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!
Search found 62 matches
- 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: 226
- Wed Mar 13, 2019 7:06 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: K and k
- Replies: 1
- Views: 208
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 ...
- Wed Mar 13, 2019 7:02 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Pre-equilibrium approach
- Replies: 4
- Views: 393
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, ...
- Wed Mar 06, 2019 4:26 pm
- Forum: General Rate Laws
- Topic: how to calculate reaction rates
- Replies: 2
- Views: 362
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 :)
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 :)
- Wed Mar 06, 2019 4:23 pm
- Forum: First Order Reactions
- Topic: Deriving these equations
- Replies: 2
- Views: 318
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 :)
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 :)
- 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: 362
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 :)
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 :)
- Sun Feb 24, 2019 3:24 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Oxidation and Reduction
- Replies: 3
- Views: 323
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...
- Sun Feb 24, 2019 3:21 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: salt bridge
- Replies: 3
- Views: 327
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...
- Sun Feb 24, 2019 3:17 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Electrode Metal
- Replies: 1
- Views: 184
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...
- Mon Feb 18, 2019 10:51 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: K=e^lnK
- Replies: 4
- Views: 610
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 :)
Hope this helps :)
- 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: 1407
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)
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)
- Mon Feb 18, 2019 10:45 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy and equilibrium
- Replies: 2
- Views: 316
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.
- 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: 331
Re: Standard Molar Entropies
Hello!
Heavier/larger atoms have more thermally accessible vibrational energy levels so thus, have higher entropy.
Hope this helps :)
Heavier/larger atoms have more thermally accessible vibrational energy levels so thus, have higher entropy.
Hope this helps :)
- Tue Feb 12, 2019 5:04 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.45
- Replies: 2
- Views: 308
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 :)
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 :)
- Tue Feb 12, 2019 5:02 pm
- Forum: Calculating Work of Expansion
- Topic: constant external pressure
- Replies: 2
- Views: 294
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). ...
- Tue Feb 05, 2019 8:40 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Ideal Gas
- Replies: 3
- Views: 401
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 :)
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 :)
- 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: 224
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 :)
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 :)
- 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: 263
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 ...
- 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: 404
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 :)
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 :)
- Tue Jan 29, 2019 5:11 pm
- Forum: Phase Changes & Related Calculations
- Topic: Properties
- Replies: 2
- Views: 316
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...
- 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: 350
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...
- 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: 321
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...
- Mon Jan 21, 2019 12:54 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Question 5H1
- Replies: 2
- Views: 287
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 ...
- Mon Jan 21, 2019 12:51 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Textbook 7th ed 5I19
- Replies: 1
- Views: 154
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...
- Mon Jan 14, 2019 12:43 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Solids and Liquids in Equilibrium [ENDORSED]
- Replies: 2
- Views: 117
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...
- Mon Jan 14, 2019 12:40 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: solids on the reactant side
- Replies: 3
- Views: 2564
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...
- 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: 452
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...
- Mon Jan 07, 2019 4:32 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Kc Value
- Replies: 4
- Views: 1462
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...
- Mon Jan 07, 2019 4:29 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.13 Reaction Quotient [ENDORSED]
- Replies: 3
- Views: 209
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 :)
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 :)
- Mon Jan 07, 2019 4:26 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Solids and Liquids
- Replies: 4
- Views: 375
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...
- Tue Dec 04, 2018 11:17 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Polydentates: HW 17.33 6th Ed
- Replies: 2
- Views: 341
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...
- Tue Dec 04, 2018 11:11 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Shape of Complex
- Replies: 1
- Views: 242
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 :)
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 :)
- Tue Dec 04, 2018 11:10 am
- Forum: Biological Examples
- Topic: Transition metals
- Replies: 3
- Views: 551
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 ...
- Tue Dec 04, 2018 11:07 am
- Forum: Naming
- Topic: Roman Numerals
- Replies: 3
- Views: 251
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 :)
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 :)
- Tue Nov 27, 2018 6:26 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 3
- Views: 165
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?
Also, Octahedral compounds have bond angles of 90 and 180. What would be the bond angles in square pyramidal and square planar shaped molecules?
- Mon Nov 26, 2018 6:55 pm
- Forum: Hybridization
- Topic: Ethene example
- Replies: 2
- Views: 314
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...
- Mon Nov 26, 2018 6:51 pm
- Forum: Hybridization
- Topic: 4.31
- Replies: 2
- Views: 135
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...
- Mon Nov 26, 2018 6:50 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Problem 2E # 1 7th edition
- Replies: 2
- Views: 289
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?
- Tue Nov 20, 2018 4:16 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 2
- Views: 154
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...
- Tue Nov 20, 2018 4:10 pm
- Forum: Hybridization
- Topic: E- Promotion vs Hybridization
- Replies: 1
- Views: 511
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 ...
- Tue Nov 20, 2018 4:05 pm
- Forum: Hybridization
- Topic: Hybrid Orbital Energy
- Replies: 1
- Views: 140
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...
- Wed Nov 14, 2018 3:47 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: The A in the VSEPR Formula [ENDORSED]
- Replies: 4
- Views: 240
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 :)
X represents the number of bonded atoms
E represents the lone pairs attached to the central atom
Hope this helps :)
- Wed Nov 14, 2018 3:45 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Question 4.5 (Determining shape from Lewis structure)
- Replies: 2
- Views: 312
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...
- Wed Nov 14, 2018 3:40 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 3
- Views: 337
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...
- Wed Nov 07, 2018 11:38 am
- Forum: Bond Lengths & Energies
- Topic: Bond Lengths [ENDORSED]
- Replies: 2
- Views: 329
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...
- Wed Nov 07, 2018 11:32 am
- Forum: Bond Lengths & Energies
- Topic: 2.25 Homework Problem
- Replies: 4
- Views: 874
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 :)
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 :)
- Wed Nov 07, 2018 11:30 am
- Forum: Dipole Moments
- Topic: Interaction Potential Energy
- Replies: 1
- Views: 158
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...
- Tue Oct 30, 2018 6:29 pm
- Forum: Resonance Structures
- Topic: How to tell when an molecule is resonant?
- Replies: 7
- Views: 1564
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,...
- Tue Oct 30, 2018 6:26 pm
- Forum: Lewis Structures
- Topic: kekule structure?
- Replies: 3
- Views: 274
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 ...
- Tue Oct 30, 2018 6:23 pm
- Forum: Ionic & Covalent Bonds
- Topic: 2B.7
- Replies: 3
- Views: 400
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...
- Tue Oct 23, 2018 5:53 pm
- Forum: Photoelectric Effect
- Topic: Question 1.33 (sixth edition)
- Replies: 2
- Views: 359
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 :)
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 :)
- Tue Oct 23, 2018 5:44 pm
- Forum: Properties of Light
- Topic: EM Energy Spectrum
- Replies: 4
- Views: 399
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 :)
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 :)
- Tue Oct 23, 2018 5:38 pm
- Forum: Photoelectric Effect
- Topic: Module Question 32.B
- Replies: 1
- Views: 325
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 ...
- Tue Oct 16, 2018 5:57 pm
- Forum: Properties of Electrons
- Topic: Wave Properties of Electrons
- Replies: 3
- Views: 353
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...
- Tue Oct 16, 2018 5:52 pm
- Forum: Properties of Light
- Topic: Wave Model
- Replies: 3
- Views: 387
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...
- Tue Oct 16, 2018 5:40 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: What the equation is used for
- Replies: 5
- Views: 463
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...
- Mon Oct 08, 2018 6:28 pm
- Forum: Properties of Light
- Topic: Formula for energy of a photon
- Replies: 4
- Views: 215
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 :)
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 :)
- Mon Oct 08, 2018 6:22 pm
- Forum: Photoelectric Effect
- Topic: Increasing photon energy [ENDORSED]
- Replies: 2
- Views: 342
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...
- Mon Oct 08, 2018 6:08 pm
- Forum: Properties of Light
- Topic: Difference between "intensity of light" and "frequency of light"
- Replies: 3
- Views: 9001
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...
- Wed Oct 03, 2018 4:29 pm
- Forum: Limiting Reactant Calculations
- Topic: HW Question M.5
- Replies: 2
- Views: 165
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...
- Wed Oct 03, 2018 4:21 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Fundamentals E.9
- Replies: 2
- Views: 171
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...
- Wed Oct 03, 2018 4:10 pm
- Forum: SI Units, Unit Conversions
- Topic: Converting between Temperatures (K,C,F)
- Replies: 4
- Views: 409
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 :)
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 :)