Hey,
For this question, you would know to find the equations of combustion because the given enthalpies are of combustion. So you would find the combustion reactions, then balance those, and then you could use Hess's Law to find the final reaction enthalpy.
Search found 54 matches
- Wed Mar 14, 2018 10:25 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.57
- Replies: 1
- Views: 265
- Wed Mar 14, 2018 10:19 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: oxidation and reduction agents
- Replies: 1
- Views: 416
Re: oxidation and reduction agents
Hey, For the cathode, the oxidizing agent is the reactant that gains the electrons. For the anode, the reducing agent is going to be on the left hand side of the reaction, opposite the side that gains the electrons. So for example, in test question 6, H+ gains the electrons so it is the oxidizing ag...
- Wed Mar 14, 2018 10:17 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: question 6 on test 2
- Replies: 2
- Views: 516
Re: question 6 on test 2
Hey,
Each pair has either an oxidizing agent and a reducing agent. For the H+ pair, H+ is oxidizing, but it is also the reduced agent. For Ni, it is the reducing agent, but it is also the oxidized agent.
Each pair has either an oxidizing agent and a reducing agent. For the H+ pair, H+ is oxidizing, but it is also the reduced agent. For Ni, it is the reducing agent, but it is also the oxidized agent.
- Sun Mar 11, 2018 8:52 pm
- Forum: First Order Reactions
- Topic: The integrated rate laws
- Replies: 3
- Views: 576
Re: The integrated rate laws
Hey,
The basic versions of the integrated rate laws for each reaction order is given during the tests. Any different versions of the laws are just algebraic manipulations that you can perform yourself if needed.
The basic versions of the integrated rate laws for each reaction order is given during the tests. Any different versions of the laws are just algebraic manipulations that you can perform yourself if needed.
- Sun Mar 11, 2018 8:51 pm
- Forum: First Order Reactions
- Topic: slope
- Replies: 3
- Views: 574
Re: slope
Hey,
The slope of the graph is -k. The graph is baed on ln(a) against time.
The slope of the graph is -k. The graph is baed on ln(a) against time.
- Sun Mar 11, 2018 8:49 pm
- Forum: First Order Reactions
- Topic: Slope of 1st order RXNs
- Replies: 9
- Views: 1260
Re: Slope of 1st order RXNs
Hey,
The slope of both 0 and 1st order reactions is -k, second order reactions have a positive k slope. Time is graphed against ln(a).
The slope of both 0 and 1st order reactions is -k, second order reactions have a positive k slope. Time is graphed against ln(a).
- Sun Mar 04, 2018 6:47 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Homogenous vs. Heterogenous Catalyst and Rate Law
- Replies: 2
- Views: 586
Re: Homogenous vs. Heterogenous Catalyst and Rate Law
Hey,
I would assume heterogenous catalysts would be included in the rate law because its purpose as a catalyst is still the same. It lowers the activation energy, increasing the reaction speed, so the rate would be affected.
I would assume heterogenous catalysts would be included in the rate law because its purpose as a catalyst is still the same. It lowers the activation energy, increasing the reaction speed, so the rate would be affected.
- Sun Mar 04, 2018 6:44 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: When forming rate laws do we include solids and liquids?
- Replies: 5
- Views: 9457
Re: When forming rate laws do we include solids and liquids?
Hey,
Most of the time we would not include solids and liquids, only gases and aqueous solutions. Water can be included if it is part of the reaction.
Most of the time we would not include solids and liquids, only gases and aqueous solutions. Water can be included if it is part of the reaction.
- Sun Mar 04, 2018 5:54 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Catalyst, Equation, and Rate Law
- Replies: 4
- Views: 5715
Re: Catalyst, Equation, and Rate Law
The catalyst itself is not consumed during the reaction and so it does not interfere with the reactants turning into the products, aside from decreasing activation energy so that the process is sped up.
- Sun Mar 04, 2018 5:03 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Catalyst, Equation, and Rate Law
- Replies: 4
- Views: 5715
Re: Catalyst, Equation, and Rate Law
Hey,
A catalyst is accounted for in the rate law because it speeds up the reaction. It isn't considered in a balanced chemical equation because it doesn't undergo any change, nor does it affect any of the reactants in anyway besides increasing the speed of the reaction.
A catalyst is accounted for in the rate law because it speeds up the reaction. It isn't considered in a balanced chemical equation because it doesn't undergo any change, nor does it affect any of the reactants in anyway besides increasing the speed of the reaction.
- Sun Feb 25, 2018 9:35 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.33 (b) Half-reactions
- Replies: 2
- Views: 444
Re: 14.33 (b) Half-reactions
Hey,
Normally you could use that half reaction, but the values for the e cell are not given. Also, the question asks if it disproportionates, which means that the element would act as both an oxidating agent and reducing agent, so you would have to use the two half reactions in the solution manual.
Normally you could use that half reaction, but the values for the e cell are not given. Also, the question asks if it disproportionates, which means that the element would act as both an oxidating agent and reducing agent, so you would have to use the two half reactions in the solution manual.
- Sat Feb 24, 2018 8:16 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Q of eqn [ENDORSED]
- Replies: 2
- Views: 409
Re: Q of eqn [ENDORSED]
Hey,
If you can recall from last year, Q can be calculated from both partial pressures and the concentrations of aqueous solutions. Because Q doesn't have any units, partial pressures and concentrations can be combined.
If you can recall from last year, Q can be calculated from both partial pressures and the concentrations of aqueous solutions. Because Q doesn't have any units, partial pressures and concentrations can be combined.
- Sat Feb 24, 2018 7:52 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Calculating n in the Nernst Equation
- Replies: 2
- Views: 1478
Re: Calculating n in the Nernst Equation
Hey,
When trying to find n in the nerst equation, and there is an uneven amount of electrons on each side, you usually have to balance out the equations so that the electrons cancel. The final amount of electrons after you balance is what you use for n.
When trying to find n in the nerst equation, and there is an uneven amount of electrons on each side, you usually have to balance out the equations so that the electrons cancel. The final amount of electrons after you balance is what you use for n.
- Sun Feb 18, 2018 5:18 pm
- Forum: Balancing Redox Reactions
- Topic: Bruincast
- Replies: 4
- Views: 637
Re: Bruincast
Hey,
As far as I know, this class isn't bruincasted. The other chem14b class might the one that is bruincasted, so if you need information from a lecture, you would probably have to get it from a classmate or a friend.
As far as I know, this class isn't bruincasted. The other chem14b class might the one that is bruincasted, so if you need information from a lecture, you would probably have to get it from a classmate or a friend.
- Sun Feb 18, 2018 5:09 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagrams
- Replies: 2
- Views: 450
Re: Cell Diagrams
Hey,
Usually for a cell diagram, the anode would be on the left, and the cathode be on the right. If there is a salt bridge in the cell, two lines are drawn between the cathode and anode. We know an element is reduced when it gains electrons. A vertical line is the interface between phases.
Usually for a cell diagram, the anode would be on the left, and the cathode be on the right. If there is a salt bridge in the cell, two lines are drawn between the cathode and anode. We know an element is reduced when it gains electrons. A vertical line is the interface between phases.
- Sun Feb 18, 2018 5:02 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Electrode potential
- Replies: 1
- Views: 267
Re: Electrode potential
Hey,
The standard potentials for strongly oxidizing half reactions are usually positive, while the standard potentials for strongly reducing reactions are negative.
The standard potentials for strongly oxidizing half reactions are usually positive, while the standard potentials for strongly reducing reactions are negative.
- Sat Feb 10, 2018 4:59 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: internal energy for isothermal expansion
- Replies: 2
- Views: 390
Re: internal energy for isothermal expansion
Hey, The book explains it best on page 274, but basically the molecules of an ideal gas move at the same average speed as it expands, meaning the total kinetic energy is the same, and since there is nothing acting against the molecules, the potential energy remains the same. Because both of these en...
- Sat Feb 10, 2018 4:55 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.101 (c) and (f)
- Replies: 3
- Views: 541
Re: 8.101 (c) and (f)
Hey,
A change in temperature is associated with the amount of heat energy transferred or absorbed. So even if temperature is constant, work energy can be done, and this would be the change in internal energy.
A change in temperature is associated with the amount of heat energy transferred or absorbed. So even if temperature is constant, work energy can be done, and this would be the change in internal energy.
- Sat Feb 10, 2018 4:15 pm
- Forum: Biological Examples (*DNA Structural Transitions, etc.)
- Topic: STP [ENDORSED]
- Replies: 8
- Views: 1928
Re: STP [ENDORSED]
Hey,
I think STP is supposed to be at 0 degrees Celsius or 273.15K
I think STP is supposed to be at 0 degrees Celsius or 273.15K
- Sun Feb 04, 2018 10:18 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: qrev
- Replies: 3
- Views: 423
Re: qrev
Hey,
Usually qrev will be given as a value in J transferred or absorbed as heat.
Usually qrev will be given as a value in J transferred or absorbed as heat.
- Sun Feb 04, 2018 9:56 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: System vs Surrounding
- Replies: 4
- Views: 613
Re: System vs Surrounding
Hey,
In that reaction in a beaker, it is assumed that the beaker is the surrounding because the system (the reaction) is specifically said to be contained inside the beaker. When a system is contained in something, like a piston, that is usually the surrounding.
In that reaction in a beaker, it is assumed that the beaker is the surrounding because the system (the reaction) is specifically said to be contained inside the beaker. When a system is contained in something, like a piston, that is usually the surrounding.
- Sun Feb 04, 2018 9:53 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Units of ΔS
- Replies: 3
- Views: 3243
Re: Units of ΔS
Hey, If J/k mol is given for enthalpy, and there is no value of mol that needs to be accounted for, then the entropy would also be in J/k mol. In some cases, such as when calculating standard reaction entropy, and you are given the amount of moles per product or reactant, then the answer would likel...
- Thu Jan 25, 2018 10:33 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Examples of work being done
- Replies: 7
- Views: 974
Re: Examples of work being done
Hey,
An example of a system doing work would be when a gas is pushing against a piston. When volume and pressure are changing, there can be work being done on the system. If there is a vacuum, work cannot be done because there is nothing to do work against.
An example of a system doing work would be when a gas is pushing against a piston. When volume and pressure are changing, there can be work being done on the system. If there is a vacuum, work cannot be done because there is nothing to do work against.
- Wed Jan 24, 2018 9:06 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Homework question 8.41
- Replies: 4
- Views: 577
Re: Homework question 8.41
Hey, You're on the right track. Heat gained by the ice cube is heat lost in the surrounding water. For the ice cube, you have to consider the enthalpy of fusion of the ice, meaning you have to calculate the amount of heat it takes to melt the ice, and then add it to the amount of heat that is gained...
- Wed Jan 24, 2018 9:01 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.101e
- Replies: 2
- Views: 340
Re: 8.101e
Hey,
This means that energy is released in terms of heat.
This means that energy is released in terms of heat.
- Thu Jan 18, 2018 7:15 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.57
- Replies: 3
- Views: 1012
Re: 8.57
Hey, The question is referring to enthalpy of combustion, so you would create a combustion reaction for each of the reactants and products. Then with the combustion reactions, you would balance them and order them so that if you would add them, you would be left with the original equation. This mean...
- Thu Jan 18, 2018 5:01 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.67b
- Replies: 1
- Views: 183
Re: 8.67b
Hey,
I believe that the value of the standard enthalpy of sublimation of carbon is given as +717 kJ/mol in the problem.
I believe that the value of the standard enthalpy of sublimation of carbon is given as +717 kJ/mol in the problem.
- Thu Jan 18, 2018 4:30 pm
- Forum: Phase Changes & Related Calculations
- Topic: Heat Capacity
- Replies: 6
- Views: 744
Re: Heat Capacity
Hey,
Table 8.2 uses Joules for specific and molar heat capacity, but it can be converted to kilojoules if heat energy is present in large values.
Table 8.2 uses Joules for specific and molar heat capacity, but it can be converted to kilojoules if heat energy is present in large values.
- Sun Jan 14, 2018 4:27 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: q and w
- Replies: 5
- Views: 622
Re: q and w
Hey, W is work and Q is heat. Work is motion against an opposing force, and heat is the energy transferred as a result of a temperature difference. When work is done on a system, w is positive, and when work is done by the system, w is negative. If heat enters the system, q is positive, and if heat ...
- Sun Jan 14, 2018 4:21 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.43 Heat Capacity (Units)
- Replies: 1
- Views: 285
Re: 8.43 Heat Capacity (Units)
Hey,
Molar heat capacity is usually given by J.K^-1.mol^-1, so I think the book left out the temperature units for 8.43.
Molar heat capacity is usually given by J.K^-1.mol^-1, so I think the book left out the temperature units for 8.43.
- Tue Jan 09, 2018 7:16 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase Changes [ENDORSED]
- Replies: 3
- Views: 274
Re: Phase Changes [ENDORSED]
Hey,
Some of the phase changes that you should know include:
Vaporization: liquid to vapor, Melting: solid to liquid, Sublimation: solid to vapor (Endothermic)
Freezing:liquid to solid, Condensation: gas to liquid, Deposition: gas to solid (Exothermic)
Some of the phase changes that you should know include:
Vaporization: liquid to vapor, Melting: solid to liquid, Sublimation: solid to vapor (Endothermic)
Freezing:liquid to solid, Condensation: gas to liquid, Deposition: gas to solid (Exothermic)
- Tue Dec 05, 2017 9:57 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Delocalization and how to identify it
- Replies: 1
- Views: 185
Re: Delocalization and how to identify it
Hey, You can find out if there is delocalization in a compound by checking to see if there are any resonance structures. If there are any resonance structures, there is going to be delocalization. Delocalization basically means that when you have a compound with resonance structures, some of the ele...
- Tue Dec 05, 2017 9:53 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: 12.63
- Replies: 1
- Views: 224
Re: 12.63
Hey,
The solutions manual doesn't use the approximation, but even if you use the approximation, you will still get the same answer for pH, so its probably okay to use the approximation in this case.
The solutions manual doesn't use the approximation, but even if you use the approximation, you will still get the same answer for pH, so its probably okay to use the approximation in this case.
- Tue Dec 05, 2017 9:43 pm
- Forum: Coordinate Covalent Bonds
- Topic: Naming - Prefixes
- Replies: 2
- Views: 631
Re: Naming - Prefixes
Hey,
You use the prefixes bis-, tris-, or tetrakis-, when the ligand is a polydentate.
You use the prefixes bis-, tris-, or tetrakis-, when the ligand is a polydentate.
- Fri Dec 01, 2017 9:20 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Change in Equilibrium for Aqueous Solutions
- Replies: 2
- Views: 479
Re: Change in Equilibrium for Aqueous Solutions
Hey,
Compression will only affect gaseous reactions. Aqueous/liquids and solids in the reaction cannot be compressed, so there would be no change in the equilibrium reaction with molecules in those phases.
Compression will only affect gaseous reactions. Aqueous/liquids and solids in the reaction cannot be compressed, so there would be no change in the equilibrium reaction with molecules in those phases.
- Fri Dec 01, 2017 8:50 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Effects with Water
- Replies: 2
- Views: 336
Re: Effects with Water
Hey,
Adding or subtracting water only creates a shift in the reaction when it is in a gas phase. When it is in an aqueous phase, the reaction doesn't change because the water is acting as a solvent on both sides.
Adding or subtracting water only creates a shift in the reaction when it is in a gas phase. When it is in an aqueous phase, the reaction doesn't change because the water is acting as a solvent on both sides.
- Sun Nov 26, 2017 5:00 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Chem equilibrium module 4, Question #13?
- Replies: 1
- Views: 207
Re: Chem equilibrium module 4, Question #13?
Hey, For part A, you have the correct reasoning, but on the right side of the reaction, you have I2(s), which is a solid. You only have to check for the most amount of gas moles in the reaction, not including solids or liquids. For part B, when the reaction is exothermic, and the heat is decreased, ...
- Fri Nov 24, 2017 8:46 am
- Forum: Hybridization
- Topic: Hypervalent Hybridization
- Replies: 2
- Views: 407
Re: Hypervalent Hybridization
Hey,
In SO4 2-, we see that the sulfur is bonded to four oxygens. This means it has four bonding densities, and therefore the hybridization would be sp3. You would have a tetrahedral shape with 4 atomic orbitals and 4 hybrid orbitals.
In SO4 2-, we see that the sulfur is bonded to four oxygens. This means it has four bonding densities, and therefore the hybridization would be sp3. You would have a tetrahedral shape with 4 atomic orbitals and 4 hybrid orbitals.
- Thu Nov 23, 2017 11:29 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: HW 4.15 (c)
- Replies: 1
- Views: 206
Re: HW 4.15 (c)
Hey,
When writing the VSEPR formula, you should count the number of bonding densities or regions instead of counting the number of bonded atoms. The double bond between the C and O is considered to be one bonding region, meaning you will have an AX3 formula.
When writing the VSEPR formula, you should count the number of bonding densities or regions instead of counting the number of bonded atoms. The double bond between the C and O is considered to be one bonding region, meaning you will have an AX3 formula.
- Wed Nov 22, 2017 1:00 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 4.81
- Replies: 1
- Views: 266
Re: 4.81
Hey,
Once you draw out the lewis structure, the question is asking you for the hybridization of the B and N atoms. Since there are three bonding regions around both the B and N atoms, the orbitals will be sp^2 hybridized.
Once you draw out the lewis structure, the question is asking you for the hybridization of the B and N atoms. Since there are three bonding regions around both the B and N atoms, the orbitals will be sp^2 hybridized.
- Sat Nov 18, 2017 11:53 am
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Methanol polarity
- Replies: 2
- Views: 520
Re: Methanol polarity
Hey, In methanol, the CH3 group is attached to OH. Oxygen is more electronegative than the rest of the molecule, so there is a slight negative charge on the OH end of the molecule, and this leaves a slight positive charge on the CH3 group. Hydrocarbons are always non-polar, so there is nothing to ca...
- Sat Nov 18, 2017 11:44 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Radicals and Shape
- Replies: 2
- Views: 761
Re: Radicals and Shape
Hey, Free radicals affect the shape of other molecules that they interact with. The atom with a free radical does not have a full octet set, so it tries to fix this by stealing an electron from another molecule. When they steal an electron from another molecule, particularly an organic molecule, it ...
- Sat Nov 18, 2017 11:38 am
- Forum: Octet Exceptions
- Topic: Radical Question
- Replies: 4
- Views: 746
Re: Radical Question
Hey, The radical electron, or the lone electron, would be placed on the least electronegative atom because the more electronegative atoms in the molecule would rather have the full set. The more electronegative an atom is, the more it pulls away other electrons, so the least electronegative atom is ...
- Mon Nov 13, 2017 6:10 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 4.5 Shape
- Replies: 2
- Views: 456
Re: 4.5 Shape
Hey, There is a trigonal planar arrangement because there are actually 3 electron densities. There is a lone pair on the chlorine and two oxygens double bonded to it. Because there are 3 electron densities, and one of them is a lone pair, this makes the shape of the molecule angular. The electron pa...
- Fri Nov 03, 2017 8:43 am
- Forum: Lewis Structures
- Topic: 3.61
- Replies: 2
- Views: 346
Re: 3.61
Hey, In response to question 61 part b, you have to first count the total number of electrons in the molecule. Once we find that, and then draw the basic lewis structure, we can start adding on electrons to the chlorines. All the chlorines will come out with full valence shells, but we are still lef...
- Thu Nov 02, 2017 4:56 pm
- Forum: Ionic & Covalent Bonds
- Topic: How to easily tell between Ionic and Covalent bonds?
- Replies: 8
- Views: 1515
Re: How to easily tell between Ionic and Covalent bonds?
Hey, Yes, the method you use is a simple way to figure out if a bond is ionic or not. This trend is because ionic bonds form when there is a nonmetal and a metal reacting, and both of these are usually on opposite sides of the periodic table. Covalent bonds form between nonmetallic elements, which a...
- Fri Oct 27, 2017 5:09 pm
- Forum: Trends in The Periodic Table
- Topic: atomic size
- Replies: 4
- Views: 683
Re: atomic size
Hey,
An atom thats in lower right group would still be smaller than a lot of the atoms on the left that are higher up. This is easily comparable through a chart that shows you the relative size of each atom in the periodic table.
For example, Radon is still slightly smaller than Sodium.
An atom thats in lower right group would still be smaller than a lot of the atoms on the left that are higher up. This is easily comparable through a chart that shows you the relative size of each atom in the periodic table.
For example, Radon is still slightly smaller than Sodium.
- Fri Oct 27, 2017 4:27 pm
- Forum: Ionic & Covalent Bonds
- Topic: Cations and Anions [ENDORSED]
- Replies: 12
- Views: 2104
Re: Cations and Anions [ENDORSED]
A cation is an ion that has a positive charge because it gave away its electron, such as Na+. An anion is an ion that has a negative charge because it took an electron, such as Cl-. It might be helpful to memorize a mnemonic, like that an anion is (A) (N)egative (ION), and a CATion is PAWSitive.
- Sat Oct 21, 2017 8:29 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Orbitals and Electron Configuration
- Replies: 3
- Views: 386
Re: Orbitals and Electron Configuration
Hey, The reason we show the 2px^1 and 2py^1 instead of just writing 2p^2 is to outline the Aufbau principle. The Aufbau principle states that electrons will occupy the lowest energy levels. Instead of having those two electrons forced into 2px, the electrons would rather occupy one orbital each, all...
- Sat Oct 21, 2017 8:18 am
- Forum: Properties of Electrons
- Topic: Electron Configuration [ENDORSED]
- Replies: 3
- Views: 601
Re: Electron Configuration [ENDORSED]
Hey,
I'm pretty sure Dr. Lavelle stated in class that the concept is good to know and understand, but you wouldn't get points marked off if you chose to not label px or py.
I'm pretty sure Dr. Lavelle stated in class that the concept is good to know and understand, but you wouldn't get points marked off if you chose to not label px or py.
- Fri Oct 13, 2017 7:58 pm
- Forum: Properties of Light
- Topic: 1.11 [ENDORSED]
- Replies: 3
- Views: 596
Re: 1.11 [ENDORSED]
Hey, The series consist of the lines that are grouped together based on their lower energy principal quantum level. Each series has the same principal quantum number associated with them, so the Lyman series has a lower energy level of n=1. The series are just a convenient way to associate the lines...
- Fri Oct 13, 2017 7:35 pm
- Forum: Photoelectric Effect
- Topic: Confused about photoelectric effect! [ENDORSED]
- Replies: 6
- Views: 939
Re: Confused about photoelectric effect! [ENDORSED]
Hey, Basically, the photoelectric effect states that electrons will be ejected from the surface of a metal when exposed to electromagnetic radiation. However, in order for an electron to be ejected, the radiation has to have a frequency above the threshold value for that particular metal. The photoe...
- Thu Oct 05, 2017 9:43 pm
- Forum: Limiting Reactant Calculations
- Topic: M5
- Replies: 2
- Views: 717
Re: M5
Hey, In part A, we found that the limiting reactant was ClO2. This means that the amount of product that is created is based on the amount of moles of ClO2 present. We have 12 moles of ClO2, and the ratio of ClO2 to Br2 is 6:1. So that means for every 6 moles of ClO2 there is, 1 mole of Br2 can be m...
- Thu Oct 05, 2017 6:28 pm
- Forum: Limiting Reactant Calculations
- Topic: M11
- Replies: 2
- Views: 500
Re: M11
Hey, You have all your numbers correct, but I think the reason you aren't getting the right answer is because the ratio of the oxygen to the product, phosphorus(V) oxide is 2:1. This means that when calculating for the final product, you have to divide the moles of oxygen remaining by two. This shou...