Search found 60 matches

by Ivy Tan 1E
Fri Jan 15, 2021 12:24 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: lecture 5 question
Replies: 13
Views: 55

Re: lecture 5 question

Hi! I think you have to memorize the strong acids and bases for this class. This is a mnemonic I learned to help me memorize the strong acids: So I Br ought No Cl ean Cl othes. So=H2SO4, I=HI, Br=HBr, No=HNO3, Cl=HCl and HClO4. Hopefully this makes it a little easier to memorize! Also, conjugate bas...
by Ivy Tan 1E
Fri Jan 15, 2021 12:13 pm
Forum: Ideal Gases
Topic: Ice method
Replies: 14
Views: 63

Re: Ice method

305405193 wrote:Do any of you have any recommendations on a good YouTube video for explaining Ice tables?

Hi!
https://youtu.be/aJ0KNQ5-KaI here's a short video describing how to use ICE tables using an example. Hope this helps you!!
by Ivy Tan 1E
Fri Jan 15, 2021 12:06 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook problem 5I.27c
Replies: 3
Views: 25

Re: Textbook problem 5I.27c

Hi! The answers that you got are correct, but they are only the values of X, not the answer to the equilibrium composition of the reaction. You should use 0.074 as your value of X because using 8.71 would give you a negative concentration for [PCl3]. To solve for the equilibrium composition, you sho...
by Ivy Tan 1E
Fri Jan 15, 2021 12:01 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: the different Ks
Replies: 3
Views: 19

Re: the different Ks

Hi! K represents the equilibrium constant of a given reaction. There are different types of Ks because they are used for different reactions involving different substances. Ka: equilibrium constant for reactions involving acids. Kb: equilibrium constant for reactions involving bases. Kw: equilibrium...
by Ivy Tan 1E
Fri Jan 15, 2021 11:53 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: q vs k
Replies: 62
Views: 161

Re: q vs k

Hi! Q is the reaction quotient, or the ratio of products to reactants at any given point in the reaction. Q may equal K (when the reaction is at equilibrium) or it may higher or lower than K (when the reaction is not at equilibrium). If Q is higher than K, the reaction favors the reactants. If Q is ...
by Ivy Tan 1E
Fri Jan 08, 2021 4:25 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Favoring
Replies: 6
Views: 39

Re: Favoring

Hi! For concentration changes: Adding more reactants or decreasing the products will shift the reaction towards the products. Adding more products or decreasing the reactants will shift the reaction towards the reactants. For pressure/volume changes: Increasing the pressure will shift the reaction t...
by Ivy Tan 1E
Fri Jan 08, 2021 4:15 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Increasing the Volume
Replies: 5
Views: 22

Re: Increasing the Volume

Hi!
Yes, you're right! If we increase the volume and decrease the pressure, the reaction would shift to the side with more moles of gas (which is the opposite of what would happen if pressure was increased).
by Ivy Tan 1E
Wed Jan 06, 2021 1:54 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Boundaries for high and low K values
Replies: 3
Views: 22

Re: Boundaries for high and low K values

Hi! I think it's because if 10^-3 < K < 10^3, the product or reactant concentration is barely greater than the other, so the small difference is not enough to say that either the product or reactant is favored in the reaction. When K < 10^-3, there is clearly a higher reactant concentration than pro...
by Ivy Tan 1E
Wed Jan 06, 2021 1:42 pm
Forum: Ideal Gases
Topic: Difference between real and ideal gas
Replies: 10
Views: 75

Re: Difference between real and ideal gas

Hi! An ideal gas has negligible volume, particles that do not collide because they are in constant motion, and particles that exhibit no attraction or repulsion because they do not interact with each other. A real gas violates all of these criteria: it has volume and has particles that collide and i...
by Ivy Tan 1E
Tue Jan 05, 2021 11:59 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Kc vs Kp
Replies: 14
Views: 69

Re: Kc vs Kp

Hi! I think the Kc and Kp values are the same once they're calculated out. The only difference is when calculating Kc, you use the concentrations of each molecule and when calculating Kp, you use the partial pressures of each gas. Hope this helps!!
by Ivy Tan 1E
Thu Dec 10, 2020 3:28 pm
Forum: Identifying Acidic & Basic Salts
Topic: Sapling Wk 10 Q7
Replies: 2
Views: 44

Re: Sapling Wk 10 Q7

Hi! If the cation or anion of the salt is a conjugate acid of a weak base , then the salt is acidic, and if the cation or anion of the salt is a conjugate base of a weak acid , then the salt is basic. If the cation or anion is the conjugate acid or base of a strong acid or base , then it doesn't aff...
by Ivy Tan 1E
Thu Dec 10, 2020 3:09 pm
Forum: Lewis Acids & Bases
Topic: Identifying Lewis Acids and Bases
Replies: 3
Views: 44

Re: Identifying Lewis Acids and Bases

Hi! One thing you could do to identify Lewis Acids and Bases is by drawing out the Lewis structures of both molecules. If the molecule already has a full octet, it's likely to be a Lewis Base because it would not want to accept electrons to add to another electron shell, as it would make it unstable...
by Ivy Tan 1E
Thu Dec 10, 2020 2:52 pm
Forum: Conjugate Acids & Bases
Topic: Can strong acids become conjugate bases?
Replies: 3
Views: 52

Re: Can strong acids become conjugate bases?

Hi! I think all acids have conjugate bases, regardless of whether they're strong or weak. In a neutralization reaction where an acid reacts with a base, the acid has a corresponding conjugate base and the base has a corresponding conjugate acid. For example, in the reaction: HCl + NaOH --> NaCl + H2...
by Ivy Tan 1E
Thu Dec 10, 2020 2:33 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Ph and PKA
Replies: 7
Views: 73

Re: Ph and PKA

Hi! Adding on to the response above, when pH < pKa, the solution is already in an acidic environment because the pH is lower, so the acid will not dissociate to make the solution even more acidic (it is neutral). When pH > pKa, the solution is in a basic environment because the pH is higher, so the ...
by Ivy Tan 1E
Thu Dec 10, 2020 2:26 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: 6B1
Replies: 3
Views: 54

Re: 6B1

Hi! Here's how I solved this problem: 1. the molarity of H+ ions is equal to the molarity of HCl because HCl is a strong acid so its ions will completely dissociate 2. pH of initial HCl solution = -log([HCl]) and pH of diluted HCl solution = -log(0.12 x [HCl]) 3. I divided the pH of the diluted HCl ...
by Ivy Tan 1E
Thu Dec 03, 2020 12:01 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Sapling Homework #5
Replies: 2
Views: 38

Re: Sapling Homework #5

Hi! To find the coordination number, you should first locate the central atom (usually a transition metal), and then count the number of atoms in the nearby molecules or ions. https://sites.google.com/site/chempendix/ligands This chart tells you all the donor atoms for different molecules or ions th...
by Ivy Tan 1E
Thu Dec 03, 2020 11:45 am
Forum: Naming
Topic: halide ion naming
Replies: 2
Views: 25

Re: halide ion naming

Hi!
I think chloro and chlorido are interchangeable, so either one can be used. In general, the suffix for the anionic ligands is -o. The exceptions are -ide --> -ido (except for chloride, both chloro or chlorido can be used), -ate --> -ato, and -ite --> -ito. Hope this helps!!
by Ivy Tan 1E
Thu Dec 03, 2020 11:21 am
Forum: Hybridization
Topic: 2F.7
Replies: 4
Views: 79

Re: 2F.7

Hi! An easy way to identify hybrid orbitals is by memorizing this: 2 bonds --> sp 3 bonds --> sp2 4 bonds --> sp3 5 bonds --> sp3d 6 bonds --> sp3d2 Since VSEPR disregards multiple bonds, a double or triple bond would still be considered one bond. For example, if a molecule has one single bond and o...
by Ivy Tan 1E
Thu Dec 03, 2020 11:15 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Textbook Problem 2E #13
Replies: 3
Views: 60

Re: Textbook Problem 2E #13

Hi!
Like the responses above stated, the answer key should say AX2E3. It's X2, not X3, because the subscript for X represents the number of bonds in the molecule. In I3-, the central Iodine atom only makes 2 bonds (with the other 2 I atoms), therefore making it X2, not X3. Hope this was helpful!!
by Ivy Tan 1E
Thu Dec 03, 2020 11:10 am
Forum: Determining Molecular Shape (VSEPR)
Topic: 2E.25 Part D
Replies: 2
Views: 30

Re: 2E.25 Part D

Hi!
Yes, SF4 would be polar because its molecular shape is trigonal bipyramidal. The single lone pair on the central atom, Sulfur, repels the electrons on the Fluorine atoms, making the overall molecule polar. Hope this helps!!
by Ivy Tan 1E
Tue Nov 24, 2020 5:55 pm
Forum: Sigma & Pi Bonds
Topic: sp3d orbitals
Replies: 4
Views: 61

Re: sp3d orbitals

Hi!
A sp3d orbital has 3 hybrid orbitals lying on the horizontal plane 120 degrees apart, with 2 more hybrid orbitals lying on the vertical plane 90 degrees from the horizontal orbitals. It looks like the trigonal bipyramidal shape. Hope this was helpful!!
by Ivy Tan 1E
Tue Nov 24, 2020 5:49 pm
Forum: Dipole Moments
Topic: Polarity
Replies: 30
Views: 152

Re: Polarity

Hi! If the polar bonds within the molecule cancel each other out, then the molecule as a whole will be nonpolar. For example, in CO2, the bonds between carbon and oxygen are polar because oxygen is more electronegative than carbon. However, the symmetrical shape of the molecule makes it nonpolar, si...
by Ivy Tan 1E
Tue Nov 24, 2020 5:42 pm
Forum: Hybridization
Topic: Sapling #18 Week7&8
Replies: 3
Views: 59

Re: Sapling #18 Week7&8

Hi!
https://images.app.goo.gl/zq3GtMUBE5AzJ5gT7 here's an image to help visualize the pi bonds in H2CCCH2. Since pi bonds are between p-orbitals (which are perpendicular to each other), the two pi bonds are also perpendicular. Hope this helps!!
by Ivy Tan 1E
Tue Nov 24, 2020 4:55 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Sapling #6
Replies: 5
Views: 57

Re: Sapling #6

Hi!
XeF2 is linear because the 3 lone pairs cancel eachother out, therefore making the molecular shape linear, not bent. Since the shape is linear, the bond angle would have to be 180 degrees. Hope this helps!!
by Ivy Tan 1E
Tue Nov 24, 2020 4:49 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: memorization
Replies: 12
Views: 114

Re: memorization

Hi!
here's another reference chart that might be helpful for you in memorizing the shape names. https://images.app.goo.gl/ETv92gyFr1m1PnUN9
by Ivy Tan 1E
Thu Nov 19, 2020 12:22 pm
Forum: Trends in The Periodic Table
Topic: Positive and Negative Electron Affinity Values
Replies: 2
Views: 52

Re: Positive and Negative Electron Affinity Values

Hi Samantha! I think you're right. A positive electron affinity value means that energy is being released when an electron is added to the atom. In other words, it is more energetically favorable because elements with high electron affinity are also highly electronegative, and, therefore, easily att...
by Ivy Tan 1E
Thu Nov 19, 2020 12:17 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: variations in electron configuration names
Replies: 3
Views: 53

Re: variations in electron configuration names

Hi! The valence shell configuration is when you start with the noble gas core, and then write out the configurations for the electrons in the outer shell. Ground state configuration is a type of valence shell configuration for an atom is in its ground state (not excited). All the electrons in the co...
by Ivy Tan 1E
Thu Nov 19, 2020 12:08 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Textbook Problem 1E.25 d
Replies: 2
Views: 30

Re: Textbook Problem 1E.25 d

Hi Shannon! Like the response above states, the exception for Cu applies to Ag and Au as well because they are in the same group of the periodic table. It is more stable for the d orbitals to be filled completely with 10 electrons (with all electrons paired), rather than 9 electrons (with 1 unpaired...
by Ivy Tan 1E
Thu Nov 19, 2020 11:58 am
Forum: Quantum Numbers and The H-Atom
Topic: under which circumstances does 4s have lower energy than 3d?
Replies: 2
Views: 40

Re: under which circumstances does 4s have lower energy than 3d?

Hi! Yes, you're right. The only circumstance in which the 4s state has less energy than the 3d state is when the d orbitals are empty. This only applies to elements K and Ca in period 4. For all elements in period 4 from groups 3-18, the 4s state has more energy than the 3d state because these eleme...
by Ivy Tan 1E
Thu Nov 19, 2020 11:51 am
Forum: Quantum Numbers and The H-Atom
Topic: quantum number f
Replies: 6
Views: 72

Re: quantum number f

Hi! I think subshell f is only used by elements in Period 6 and on. If you look at the periodic table, the f block is elements 57-71 and 89-103 (which is shown in periods 6 and 7). For example, if you were writing the period 6 element lead (Pb), the configuration would be [Xe]4f14 5d10 6s2 6p2. Alth...
by Ivy Tan 1E
Fri Nov 13, 2020 3:31 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Definition of valence electrons
Replies: 2
Views: 19

Re: Definition of valence electrons

Hi! Valence electrons are outer shell electrons that participate in bonding. For some elements, the electrons in the d-orbital are able to participate in bonding so the d-orbital electrons would be considered valence electrons. For example, Manganese (configuration: [Ar]3d5 4s2) has 7 valence electr...
by Ivy Tan 1E
Fri Nov 13, 2020 3:23 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Textbook Exercise 2A.11
Replies: 2
Views: 19

Re: Textbook Exercise 2A.11

Hi! Adding on to the response above, the general process for figuring out the elements in this question is by rewriting the electron configuration with 3 extra electrons to account for the ground state of the M3+ ions. For example, for b) adding 3 electrons to the electron configuration would make i...
by Ivy Tan 1E
Fri Nov 13, 2020 3:06 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Textbook Problem 1.E.25
Replies: 7
Views: 73

Re: Textbook Problem 1.E.25

The alkali metals have a valence shell with the configuration ns 1 , with n being the principal quantum number of the atom. Similarly, a valence shell for a group 15 atom would have the configuration ns 2 np 3 . A group 5 transition metal would have a configuration of (n-1)d 3 ns 2 . The coinage me...
by Ivy Tan 1E
Fri Nov 13, 2020 12:18 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Electron Configuration for Transition Metals
Replies: 3
Views: 19

Re: Electron Configuration for Transition Metals

Hi! Nickel's electron configuration is [Ar]3d84s2. In the fourth period, the only two elements that have the configuration exception you're referring to are chromium, [Ar]3d54s1, and copper, [Ar]3d104s1. This is because these atoms are more stable when their d-orbital shells are half-filled (for chr...
by Ivy Tan 1E
Fri Nov 13, 2020 10:58 am
Forum: Lewis Structures
Topic: textbook problem 2B #11
Replies: 3
Views: 36

Re: textbook problem 2B #11

Hi Kaylee! Adding on to the response above, another helpful tip is that if the compound has an OH (like in CH3OH and H2C(NH2)(COOH)), the oxygen and hydrogen atom will always be bonded to each other. For example, in CH3OH, carbon would be the central atom with 3 hydrogens and an O-H bonded to it. Ho...
by Ivy Tan 1E
Fri Nov 06, 2020 2:35 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: valence electrons
Replies: 3
Views: 39

Re: valence electrons

Hi Emma! I think the easiest way to find the valence electrons for elements in the d block is to count the electrons in the s and d orbitals, because the d orbital electrons also participate in bonding. In addition to the example provided in the response above, Manganese has an electron configuratio...
by Ivy Tan 1E
Fri Nov 06, 2020 2:27 pm
Forum: Ionic & Covalent Bonds
Topic: Polarity
Replies: 9
Views: 89

Re: Polarity

Hi! Yes, I think polarity and the ionic characteristics of covalent bonds are related. Polar covalent bonds have ionic character because the electrons are not shared equally, whereas nonpolar covalent bonds (when electrons are shared between the same atom) don't have ionic characteristics because th...
by Ivy Tan 1E
Fri Nov 06, 2020 12:44 pm
Forum: Lewis Structures
Topic: 2A.1
Replies: 4
Views: 42

Re: 2A.1

Hi Chloe! An easy way to figure out the valence electrons is by counting the electrons in the s, p, and d orbitals (in certain elements) in the electron configuration. For example, a) Sb: the electron configuration is [Kr]4d10 5s2 6p3, and by counting the electrons in the s and p orbitals (2 and 3, ...
by Ivy Tan 1E
Fri Nov 06, 2020 12:35 pm
Forum: Lewis Structures
Topic: Ionic vs. Covalent bond concepts
Replies: 3
Views: 27

Re: Ionic vs. Covalent bond concepts

Hi! Yes, a polar covalent bond has ionic characteristic. Adding on to the above response, the only type of covalent bond that wouldn't show ionic characteristics is a bond between two of the same nonmetals, or a non polar covalent bond, where the electronegativities of the atoms are exactly the same...
by Ivy Tan 1E
Fri Nov 06, 2020 12:21 pm
Forum: Ionic & Covalent Bonds
Topic: Textbook Problem 2A.11
Replies: 4
Views: 25

Re: Textbook Problem 2A.11

Hi! The exception for d orbitals is that they are lower in energy when occupied with electrons compared to the s orbital, but higher in energy when empty compared to the s orbital . For example, the 3d orbital is lower in energy than the 4s orbital, the 4d orbital is lower in energy than the 5s orbi...
by Ivy Tan 1E
Fri Oct 30, 2020 4:09 pm
Forum: Photoelectric Effect
Topic: UA Session Question
Replies: 2
Views: 46

Re: UA Session Question

Hi Nina! To solve this problem, you need to use the general equation for the photoelectric effect, which is: energy of photon = work function + kinetic energy of electron 1. Convert work function of 2.3 eV to joules: 1.436 x 10^19 J 2. Use the De Broglie equation to solve for the velocity of the ele...
by Ivy Tan 1E
Fri Oct 30, 2020 3:50 pm
Forum: Properties of Electrons
Topic: Sapling #11
Replies: 6
Views: 81

Re: Sapling #11

Hi Emily! The response above answered how to find values for n and l. The quantum number m tells you the specific orbital that the electron belongs in. Values of m depend on the values of l: for example, if l=2, m=-2, -1, 0, 1, and 2 and if l=3, m=-3, -2, -1, 0, 1, 2, and 3. In general, the values o...
by Ivy Tan 1E
Fri Oct 30, 2020 3:43 pm
Forum: Properties of Light
Topic: 1A.3 Question
Replies: 3
Views: 52

Re: 1A.3 Question

Hi Stella! Answer C states that the change in an electrical field at a given point is smaller when the frequency decreases. This answer is saying that when the frequency is smaller, the electrical field undergoes less change at a certain point because the wave goes through fewer cycles in one second...
by Ivy Tan 1E
Fri Oct 30, 2020 3:26 pm
Forum: Trends in The Periodic Table
Topic: Trends
Replies: 14
Views: 88

Re: Trends

Hi Lexy! An easy way for me to remember the atomic radius trend is that when you progress down a group, the value of n is increasing (meaning more shells are being added to the atom) so its radius increases. When you go across a period, the atomic numbers are increasing (meaning the number of proton...
by Ivy Tan 1E
Fri Oct 30, 2020 3:21 pm
Forum: Ionic & Covalent Bonds
Topic: Non Metals
Replies: 10
Views: 92

Re: Non Metals

Hi! Nonmetals don't lose electrons because they have very high ionization energy due to the electrons feeling more pull and attraction towards the positively charged nucleus. So, it is more favorable for the nonmetal to gain electrons to fill its shell rather than to lose electrons. Hope this was he...
by Ivy Tan 1E
Sat Oct 24, 2020 3:41 pm
Forum: Photoelectric Effect
Topic: Intensity vs. Frequency
Replies: 22
Views: 140

Re: Intensity vs. Frequency

Hi! I think intensity corresponds to the amplitude of the wave and frequency is the number of wave cycles per second. This is why in the atomic spectra experiment, increasing the intensity of the light would only increase the amplitude of the wave, not its energy. Increasing the frequency of a wave ...
by Ivy Tan 1E
Sat Oct 24, 2020 3:33 pm
Forum: Properties of Electrons
Topic: How to convert from Energy per photon to Energy per electron
Replies: 2
Views: 42

Re: How to convert from Energy per photon to Energy per electron

Hi Lucy!
To find the energy per electron using the energy per photon you should:
1. use E=(hc)/lambda to solve for lambda (the wavelength of the light)
2. use lambda=h/(mv) to find the velocity of the electron
3. use Ke=0.5mv^2 to find the kinetic energy of the electron
Hope this was helpful!!
by Ivy Tan 1E
Sat Oct 24, 2020 3:28 pm
Forum: Properties of Light
Topic: Textbook question 1.A.1
Replies: 2
Views: 26

Re: Textbook question 1.A.1

Hi!
I think the cathode ray experiment only showed that all atoms contain negative particles (electrons). The support for a wave model of electrons comes from the diffraction experiment and the De Broglie equation. Hope this was helpful!!
by Ivy Tan 1E
Sat Oct 24, 2020 3:19 pm
Forum: Student Social/Study Group
Topic: How are you studying?
Replies: 203
Views: 1267

Re: How are you studying?

Hi! I usually try to read the textbook sections that correspond to what we're learning before the lecture so I know what's going on when Professor Lavelle is going over it in class. Then, I'll take notes on the lecture and reread the textbook and take notes to solidify the information. For the midte...
by Ivy Tan 1E
Sat Oct 24, 2020 3:13 pm
Forum: Properties of Light
Topic: Wave Particle Duality: Experiments/Equations
Replies: 2
Views: 30

Re: Wave Particle Duality: Experiments/Equations

Hi Joey! I think the categories you made are great. The Bohr Frequency condition demonstrates the quantum mechanics of electrons and how they have discrete levels of energy, as it shows that only certain frequencies of light can be absorbed or emitted by electrons in an atom. The De Broglie equation...
by Ivy Tan 1E
Thu Oct 15, 2020 11:13 pm
Forum: Properties of Electrons
Topic: Energy Gaps
Replies: 8
Views: 90

Re: Energy Gaps

Hi Anna! The energy gaps in an electron get smaller because of Coloumbic attraction; when an electron is farther away from the nucleus, it feels less attraction towards the protons in the positively charged nucleus, so less energy is required to remove the electron and the energy gaps get smaller. H...
by Ivy Tan 1E
Thu Oct 15, 2020 11:01 pm
Forum: Photoelectric Effect
Topic: finding Ek
Replies: 4
Views: 46

Re: finding Ek

Hi! The equation E=hv tells you the energy of the photon, not the energy of the electron. The energy of an ejected electron can only be found through the kinetic energy equation (Ek=1/2mv^2). A question wouldn't give you the wavelength of an ejected electron, but rather the wavelength of the light s...
by Ivy Tan 1E
Thu Oct 15, 2020 10:53 pm
Forum: Properties of Electrons
Topic: Sapling 2 #6
Replies: 4
Views: 47

Re: Sapling 2 #6

Hi Michael!
In high school, I learned to use 1/wavelength in the Rydberg formula. Both ways are okay to use, because 1/wavelength just shows the inverse relationship between wavelength and frequency in the Rydberg equation. Both should give you the same answer. Hope this helps!!
by Ivy Tan 1E
Thu Oct 15, 2020 10:50 pm
Forum: Properties of Light
Topic: Rydberg Formula
Replies: 2
Views: 45

Re: Rydberg Formula

Hi! Rydberg's formula helps you calculate the frequency of light that is emitted or absorbed when an electron moves between energy levels. The negative sign refers to how the energy of a bound electron is negative compared to an electron in the n=infinity level. R represents Rydberg's constant (3.20...
by Ivy Tan 1E
Thu Oct 15, 2020 10:20 pm
Forum: Properties of Electrons
Topic: Wave Prop. of electrons post assessment
Replies: 2
Views: 28

Re: Wave Prop. of electrons post assessment

Hi Kayko! To solve this question, use the De Broglie equation, lambda=h/(mv). 1. Convert 125 km/hr to m/s: 125 km/hr x (1hr/60min) x (1min/60sec) x (1000m/1km) = 34.72m/s 2. Plug in the values for m and v into the De Broglie equation: lambda=(6.626 x 10^-34)/(275kg x 34.72m/s) = 6.94 x 10^-38m (answ...
by Ivy Tan 1E
Thu Oct 08, 2020 2:45 pm
Forum: Significant Figures
Topic: Significant Figures about Logarithms
Replies: 4
Views: 46

Re: Significant Figures about Logarithms

Hi Victor! Yes, the rule for logarithms and exponents is that the number of figures after the decimal point should be equal to the number of significant figures in the original answer. So, for the example you gave, ln(2.17)=0.775 (0.775 has 3 figures after the decimal point because the original numb...
by Ivy Tan 1E
Thu Oct 08, 2020 2:27 pm
Forum: Limiting Reactant Calculations
Topic: Fundamentals Self-test M.2A
Replies: 2
Views: 39

Re: Fundamentals Self-test M.2A

Hi Savannah!! The response above answered part b. For part c, here's what I did: 5.52g Na x (1 mol Na / 22.99g) x (1 mol Al2O3 / 6 mols Na) x (101.96g / 1 mol Al2O3) = 4.0802g Al2O3 (this equation is used to find the amount of Al2O3, the excess reactant, that was used up in the reaction). 5.10g - 4....
by Ivy Tan 1E
Thu Oct 08, 2020 2:10 pm
Forum: Molarity, Solutions, Dilutions
Topic: Molar Mass of a Sulfide of a Metal.
Replies: 3
Views: 48

Re: Molar Mass of a Sulfide of a Metal.

Hi!! When the question says "metal hydroxide", it means that an unknown metal is combined with hydroxide (you have to figure out which metal it is). When it asks for the molar mass for the "sulfide of the metal", you have to calculate the mass of the unknown metal you just found ...
by Ivy Tan 1E
Tue Oct 06, 2020 11:12 pm
Forum: Limiting Reactant Calculations
Topic: Dis 1L Week 1 WS #6
Replies: 4
Views: 44

Re: Dis 1L Week 1 WS #6

Hi!! To determine the limiting reactant, the first thing you need to do is convert the 5g of glucose and 5g of oxygen to moles of glucose and oxygen, using the molar mass (you should get 0.0278 moles of glucose and 0.1563 moles of oxygen). In the given reaction, the moles of glucose to moles of oxyg...
by Ivy Tan 1E
Mon Oct 05, 2020 12:00 pm
Forum: SI Units, Unit Conversions
Topic: Fundamentals F.23 [ENDORSED]
Replies: 7
Views: 163

Re: Fundamentals F.23 [ENDORSED]

Hi!! For each fuel you have to: 1. calculate the molar mass 2. calculate the mass of carbon within the compound 3. divide the mass of carbon within the compound by the molar mass and multiply by 100 to find the mass percentage of carbon 4. rank in order from lowest percentage to highest percentage E...

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