Search found 67 matches

by Asia Yamada 2B
Wed Jan 20, 2021 12:49 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Ka correlation to strength of an acid
Replies: 17
Views: 51

Re: Ka correlation to strength of an acid

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...
by Asia Yamada 2B
Tue Jan 19, 2021 11:56 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: HW 5j #5
Replies: 5
Views: 58

Re: HW 5j #5

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.
by Asia Yamada 2B
Tue Jan 19, 2021 11:47 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 5J.1
Replies: 2
Views: 45

Re: 5J.1

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...
by Asia Yamada 2B
Tue Jan 19, 2021 11:38 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Pressure Changes
Replies: 3
Views: 20

Re: Pressure Changes

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.
by Asia Yamada 2B
Tue Jan 19, 2021 9:30 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Equilibrium Adjustments
Replies: 9
Views: 32

Re: Equilibrium Adjustments

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.
by Asia Yamada 2B
Tue Jan 12, 2021 1:27 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Reaction Quotient
Replies: 10
Views: 465

Re: Reaction Quotient

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...
by Asia Yamada 2B
Tue Jan 12, 2021 1:24 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: 5I.27
Replies: 6
Views: 36

Re: 5I.27

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.
by Asia Yamada 2B
Tue Jan 12, 2021 1:19 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: 5.39
Replies: 3
Views: 72

Re: 5.39

You can refer to https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf for the error in the value of Kc.
by Asia Yamada 2B
Tue Jan 12, 2021 1:14 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: aqueous solutions
Replies: 9
Views: 25

Re: aqueous solutions

Aqueous solutions are included in the equilibrium constant equation. However, pure substances like liquids and solids are not.
by Asia Yamada 2B
Tue Jan 12, 2021 1:10 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: q vs k
Replies: 62
Views: 162

Re: q vs k

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.
by Asia Yamada 2B
Tue Jan 05, 2021 1:23 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Changes in Concentration
Replies: 8
Views: 339

Re: Changes in Concentration

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...
by Asia Yamada 2B
Tue Jan 05, 2021 1:15 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: How does K change during compression/expansion?
Replies: 6
Views: 163

Re: How does K change during compression/expansion?

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,...
by Asia Yamada 2B
Tue Jan 05, 2021 1:02 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: temperature change in reactions
Replies: 11
Views: 84

Re: temperature change in reactions

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...
by Asia Yamada 2B
Tue Jan 05, 2021 12:51 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Temperature/Pressure Effect on Equilibirum (Le Chatlier)
Replies: 4
Views: 35

Re: Temperature/Pressure Effect on Equilibirum (Le Chatlier)

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...
by Asia Yamada 2B
Tue Jan 05, 2021 12:42 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Part 2 module review #19
Replies: 3
Views: 39

Re: Part 2 module review #19

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.
by Asia Yamada 2B
Mon Dec 07, 2020 3:21 pm
Forum: Electronegativity
Topic: Determining Difference in Electronegativity
Replies: 2
Views: 394

Re: Determining Difference in Electronegativity

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...
by Asia Yamada 2B
Mon Dec 07, 2020 3:15 pm
Forum: Electronegativity
Topic: Nonmetals?
Replies: 2
Views: 146

Re: Nonmetals?

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...
by Asia Yamada 2B
Mon Dec 07, 2020 3:08 pm
Forum: Electronegativity
Topic: Defining electronegativity
Replies: 4
Views: 71

Re: Defining electronegativity

Electronegativity is how strongly an atom attracts electrons from other atoms.
by Asia Yamada 2B
Mon Dec 07, 2020 3:04 pm
Forum: Electronegativity
Topic: Differences
Replies: 3
Views: 52

Re: Differences

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.
by Asia Yamada 2B
Mon Dec 07, 2020 2:24 pm
Forum: Electronegativity
Topic: HW question 3.77
Replies: 4
Views: 381

Re: HW question 3.77

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...
by Asia Yamada 2B
Mon Dec 07, 2020 2:18 pm
Forum: Amphoteric Compounds
Topic: Recognizing Amphoteric Compounds
Replies: 9
Views: 154

Re: Recognizing Amphoteric Compounds

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...
by Asia Yamada 2B
Tue Dec 01, 2020 2:22 am
Forum: Hybridization
Topic: 2F.7
Replies: 4
Views: 79

Re: 2F.7

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.
by Asia Yamada 2B
Tue Dec 01, 2020 2:02 am
Forum: Hybridization
Topic: Electronegativity
Replies: 4
Views: 73

Re: Electronegativity

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.
by Asia Yamada 2B
Tue Dec 01, 2020 1:56 am
Forum: Sigma & Pi Bonds
Topic: HW 2F.3.
Replies: 5
Views: 100

Re: HW 2F.3.

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...
by Asia Yamada 2B
Tue Dec 01, 2020 1:55 am
Forum: Hybridization
Topic: 2F.3
Replies: 2
Views: 66

Re: 2F.3

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.
by Asia Yamada 2B
Tue Dec 01, 2020 1:49 am
Forum: Hybridization
Topic: unhybridized orbitals
Replies: 4
Views: 49

Re: unhybridized orbitals

I believe that sigma bonds are formed from hybridized orbitals.
by Asia Yamada 2B
Tue Nov 24, 2020 7:25 pm
Forum: Hybridization
Topic: Electron Configuration
Replies: 7
Views: 394

Re: Electron Configuration

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
by Asia Yamada 2B
Tue Nov 24, 2020 7:23 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Nitric Oxide
Replies: 5
Views: 82

Re: Nitric Oxide

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.
by Asia Yamada 2B
Tue Nov 24, 2020 6:47 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Important aspects of Radicals
Replies: 6
Views: 74

Re: Important aspects of Radicals

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.
by Asia Yamada 2B
Tue Nov 24, 2020 6:26 pm
Forum: Hybridization
Topic: Sp3d or dsp3
Replies: 22
Views: 863

Re: Sp3d or dsp3

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.
by Asia Yamada 2B
Tue Nov 24, 2020 6:24 pm
Forum: Hybridization
Topic: d hybridized orbital confusion
Replies: 7
Views: 57

Re: d hybridized orbital confusion

In Dr. Lavelle’s most recent lecture, I believe he said you can write it either way.
by Asia Yamada 2B
Wed Nov 18, 2020 1:03 am
Forum: Electronegativity
Topic: Electronegativity Ex.
Replies: 3
Views: 71

Re: Electronegativity Ex.

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...
by Asia Yamada 2B
Wed Nov 18, 2020 12:55 am
Forum: Electronegativity
Topic: Electronegativity and Bond Strength
Replies: 4
Views: 1895

Re: Electronegativity and Bond Strength

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 ...
by Asia Yamada 2B
Wed Nov 18, 2020 12:49 am
Forum: Bond Lengths & Energies
Topic: 3.87 Homework Problem
Replies: 3
Views: 921

Re: 3.87 Homework Problem

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...
by Asia Yamada 2B
Wed Nov 18, 2020 12:48 am
Forum: Electronegativity
Topic: Strongest CX bond, where X is a halogen
Replies: 2
Views: 1848

Re: Strongest CX bond, where X is a halogen

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...
by Asia Yamada 2B
Wed Nov 18, 2020 12:46 am
Forum: Coordinate Covalent Bonds
Topic: Question 3.87 (Sixth Edition)
Replies: 2
Views: 357

Re: Question 3.87 (Sixth Edition)

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...
by Asia Yamada 2B
Wed Nov 11, 2020 2:18 am
Forum: Einstein Equation
Topic: Textbook Problem 1B.9
Replies: 5
Views: 76

Re: Textbook Problem 1B.9

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...
by Asia Yamada 2B
Wed Nov 11, 2020 2:06 am
Forum: Einstein Equation
Topic: Question 1.27 (Sixth Edition)
Replies: 3
Views: 500

Re: Question 1.27 (Sixth Edition)

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...
by Asia Yamada 2B
Wed Nov 11, 2020 1:58 am
Forum: Einstein Equation
Topic: Discussion Section Problem
Replies: 3
Views: 300

Re: Discussion Section Problem

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.
by Asia Yamada 2B
Wed Nov 11, 2020 1:50 am
Forum: Einstein Equation
Topic: Einstein Equation
Replies: 4
Views: 92

Re: Einstein Equation

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.
by Asia Yamada 2B
Wed Nov 11, 2020 1:46 am
Forum: Einstein Equation
Topic: Sapling Week 4 #23
Replies: 3
Views: 118

Re: Sapling Week 4 #23

In this case, you wouldn’t need to multiply by Avogadro’s number because the electron affinity you found is already for one atom.
by Asia Yamada 2B
Mon Nov 02, 2020 11:07 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Textbook 1B 25
Replies: 4
Views: 72

Re: Textbook 1B 25

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...
by Asia Yamada 2B
Mon Nov 02, 2020 11:01 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: delta V. in. Heisenberg questions
Replies: 6
Views: 129

Re: delta V. in. Heisenberg questions

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...
by Asia Yamada 2B
Mon Nov 02, 2020 10:54 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Heisenberg Uncertainty Module Example Question
Replies: 3
Views: 76

Re: Heisenberg Uncertainty Module Example Question

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...
by Asia Yamada 2B
Mon Nov 02, 2020 10:43 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Heisenberg Indeterminacy Principle Assessment #18
Replies: 2
Views: 53

Re: Heisenberg Indeterminacy Principle Assessment #18

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...
by Asia Yamada 2B
Mon Nov 02, 2020 10:30 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: heisenberg module #18
Replies: 3
Views: 76

Re: heisenberg module #18

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...
by Asia Yamada 2B
Tue Oct 27, 2020 8:20 pm
Forum: Einstein Equation
Topic: 1B.7 b)
Replies: 2
Views: 73

Re: 1B.7 b)

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.
by Asia Yamada 2B
Tue Oct 27, 2020 8:09 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Heisenberg Indeterminacy (Uncertainty) Equation
Replies: 5
Views: 106

Re: Heisenberg Indeterminacy (Uncertainty) Equation

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 ...
by Asia Yamada 2B
Tue Oct 27, 2020 3:23 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Heisenberg Uncertainty Principle Post Assessment Q#18
Replies: 2
Views: 53

Re: Heisenberg Uncertainty Principle Post Assessment Q#18

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...
by Asia Yamada 2B
Tue Oct 27, 2020 3:12 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Multiplying Uncertainty of Speed by two to find deltaV
Replies: 4
Views: 51

Re: Multiplying Uncertainty of Speed by two to find deltaV

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.
by Asia Yamada 2B
Tue Oct 27, 2020 3:08 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Uncertainty in speed
Replies: 5
Views: 2013

Re: Uncertainty in speed

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
by Asia Yamada 2B
Wed Oct 21, 2020 3:09 am
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: 1B.25
Replies: 2
Views: 87

Re: 1B.25

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...
by Asia Yamada 2B
Wed Oct 21, 2020 3:02 am
Forum: Einstein Equation
Topic: 1B.9 (7th edition)
Replies: 8
Views: 445

Re: 1B.9 (7th edition)

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...
by Asia Yamada 2B
Wed Oct 21, 2020 2:46 am
Forum: Einstein Equation
Topic: Homework Problem 1B.7
Replies: 2
Views: 56

Re: Homework Problem 1B.7

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 ...
by Asia Yamada 2B
Wed Oct 21, 2020 2:30 am
Forum: DeBroglie Equation
Topic: Measureable-wavelike properties
Replies: 5
Views: 66

Re: Measureable-wavelike properties

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.
by Asia Yamada 2B
Wed Oct 21, 2020 2:23 am
Forum: DeBroglie Equation
Topic: Use
Replies: 4
Views: 143

Re: Use

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.
by Asia Yamada 2B
Tue Oct 13, 2020 5:44 pm
Forum: SI Units, Unit Conversions
Topic: Chemistry Review Section E
Replies: 4
Views: 247

Re: Chemistry Review Section E

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.
by Asia Yamada 2B
Tue Oct 13, 2020 5:34 pm
Forum: Photoelectric Effect
Topic: Photoelectric Effect Energy and Excess Energy
Replies: 9
Views: 80

Re: Photoelectric Effect Energy and Excess Energy

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.
by Asia Yamada 2B
Tue Oct 13, 2020 5:12 pm
Forum: Einstein Equation
Topic: Wavelength and KE
Replies: 5
Views: 206

Re: Wavelength and KE

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...
by Asia Yamada 2B
Mon Oct 12, 2020 9:02 pm
Forum: Molarity, Solutions, Dilutions
Topic: Homework problem Fundamentals E.15
Replies: 3
Views: 458

Re: Homework problem Fundamentals E.15

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 ...
by Asia Yamada 2B
Mon Oct 12, 2020 8:37 pm
Forum: Properties of Light
Topic: Photons
Replies: 3
Views: 227

Re: Photons

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.
by Asia Yamada 2B
Thu Oct 08, 2020 2:00 am
Forum: Balancing Chemical Reactions
Topic: Fundamental Exercise H7
Replies: 5
Views: 55

Re: Fundamental Exercise H7

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.
by Asia Yamada 2B
Thu Oct 08, 2020 1:52 am
Forum: Significant Figures
Topic: Inaccurate Answer
Replies: 4
Views: 101

Re: Inaccurate Answer

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...
by Asia Yamada 2B
Thu Oct 08, 2020 1:43 am
Forum: Accuracy, Precision, Mole, Other Definitions
Topic: Textbook L. 35
Replies: 3
Views: 77

Re: Textbook L. 35

Yes, instead of FeBr2, it should be Fe3Br8.
by Asia Yamada 2B
Wed Oct 07, 2020 2:29 am
Forum: SI Units, Unit Conversions
Topic: Formula Units?
Replies: 6
Views: 181

Re: Formula Units?

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.
by Asia Yamada 2B
Wed Oct 07, 2020 2:11 am
Forum: SI Units, Unit Conversions
Topic: Prefixes
Replies: 6
Views: 141

Re: Prefixes

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.
by Asia Yamada 2B
Wed Oct 07, 2020 1:56 am
Forum: Empirical & Molecular Formulas
Topic: Empirical to Molecular Formulas [ENDORSED]
Replies: 6
Views: 153

Re: Empirical to Molecular Formulas [ENDORSED]

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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