.

Asia Yamada 2B

In H2S, there are 2 sigma bonds and no pi bonds because there are just two single bonds in this molecule. There is only one sigma bond in a single bond. In SO2, there are 2 sigma bonds and 2 pi bonds because there are two double bonds in this molecule. In a double bond, there is one sigma bond and o...

Asia Yamada 2B

In H2S, there are 2 sigma bonds and no pi bonds because there is just two single bonds in this molecule. In SO2, there are 2 sigma bonds and 2 pi bonds because there are two double bonds in this molecule.

Asia Yamada 2B

I believe that sigma bonds are formed from hybridized orbitals.

Asia Yamada 2B

I believe the two exceptions that are relevant to this class are chromium and copper.
chromium = [Ar] 3d^5 4s^1
copper = [Ar] 3d^10 4s^1

Asia Yamada 2B

NO is linear and also polar because oxygen is slightly more electronegative than nitrogen. I agree that in this case, the lone electron doesn’t affect the shape because a two atom molecule is always linear, but I’m not too sure as to how it would affect molecules with more than two atoms.

Asia Yamada 2B

I think the most important thing to know about radicals is that they have an odd number of valence electrons which means there is an unpaired electron, making them much more reactive because they want an electron.

Asia Yamada 2B

In Dr. Lavelle’s most recent lecture, I believe he said you can write it either way, but I prefer to write it as sp3d.

Asia Yamada 2B

In Dr. Lavelle’s most recent lecture, I believe he said you can write it either way.

Asia Yamada 2B

To determine which molecule has greater ionic character, you have to look at electronegativity. The greater the difference in electronegativity, the greater ionic character the molecule has because the electrons are being unequally shared. The electronegativity difference between Hydrogen and Chlori...

Asia Yamada 2B

C—F bonds have the smallest electronegativity difference which means that the C—F bond is the least polar out of the three different bonds. This means that CF4 has the least ionic character while CBr4 has the greatest ionic character. CBr4 has the most unequal sharing of electrons while CF4 has the ...

Asia Yamada 2B

You can approach this problem by thinking about the atomic radius of each halogen. As the distance between the nuclei of the bonded atoms decrease, then the strength of that bond increases. Out of these three halogens, fluorine has the smallest atomic radius, so the nucleus of Carbon and Fluorine ar...

Asia Yamada 2B

You can approach this problem by thinking about the atomic radius of each halogen. As the distance between the nuclei of the bonded atoms decrease, then the strength of that bond increases. Out of these three halogens, fluorine has the smallest atomic radius, so the nucleus of Carbon and Fluorine ar...

Asia Yamada 2B

You can approach this problem by thinking about the atomic radius of each halogen. As the distance between the nuclei of the bonded atoms decreases, then the strength of that bond increases. Out of these three halogens, Fluorine has the smallest atomic radius, so the nucleus of Carbon and Fluorine a...

Asia Yamada 2B

First, you find the total energy by multiplying 2.0 s by the conversion factor 32 W (32 J/s). The seconds will cancel and you will get 64 J as the total energy. Then you have to find the energy per photon. You have to convert given wavelength into meters which is 4.2 x 10^-7 m. Plug this into the eq...

Asia Yamada 2B

First, you find the total energy by multiplying 2.0 s by the conversion factor 32 W (32 J/s). The seconds will cancel and you will get 64 J as the total energy. Then you have to find the energy per photon. You have to convert given wavelength into meters which is 4.2 x 10^-7 m. Plug this into the eq...

Asia Yamada 2B

When you are given the frequency and are asked to find the energy, all you have to do is plug that frequency into the equation, E=hv, where h (Planck’s constant) = 6.63 x 10^-34. When you solved for wavelength and simplified the equation, it resulted in E=h(v/wavelength) which is untrue.

Asia Yamada 2B

Yes, the equation E=hv is an equation for the energy of an individual photon. If you wanted to find the amount of photons, then you divide the total energy by the energy of a single photon.

Asia Yamada 2B

In this case, you wouldn’t need to multiply by Avogadro’s number because the electron affinity you found is already for one atom.

Asia Yamada 2B

For this problem I follow the logic and step/process, but I am confused as to why the delta(v) is only 5 instead of 10. In the equation it says that the speed is 5.0 m/s + or - 5, so wouldn't that mean the uncertainty is 10 not 5? (5 for +5 and 5 for -5) I guess I am confused because I thought in l...

Asia Yamada 2B

I agree that the uncertainty in velocity would be 10 because you multiply by two when there’s a ± because that represents the range of values that the speed could be. I believe that there’s an error in the answer key. You can find the correct solution here. https://lavelle.chem.ucla.edu/wp-content/s...

Asia Yamada 2B

The process you went through was correct, except you divided by the uncertainty in position again when you were supposed to divide by the mass of the car instead. Another way you could have approached the problem is by isolating Δv from Heisenberg’s Uncertainty Equation (Δx • Δp ≥ h/4pi) first, sinc...

Asia Yamada 2B

First, you would convert the radius into meters which would be 5x10^-11m. You have to multiply the radius by 0.01 because the radius is known to an accuracy of 1%. You should get 5x10^-13m. This is the uncertainty in position and from there you can just plug the uncertainty in position into Heisenbe...

Asia Yamada 2B

First, you would convert the radius into meters which would be 5x10^-11m. You have to multiply the radius by 0.01 because the radius is known to an accuracy of 1%. You should get 5x10^-13m. This is the uncertainty in position and from there you can just plug the uncertainty in position into Heisenbe...

Asia Yamada 2B

In part a, you found the energy emitted by one atom of sodium which was 3.37x10^-19 J. Since there are 1.31^20 atoms in 5.00mg of Sodium, it emits 1.31x^20 times more energy than one Sodium atom alone. This is why you multiply instead of divide.

Asia Yamada 2B

Heisenberg’s Indeterminacy Equation is Δx • Δp ≥ h/4pi. Δx and Δp are inversely related because they are on the same side of the equation and h/4pi needs to stay constant. For example, if uncertainty in position increases, uncertainty in momentum must decrease to keep the right side of the equation ...

Asia Yamada 2B

First, you would convert the radius into meters which would be 5x10^-11m. You have to multiply the radius by 0.01 because the radius is known to an accuracy of 1%. You should get 5x10^-13m. This is the uncertainty in position and from there you can just plug the uncertainty in position into Heisenbe...

Asia Yamada 2B

I believe that you multiply by two when there’s a ± because that represents the range of values that the speed could be. In this case, the uncertainty in velocity should be 10 seconds.

Asia Yamada 2B

Velocity is related to momentum because the product of velocity and mass is equal to momentum. Therefore, uncertainty in velocity is related to uncertainty in position through Heisenberg’s Indeterminacy Equation which is Δx • Δp ≥ h/4pi or Δx • Δv• m ≥ h/4pi

Asia Yamada 2B

The diameter represents the uncertainty in position because the electron is confined to the space within the lead atom. Since you know Δx and ℏ (“h bar”), you can plug these values into Heisenberg’s Indeterminacy Equation (Δp x Δx ≥ 1/2 ℏ) to solve for Δp. Because the mass of an electron is 9.109x10...

Asia Yamada 2B

First, you calculate 2s•32J/s=64J (total energy). We are also given the wavelength (420nm) which we need to convert to meters in order to plug into the equation (c= λv) and solve for frequency. Once we have the frequency, we can solve for the energy of a photon by multiplying the frequency by Planck...

Asia Yamada 2B

For part b, you find the atoms of sodium in 5.00 mg of sodium by converting from milligrams to grams, then from grams to moles (molar mass of sodium), then from moles to atoms (Avogadro's number). Since you know the energy emitted by one excited sodium atom (part a), you multiply that energy by the ...

Asia Yamada 2B

To be detectable, a particle must have a wavelength of at least 10^-15 m. As the mass of an object increases, the wavelength decreases which makes it more likely that the object has particle-like properties rather than wave-like properties.

Asia Yamada 2B

I believe that the De Broglie equation works for any particle that has momentum (p or mv). It’s helpful in describing smaller particles because their wavelengths can be detected while objects with a larger mass have smaller wavelengths that cannot be detected.

Asia Yamada 2B

A mole of an ionic compound corresponds to 6.02x10^23 formula units. I think the answers are the same as molecules because the calculations are the same, but the type of particles are different.

Asia Yamada 2B

Amplitude is related to the intensity of light. However, because light isn’t characterized by wave properties only, the intensity of the light (and its amplitude) doesn’t affect whether electrons are ejected or not. This was discovered in the photoelectric experiments.

Asia Yamada 2B

From the Kinetic Energy equation, you can isolate velocity and then substitute it into De Broglie's equation. This will give you: λ = h/√(2mKE). To solve for the wavelength, you can then plug in the value that you’re given for kinetic energy along with Planck’s constant and the mass of an electron w...

Asia Yamada 2B

You are trying to figure out the identity of the metal in the metal hydroxide. In order to do that, you have to subtract the molar mass of (OH)₂ from the total molar mass of the metal hydroxide which is 74.10 g/mol. 74.10-34=40.1 g/mol. Using your periodic table, you will find that Calcium (Ca) has ...

Asia Yamada 2B

Although photons are extremely small particles of light, research suggests that photons are actually detectable by the human eye. Scientists discovered that once you are able to see a photon, there is a higher chance that you’ll be able to detect another photon shortly after that.

Asia Yamada 2B

Catalysts don’t affect chemical equations since they are not consumed in reactions. Catalysts aren’t considered reactants or products, so they just appear above the yield arrow, if you were to include it in the chemical equation.

Asia Yamada 2B

The book could just be rounding to a different number of sig figs than you are. Because the difference between the book’s values and your values are so small, it doesn’t cause a major discrepancy. I would continue to use your atomic masses because they include up to the thousandths place which is mo...

Asia Yamada 2B

Yes, instead of FeBr2, it should be Fe3Br8.

Asia Yamada 2B

A mole of an ionic compound corresponds to 6.02x10^23 formula units. To find the amount of formula units in 5.15g of Magnesium Sulfate Heptahydrate, you divide by the molar mass of this compound, then multiply by Avogadro’s number.

Asia Yamada 2B

I think that it would be best to just memorize all of the prefixes so it’s easier when you’re doing conversions. It’s especially important to have the most commonly used prefixes memorized.

Asia Yamada 2B

Once you’ve found the empirical formula, you need to calculate its molar mass. The molar mass of the molecular formula will be given. You divide the molecular molar mass by the empirical molar mass and this will result in a whole number, approximately. To find the molecular formula, you multiply eac...

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