Search found 62 matches

by Fayez Kanj
Wed Mar 13, 2019 7:11 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Reduction of Oxygen half reaction?
Replies: 1
Views: 43

Reduction of Oxygen half reaction?

Hello!

Can someone please explain to me how to come up with the half reactions for this cell diagram?

Pt(s)|O2(g)|H+(aq)∥OH−(aq)|O2(g)|Pt(s)

Thanks!
by Fayez Kanj
Wed Mar 13, 2019 7:06 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: K and k
Replies: 1
Views: 42

Re: K and k

Hello! if we have A + B ⇄ C + D, we can write these 2 reactions: A + B → C + D . forward rate = k[A][B] C + D → A + B backward rate = k'[C][D] At equilibrium, since the forward rate = backwards rate, then k[A][B] = k'[C][D]. Rearranging this makes [C][D]/[A][B] = k/k' . This also equals K. Thus, at ...
by Fayez Kanj
Wed Mar 13, 2019 7:02 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Pre-equilibrium approach
Replies: 4
Views: 73

Re: Pre-equilibrium approach

Hello! We use the pre-equilibrium approach when doing problems with mechanisms, if the observed rate expression and the rate expression you obtain from the mechanisms do not match. You can make the arrow and equilibrium sign for the fast step, determine the expression of the equilibrium constant K, ...
by Fayez Kanj
Wed Mar 06, 2019 4:26 pm
Forum: General Rate Laws
Topic: how to calculate reaction rates
Replies: 2
Views: 52

Re: how to calculate reaction rates

Hello!

You could calculate the average rate (total change in concentration/change in time), or instantaneous rate (derivative, or slope of the tangent line to a point), or the unique rate, which helps you solve for the rate expressions (differentiated and integrated)

Hope this helps :)
by Fayez Kanj
Wed Mar 06, 2019 4:23 pm
Forum: First Order Reactions
Topic: Deriving these equations
Replies: 2
Views: 54

Re: Deriving these equations

Hello!

Dr Lavelle derived the equations so that we could have a better understanding of why there are true and where they come from, rather than just giving it to us. However, for the exam, you could just memorize them and use them as is.

Hope this helps :)
by Fayez Kanj
Wed Mar 06, 2019 4:22 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: first and second order
Replies: 2
Views: 47

Re: first and second order

Hello!

First order means that a change in reactant concentration will change the rate by that same amount. It value of "n" would be 1.
Second order means that a change in reactant concentration will result in a 4x change in rate

Hope this helps :)
by Fayez Kanj
Sun Feb 24, 2019 3:24 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Oxidation and Reduction
Replies: 3
Views: 49

Re: Oxidation and Reduction

Hello, In oxidation reactions, since electrons are being lost, they will be present in the products side of the oxidation half reaction. In reduction reactions, since electrons are being gained, they will be present in the reactants side of the half reaction. When combining the . half reactions into...
by Fayez Kanj
Sun Feb 24, 2019 3:21 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: salt bridge
Replies: 3
Views: 46

Re: salt bridge

Hello In the anode (where oxidation occurs), species lose electrons and thus become positive ions. If this is allowed to keep going on, there would be a buildup on positive ions at the anode half of the cell. This would prevent any additional oxidation from occurring, since it is already too positiv...
by Fayez Kanj
Sun Feb 24, 2019 3:17 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Electrode Metal
Replies: 1
Views: 33

Re: Electrode Metal

Hello, The most common inert metal used as an electrode for batteries that do not have any solids would be Platinum. Graphite is also an option, however, I would recommend sticking with Pt. It is important to note that when writing a write cell diagram with a metal electrode, instead of putting vert...
by Fayez Kanj
Mon Feb 18, 2019 10:51 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: K=e^lnK
Replies: 4
Views: 88

Re: K=e^lnK

e is Euler's number, which is a special number similar to "pi" and is roughly equal to 2.71828. Since e raised to the power of ln (which has base e) cancel out, you are left with simply K.

Hope this helps :)
by Fayez Kanj
Mon Feb 18, 2019 10:49 pm
Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
Topic: Entropy at 0 K
Replies: 6
Views: 199

Re: Entropy at 0 K

Hello.

when temperature reaches 0K, then entropy tends to zero. At 0K theoretically, a molecule/compound can still have positional entropy (number of micro-states). If it is perfectly ordered, then yes, entropy would theoretically be 0)
by Fayez Kanj
Mon Feb 18, 2019 10:45 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Gibbs Free Energy and equilibrium
Replies: 2
Views: 52

Re: Gibbs Free Energy and equilibrium

It is also important to note that if deltaG = 0, then the rate of the forward reaction equals the rate of the backwards reaction, meaning that the reaction is at equilibrium.
by Fayez Kanj
Tue Feb 12, 2019 5:06 pm
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: Standard Molar Entropies
Replies: 2
Views: 55

Re: Standard Molar Entropies

Hello!

Heavier/larger atoms have more thermally accessible vibrational energy levels so thus, have higher entropy.

Hope this helps :)
by Fayez Kanj
Tue Feb 12, 2019 5:04 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: 8.45
Replies: 2
Views: 51

Re: 8.45

Hello!

Delta H is usually assumed to be KJ/mol. However, the questions in the textbook, and most likely the ones on the midterm and tests will have this specified. Always make sure that your units are matching each other in order to add/subtract two values.

Hope this helps :)
by Fayez Kanj
Tue Feb 12, 2019 5:02 pm
Forum: Calculating Work of Expansion
Topic: constant external pressure
Replies: 2
Views: 54

Re: constant external pressure

Hello! In most cases, this is explicitly stated in the problem ("against a constant pressure" of ...). However, if it is not, then you could try to infer. If the volume changes, then this is most likely going to be a reaction that involves a change in pressure (since we use w=nRTlnV2/V1). ...
by Fayez Kanj
Tue Feb 05, 2019 8:40 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Ideal Gas
Replies: 3
Views: 75

Re: Ideal Gas

Hello!

An ideal gas is a gas that:
1. takes up no volume
2. has negligible interactions between different particles

This is just an assumption, as not real gas is actually an ideal gas. However, it is used for calculations, since they obey the ideal gas law Pv=nRT.

Hope this helps :)
by Fayez Kanj
Tue Feb 05, 2019 8:34 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Total bond energy
Replies: 2
Views: 39

Re: Total bond energy

Hello!

To remember:

higher energy = less stable bone
lower energy = more stable bond

whenever bonds form, the stabilize the molecule by lowering their energy.

Hope this helps :)
by Fayez Kanj
Tue Feb 05, 2019 8:32 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Bond enthalpy vs Bond Enregy
Replies: 3
Views: 58

Re: Bond enthalpy vs Bond Enregy

Hello! Bond enthalpy is the amount of energy needed to break 1 mol of a specific bond for gaseous atoms under standard conditions. These always have positive values, since bond breaking requires energy and is thus endothermic. Bond energy is more ambiguous. It could mean the amount of energy a bond ...
by Fayez Kanj
Tue Jan 29, 2019 5:13 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Signs for Bond Enthalpies
Replies: 4
Views: 69

Re: Signs for Bond Enthalpies

Hello!

Yes, it always takes energy to break a bond, so bond enthalpies are positive values. To break a bond, energy is released, so they are negative values.

Bond making requires energy (+ value) whereas bond breaking releases energy (- value)

Hope this helps :)
by Fayez Kanj
Tue Jan 29, 2019 5:11 pm
Forum: Phase Changes & Related Calculations
Topic: Properties
Replies: 2
Views: 51

Re: Properties

Extensive Properties depend on the amount of substance. For example, Heat capacity is one. We could determine how much heat is needed to raise a volume of water by 1 degree Celcius, but this will change depending on how much water there is. Intensive properties such as Specific Heat Capacities do no...
by Fayez Kanj
Tue Jan 29, 2019 5:04 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Method 2: Using bond enthalpies
Replies: 3
Views: 48

Re: Method 2: Using bond enthalpies

Hey! So basically this method uses this formula: Sum of the energies required to break all bonds of reactants - Sum of energies required to break all bonds of products. These values will have to be given to you Alternatively, you could do the same for only the bonds that change. This method will giv...
by Fayez Kanj
Mon Jan 21, 2019 12:57 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Change in temperature's effect on K
Replies: 4
Views: 48

Re: Change in temperature's effect on K

Hello! Le Chatelier's Principle says that if a chemical equilibrium is subject to change, the equilibrium position will shift so as to minimize the effect of the change. When heating a reaction: The new equilibrium position will shift to the endothermic side, so as to absorb/use up all that extra he...
by Fayez Kanj
Mon Jan 21, 2019 12:54 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Question 5H1
Replies: 2
Views: 56

Re: Question 5H1

Hello! Notice that in part (b), all the stoichiometric coefficients have been halved. This means that in the expression for K, all the exponents will also be halved (since exponents in the expression of K correspond to the stoichiometric coefficients.) Halving the exponents essentially means taking ...
by Fayez Kanj
Mon Jan 21, 2019 12:51 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook 7th ed 5I19
Replies: 1
Views: 30

Re: Textbook 7th ed 5I19

Hello! In the question, it says that 60% of the H2 was used/had reacted at equilibrium. This means that when equilibrium is reached, only 40% of the H2 remains. This corresponds with the molarity of H2 at equilibrium from the ICE table, which is 0.133-x. As such, when you solve: (0.40)(0.133M) = 0.1...
by Fayez Kanj
Mon Jan 14, 2019 12:43 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Solids and Liquids in Equilibrium [ENDORSED]
Replies: 2
Views: 53

Re: Solids and Liquids in Equilibrium [ENDORSED]

Hello What is included in the calculation of K/Kc/Kp are Gases and Aqueous compounds. What are NOT included in the calculations of K are Solids and Liquids, for the same reasons you stated in the question. Keep in mind that we do not just ignore them. They have a value of [1]. This means that if all...
by Fayez Kanj
Mon Jan 14, 2019 12:40 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: solids on the reactant side
Replies: 3
Views: 2265

Re: solids on the reactant side

Hello Yes, assume that the solid is added to a beaker/flask/test tube of water. You can also imaging having a +H2O on both the reactant and product side. In either way, solids and liquids (water) are never included in the calculation of the equilibrium constant, so should have no impact on your fina...
by Fayez Kanj
Mon Jan 14, 2019 12:35 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: ICE Table [ENDORSED]
Replies: 3
Views: 80

Re: ICE Table [ENDORSED]

Hello, No, the subscripts do not matter for calculating the changes in concentration "x." Only the stoichiometric coefficients determine the values of x. For example, if 1 mol of reactant forms 2 moles of products, then the change for reactant concentration is -x, and the change in product...
by Fayez Kanj
Mon Jan 07, 2019 4:32 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Kc Value
Replies: 4
Views: 87

Re: Kc Value

Hello! Basically, the general rule is that if Kc is larger than 1, the equilibrium lies to the right (products are favored) and the opposite if Kc is less than 1. However, if you have 10^7 and 10^9, both are larger than 1. However, 10^9 is 100x larger than 10^7, so the equilibrium for the reaction w...
by Fayez Kanj
Mon Jan 07, 2019 4:29 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 11.13 Reaction Quotient [ENDORSED]
Replies: 3
Views: 73

Re: 11.13 Reaction Quotient [ENDORSED]

Hello!

The reaction quotient is the same expression as the equilibrium constant (calculated the same way), rather it is for a reaction NOT in equilibrium (ie: in any point in time)

Hope this helps :)
by Fayez Kanj
Mon Jan 07, 2019 4:26 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Solids and Liquids
Replies: 4
Views: 71

Re: Solids and Liquids

Hello! Dr Lavelle said in class that solids and liquids are not included in the equilibrium constants, not because they are in excess, rather because the molar concentrations of pure substances (ie: solids and liquids) does not change in a reaction. As such, in all cases, whether limiting or not, th...
by Fayez Kanj
Tue Dec 04, 2018 11:17 am
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Polydentates: HW 17.33 6th Ed
Replies: 2
Views: 64

Re: Polydentates: HW 17.33 6th Ed

Hey! Ligands must have atleast 1 lone pair to be considered a ligand. So to determine whether a ligand is polydentate, you count how many atoms have atleast 1 lone pair Follow this guide: Bind at 1 site (donate 1 pair of electrons) → Monodentate Binds at 2 sites (donates 2 pairs of electrons) → Bide...
by Fayez Kanj
Tue Dec 04, 2018 11:11 am
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Shape of Complex
Replies: 1
Views: 68

Re: Shape of Complex

Hey!

Dr Lavelle said in class that we don't need to be able to determine whether it is square planar or tetrahedral, since the experimental techniques of doing so are too complicated.

Hope this helps :)
by Fayez Kanj
Tue Dec 04, 2018 11:10 am
Forum: Biological Examples
Topic: Transition metals
Replies: 3
Views: 163

Re: Transition metals

Hey! Yes, on the outline for the coordination compounds unit, it says to know some common biological examples and their functions. I think we need to know these: 1. EDTA (removes mineral ions from solution by reacting with them) 2. hemoglobin and myoglobin (oxygen transport in the blood and muscles ...
by Fayez Kanj
Tue Dec 04, 2018 11:07 am
Forum: Naming
Topic: Roman Numerals
Replies: 3
Views: 53

Re: Roman Numerals

Hi!

Roman numerals indicate the oxidation state of the transition metal in the coordination compound (inside the [square brackets]). If the oxidation state is +1, then use (I), if +2, use (II) and so on.

Hope this helps :)
by Fayez Kanj
Tue Nov 27, 2018 6:26 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Bond Angles
Replies: 3
Views: 50

Bond Angles

In trigonal bipyramidal compounds, I know there are 120, 90, and 180 degree bond angles. What the are bond angles in See-saw and T shaped molecules?

Also, Octahedral compounds have bond angles of 90 and 180. What would be the bond angles in square pyramidal and square planar shaped molecules?
by Fayez Kanj
Mon Nov 26, 2018 6:55 pm
Forum: Hybridization
Topic: Ethene example
Replies: 2
Views: 63

Re: Ethene example

Hello! For Ethene, notice that each C has 3 regions of electron density. This means that the hybridization must be sp2 (by the conservation of orbitals law). Regular ethene has configuration 2s2, 2px1, 2px2. When it gets hybridized, an electron gets promoted from the 2s to the 2pz. The new configura...
by Fayez Kanj
Mon Nov 26, 2018 6:51 pm
Forum: Hybridization
Topic: 4.31
Replies: 2
Views: 52

Re: 4.31

I think by orientation, they mean electron arrangement/geometry. This would be the basic shapes (ie: linear, trigonal planar, tetrahedral, trigonal bipyramidal or octahedral) So for example, sp3 would be tetrahedral, since there are 4 hybrid orbitals, and there is the conservation of orbitals Hope t...
by Fayez Kanj
Mon Nov 26, 2018 6:50 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Problem 2E # 1 7th edition
Replies: 2
Views: 49

Problem 2E # 1 7th edition

The question asks whether a molecule with 180 degree bond angles and only 2 bonded atoms, must have, may have, or cannot have lone pairs. The answer is "may have lone pairs." Why is that?
by Fayez Kanj
Tue Nov 20, 2018 4:16 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Bond Angles
Replies: 2
Views: 44

Re: Bond Angles

Hello! For molecules that are linear, trigonal planar and tetrahedral, all bond angles are more or less the same, so no need to worry about them. For molecules that are trigonal pyramidal and octahedral: For trigonal pyramidal, there are 3 atoms bonded in a trigonal planar plane, so would have 120 d...
by Fayez Kanj
Tue Nov 20, 2018 4:10 pm
Forum: Hybridization
Topic: E- Promotion vs Hybridization
Replies: 1
Views: 78

Re: E- Promotion vs Hybridization

Hello! Electron promotion is when an electron is "promoted" to a higher energy level, or a higher subshell. For example, is there was an electron in the 2s orbital, and it can get "promoted" and go up to the 2p orbital. This is also called "excitation" or "excited ...
by Fayez Kanj
Tue Nov 20, 2018 4:05 pm
Forum: Hybridization
Topic: Hybrid Orbital Energy
Replies: 1
Views: 35

Re: Hybrid Orbital Energy

Hello! The energy of a hybrid orbital is indeed in between the energies of the regular orbitals (ex: energy of sp3 is in between the energy of s and that of p). However, I don't think its an average (ie: in the middle) because in sp3 for example, there are 3 p's, and 1 s. But all you need to know is...
by Fayez Kanj
Wed Nov 14, 2018 3:47 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: The A in the VSEPR Formula [ENDORSED]
Replies: 4
Views: 53

Re: The A in the VSEPR Formula [ENDORSED]

As mentioned above, A represents the central atom
X represents the number of bonded atoms
E represents the lone pairs attached to the central atom

Hope this helps :)
by Fayez Kanj
Wed Nov 14, 2018 3:45 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Question 4.5 (Determining shape from Lewis structure)
Replies: 2
Views: 45

Re: Question 4.5 (Determining shape from Lewis structure)

Hey! here is a good step-by-step way: 1. draw the lewis structure as we learned from class 2. count how many regions of electron density there are (each double and triple bond counts as 1 region, and each lone pair counts as 1 region) 2 regions = linear 3 regions = trigonal planar 4 regions = tetrah...
by Fayez Kanj
Wed Nov 14, 2018 3:40 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Bond Angles
Replies: 3
Views: 58

Re: Bond Angles

Hey! Yes, you need to memorize the bond angle of 109.5 for any tetrahedral molecule geometry. However, when there are shapes that are trigonal pyramidal ( 3 bonding pairs and 1 lone pair), you do not need to know the angles, just that they will be slightly less that 109.5. A good rule of thumb is th...
by Fayez Kanj
Wed Nov 07, 2018 11:38 am
Forum: Bond Lengths & Energies
Topic: Bond Lengths [ENDORSED]
Replies: 2
Views: 92

Re: Bond Lengths [ENDORSED]

hello! Single bonds are weaker than double bonds (attraction between 2 electrons and the nucleus vs attraction of 4 electrons and the nucleus which is stronger). Since they are weaker, the electrons are less tightly pulled towards the atoms (they are further apart in space) and thus, have a longer l...
by Fayez Kanj
Wed Nov 07, 2018 11:32 am
Forum: Bond Lengths & Energies
Topic: 2.25 Homework Problem
Replies: 4
Views: 93

Re: 2.25 Homework Problem

Hello!

P has a larger atomic radius than N, and, as a result, the distance between the P and F atoms is larger than the distance between the N and F atoms. Therefore, the bond lengths in PF3, although being single bonds just like in NF3, are longer.

Hope this helps :)
by Fayez Kanj
Wed Nov 07, 2018 11:30 am
Forum: Dipole Moments
Topic: Interaction Potential Energy
Replies: 1
Views: 48

Re: Interaction Potential Energy

Hello! Interaction potential energy is basically the attractive force (between different molecules). It is calculated using the equation: Ep = - (polarizability of molecule 1)(polarizability of molecule 2) / (distance between atoms)^6 It is always negative because it is a force of ATTRACTION (energy...
by Fayez Kanj
Tue Oct 30, 2018 6:29 pm
Forum: Resonance Structures
Topic: How to tell when an molecule is resonant?
Replies: 7
Views: 305

Re: How to tell when an molecule is resonant?

Hello! A molecule is resonant if it has double/triple bonds that can be placed elsewhere (between different atoms of the same elements) without changing the chemical bonds in the molecule. A resonance structure could also simply be changing the central atom (there was an example of this in the book,...
by Fayez Kanj
Tue Oct 30, 2018 6:26 pm
Forum: Lewis Structures
Topic: kekule structure?
Replies: 3
Views: 67

Re: kekule structure?

Hello! What i know is that the kekule structure of benzene (C6H6) is the one where the carbons form a hexagon ring and there are orienting double and single bonds. This model has however, been disproven experimentally, which is why a better representation of the benzene compound is the one with the ...
by Fayez Kanj
Tue Oct 30, 2018 6:23 pm
Forum: Ionic & Covalent Bonds
Topic: 2B.7
Replies: 3
Views: 98

Re: 2B.7

The element is a period 3 element, meaning it could have a total of 18 valence electrons (2 from the s orbital, 6 from the p orbitals and 10 from the d orbitals). To solve this problem, you would subtract the total number of valence electrons that the molecule has, by the valence electrons of all th...
by Fayez Kanj
Tue Oct 23, 2018 5:53 pm
Forum: Photoelectric Effect
Topic: Question 1.33 (sixth edition)
Replies: 2
Views: 95

Re: Question 1.33 (sixth edition)

Hi!

So for b), since you have the frequency "v" you can calculate the energy E using E=hv.

For c), since you have the frequency "v" you can calculate the wavelength of the light "lambda" using C= lambda/wavelength x frequency

Hope that helps :)
by Fayez Kanj
Tue Oct 23, 2018 5:44 pm
Forum: Properties of Light
Topic: EM Energy Spectrum
Replies: 4
Views: 144

Re: EM Energy Spectrum

Hello!

I don't think you need to know the exact values. However, I know for a fact that we need to know the trends in the EM spectrum, which are:

Longer wavelength = lower frequency = lower energy

Shorter wavelength = higher frequency = higher energy

Hope this helps :)
by Fayez Kanj
Tue Oct 23, 2018 5:38 pm
Forum: Photoelectric Effect
Topic: Module Question 32.B
Replies: 1
Views: 92

Re: Module Question 32.B

Hi! So Basically, you use the equation E - (work function) = E(Kinetic Energy) You have the value of the work function, and the kinetic energy, so you determine the energy E: E - 3.607x10^-19 = 4.200x10^-19 You get E= 7.807x10^-19 J Now, you use the equation E=h/v to figure out "v" (since ...
by Fayez Kanj
Tue Oct 16, 2018 5:57 pm
Forum: Properties of Electrons
Topic: Wave Properties of Electrons
Replies: 3
Views: 95

Re: Wave Properties of Electrons

Hello! When an atom is given an electrical discharge or is exposed to light, an electron gets excited and goes up to a higher energy level. So, in this sense, yes, an electron could go from n=1 to n=3. The electron is unstable at that higher energy so falls back down, and the energy released when th...
by Fayez Kanj
Tue Oct 16, 2018 5:52 pm
Forum: Properties of Light
Topic: Wave Model
Replies: 3
Views: 90

Re: Wave Model

Hello! Here are the main Ideas: Light is electromagnetic radiation (wave of electric and magnetic fields) When light is passed through a crystal, a diffraction pattern arises Speed of light in a vacuum (c): 3.00 x 108 ms-1 Frequency (v) How many cycles per second Hertz (Hz) 1Hz = 1s-1 Determines the...
by Fayez Kanj
Tue Oct 16, 2018 5:40 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: What the equation is used for
Replies: 5
Views: 95

Re: What the equation is used for

Hello! Basically, because particles like electrons are so small, there is a great uncertainty in determining their positions and/or momentums. The Heisenberg Uncertainty principle says that, if the location (x) of a particle is known, then the momentum of the particle (p) can be known simultaneously...
by Fayez Kanj
Mon Oct 08, 2018 6:28 pm
Forum: Properties of Light
Topic: Formula for energy of a photon
Replies: 4
Views: 61

Re: Formula for energy of a photon

Hi!

This is the formula:

Energy per photon = hv

Where h is Planck's constant of 6.63 x 10^-34
and v is the frequency (in Hertz)

Hope this helps :)
by Fayez Kanj
Mon Oct 08, 2018 6:22 pm
Forum: Photoelectric Effect
Topic: Increasing photon energy [ENDORSED]
Replies: 2
Views: 60

Re: Increasing photon energy [ENDORSED]

Hello! Basically, frequency is directly proportional to energy, meaning that the higher the frequency of a light wave, the more energy is carries. Thus, a higher frequency = more energy = electrons will be displaced from the metal. Intensity is directly proportional to the number of photons. So, a h...
by Fayez Kanj
Mon Oct 08, 2018 6:08 pm
Forum: Properties of Light
Topic: Difference between "intensity of light" and "frequency of light"
Replies: 3
Views: 105

Re: Difference between "intensity of light" and "frequency of light"

Hey, From what I understood, the frequency of light is how many cycles of the wave there are in 1 second (or per unit time). For example, if in 1 second, 10 waves pass, then the frequency is 10Hz (Hertz). Also, the frequency of light is directly proportional to the energy of its photons ie: the high...
by Fayez Kanj
Wed Oct 03, 2018 4:29 pm
Forum: Limiting Reactant Calculations
Topic: HW Question M.5
Replies: 2
Views: 76

Re: HW Question M.5

Hello! A quick and easy way to determine the limiting reactant is to first calculate the number of moles of each reactant, then divide the number of moles of each reactant by its respective stoichiometric coefficient. The smallest number will indicate which reactant is limiting and is not present in...
by Fayez Kanj
Wed Oct 03, 2018 4:21 pm
Forum: Molarity, Solutions, Dilutions
Topic: Fundamentals E.9
Replies: 2
Views: 75

Re: Fundamentals E.9

Hello! This is a quick and easy way to do problems like these. If you multiply the number of moles of molecules by the number of a specific atom in that molecule, you get the number of moles of the specific atom/element. For example, If you have 2 moles of H2SO4, then to get the number of moles of O...
by Fayez Kanj
Wed Oct 03, 2018 4:10 pm
Forum: SI Units, Unit Conversions
Topic: Converting between Temperatures (K,C,F)
Replies: 4
Views: 139

Re: Converting between Temperatures (K,C,F)

Hey!

Here are the formulas:

To convert between Celsius and Fahrenheit, use this: Fahrenheit = (1.8 x Celsius) + 32 or the inverse to convert in the other direction
To convert between Kelvin and Celsius, use this: Kelvin = Celsius + 273.15

Hope this helps :)

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