Search found 61 matches

by ThomasLai1D
Thu Mar 14, 2019 1:30 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Nernst Equation Problem.
Replies: 1
Views: 50

Nernst Equation Problem.

In question 4b (see attached photo), the solution solves the problem using n in the Nernst Equation equal to one. However, in the previous part, it was determined that two electrons were transferred in the cell. How do we know that n is supposed to be one in 4b?
by ThomasLai1D
Mon Mar 11, 2019 8:37 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: 7E3
Replies: 1
Views: 29

Re: 7E3

I don't think it matters. As long as you put everything where it should be, the ratios will work out to the right answer.
by ThomasLai1D
Mon Mar 11, 2019 8:24 pm
Forum: First Order Reactions
Topic: Units of Rate Constant
Replies: 5
Views: 93

Re: Units of Rate Constant

It depends upon the context of the question and what's given. If you can solve it with the time units as minutes, then you don't necessarily have to convert it to seconds. What is important is that all units are the same on both sides of the equation and that they cancel.
by ThomasLai1D
Mon Mar 11, 2019 8:22 pm
Forum: First Order Reactions
Topic: Integrated rate law confusion
Replies: 5
Views: 102

Re: Integrated rate law confusion

The difference is the placement of the minus signs. Both will net the same answer, but because of the treatment of the minus signs and properties of natural logs, the equations look different.
by ThomasLai1D
Thu Mar 07, 2019 7:39 am
Forum: General Rate Laws
Topic: 7A.15
Replies: 1
Views: 42

Re: 7A.15

Focus on only one reactant at a time. Usually there will only be one reactant whose initial concentrations have been changed, and by dividing the rates of that reaction and setting that equal to the division of the modified concentration of reactants to an unknown power and solving for the unknown p...
by ThomasLai1D
Thu Mar 07, 2019 7:35 am
Forum: First Order Reactions
Topic: 7B.5 7th edition
Replies: 1
Views: 56

Re: 7B.5 7th edition

For part a, you would use the equation [A] = [A]initial e^-kt, and plug in 1/2 [A] initial for [A].
For part c, you would use the same equation, but solve for t by plugging in the values for initial and final concentration into [A]initial and [A] respectively.
by ThomasLai1D
Thu Mar 07, 2019 7:30 am
Forum: First Order Reactions
Topic: 7B.7
Replies: 1
Views: 47

Re: 7B.7

For part A, because 1/8 = (1/2)^3, you can assume that substance A divided in half three times. Therefore, you can just multiply 355s by a factor of 3. Similarly, in part B, because 1/4 = (1/2)^2, substance A divided in half two times. Therefore you can multiply 355s by a factor of 3. For C and D, y...
by ThomasLai1D
Tue Feb 26, 2019 5:12 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Reducing Agent/ Oxidizing Agent Strength in Metals
Replies: 3
Views: 76

Re: Reducing Agent/ Oxidizing Agent Strength in Metals

The more negative the standard reduction potential, the stronger the metal is as a reducing agent.
by ThomasLai1D
Tue Feb 26, 2019 5:01 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Contradictory Ecell equations
Replies: 2
Views: 50

Contradictory Ecell equations

When is the equation Ecell=Eright-Eleft used instead of Ecell=Ecathode-Eanode. It seems these equations contradict themselves if the anode is always on the right and the cathode is always on the left.
by ThomasLai1D
Tue Feb 26, 2019 4:45 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L. 9
Replies: 1
Views: 63

6L. 9

Write balanced half reactions for the redox reaction of an acidified solution of potassium permanganate and iron (II) chloride. Write the balanced equation for the cell reaction and devise a galvanic cell to study the reaction. Where do you start in determining the half reactions? Is it just based u...
by ThomasLai1D
Tue Feb 26, 2019 4:20 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6L.7b
Replies: 1
Views: 57

6L.7b

How do you determine the half reactions of a reaction involving water? The question gives the reaction H+(aq) + OH-(aq) ---> H2O(l) and asks for balanced half reactions and the cell diagram.

Thanks
by ThomasLai1D
Thu Feb 21, 2019 2:09 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: equation explanation
Replies: 2
Views: 37

Re: equation explanation

RTlnQ is derived from the deltaG^0 = -RTlnK. K represents the system in the standard state at equilibrium, so if you want to find deltaG not at equilibrium, you must use Q in the equation and subtract it from -RTlnK.
by ThomasLai1D
Thu Feb 21, 2019 2:02 pm
Forum: Balancing Redox Reactions
Topic: Oxidation States
Replies: 10
Views: 147

Re: Oxidation States

I think Dr. Lavelle said that we would not need to memorize oxidation states of elements, but that they would be easy to figure out from the reaction and the compound it is in.
by ThomasLai1D
Thu Feb 21, 2019 2:01 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Flow of electrons
Replies: 4
Views: 84

Re: Flow of electrons

Yes, however Dr. Lavelle mentioned the process of charging a battery, which in response to an external electric current, the electrons flow in the reverse, towards the anode.
by ThomasLai1D
Tue Feb 12, 2019 5:19 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 2nd law of Thermodynamics
Replies: 1
Views: 82

Re: 2nd law of Thermodynamics

q refers to the heat you are adding to or taking out of the system. A positive q means you are adding heat to the system, which increases energy. I think you are confusing this with when delta H is negative, which is not the same as removing heat from the system. When delta H is given, spontaneity i...
by ThomasLai1D
Tue Feb 12, 2019 5:04 pm
Forum: Calculating Work of Expansion
Topic: 6th edition self test 8.1A
Replies: 1
Views: 44

Re: 6th edition self test 8.1A

Since w=-PdeltaV, and pressure is given, you need to find delta V. Given the densities of water, we can calculate the volume of water by multiplying 100g x 1 cm3/1g and the volume of ice by multiplying 100g x 1 cm3/0.92g. From there, you should have your change in volume (volume ice - volume water) ...
by ThomasLai1D
Tue Feb 12, 2019 4:54 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Test 1 Question
Replies: 3
Views: 84

Re: Test 1 Question

Because delta H is positive at equilibrium, the reaction is endothermic. When temperature is increased on an endothermic reaction, according to Le Chatelier's principle, the reaction will favor the products.
by ThomasLai1D
Thu Feb 07, 2019 10:31 am
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: 4G.5 7th edition
Replies: 1
Views: 55

Re: 4G.5 7th edition

Your explanation is correct. Because the atoms in the trans compound must be in opposite orientations, there are only three different configurations where atoms can be directly opposite of each other, while the atoms in the cis compound can be in adjacent configurations, allowing more permutations a...
by ThomasLai1D
Thu Feb 07, 2019 10:28 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 7th edition 4F.9
Replies: 1
Views: 44

Re: 7th edition 4F.9

From the equation delta S= nRln(V2/V1), using Boyle's law P1V1=P2V2, derive the equation delta S =nRln(P1/P2), and plug the values in.

1.50mol x 8.314 J/mol/K x ln(15.0atm/0.500atm)=42.4 J/K
by ThomasLai1D
Thu Feb 07, 2019 10:24 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy
Replies: 1
Views: 40

Re: Enthalpy

delta H in Hess's law used are the heats of component reactions while delta H values in standard enthalpy of formation are the enthalpies of formation of the reactants and products.
by ThomasLai1D
Thu Jan 31, 2019 3:27 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: HW problem 4B. 13
Replies: 5
Views: 270

Re: HW problem 4B. 13

Because PV=nRT, you can calculate the amount of moles by plugging in the given information into the equation n=PV/RT. Using the value you found for n, you should be able to use the w=-nRTln(V1/V2) formula.
by ThomasLai1D
Thu Jan 31, 2019 2:56 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: HW 8.11 part b
Replies: 2
Views: 66

Re: HW 8.11 part b

The second experiment does more work because all energy is going into moving the piston (none is going into increasing the temperature because the experiment is an isothermic expansion). You can determine this quantitatively by calculating the work done by the first experiment using the equation w=P...
by ThomasLai1D
Sun Jan 27, 2019 3:54 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 6th edition 8.11
Replies: 2
Views: 67

Re: 6th edition 8.11

Recall that P=nRT/V. A reversible isothermic expansion means that temperature remains constant, so for pressure to change, volume must change.
This requires the equation w= -nRT ln (V2/V1).
by ThomasLai1D
Tue Jan 22, 2019 5:55 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Outline 1, last bullet
Replies: 2
Views: 59

Re: Outline 1, last bullet

-Adding reagents, Q will become smaller than K, causing more product to be created to compensate and increase Q such that Q=K at equilibrium -Removing reagent, Q will become large than K, causing reaction to favor reactants to compensate and decrease Q such that Q=K at equilibrium -Compressing gas m...
by ThomasLai1D
Tue Jan 22, 2019 5:42 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: determining x for weak acids and bases
Replies: 6
Views: 273

Re: determining x for weak acids and bases

I would assume so, because any K < 10^-3 is an indication that such little product is formed to have any significant impact on the final concentrations.
by ThomasLai1D
Tue Jan 22, 2019 5:40 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: pH vs pKa
Replies: 1
Views: 200

Re: pH vs pKa

pH is a factor of pKa. pKa is an arbitrary number used to determine where an acid reaction sits in equilibrium and is calculated with -log([H3O+][A-]/[HA]) whereas pH is used to determine the acidity of a solution based upon number of H3O+, calculated with -log[H3O+]. The relationship between pKb an...
by ThomasLai1D
Tue Jan 15, 2019 9:48 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: help with equilibrium equations
Replies: 1
Views: 37

Re: help with equilibrium equations

By definition, ClO2- is the conjugate base of HClO2. Bases react with compounds as proton acceptors, so in aqueous solution, ClO2- accepts a proton from H2O, forming a conjugate acid, HClO2, and OH-.
by ThomasLai1D
Tue Jan 15, 2019 9:41 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Concentration of H30+ = A- ??
Replies: 1
Views: 44

Re: Concentration of H30+ = A- ??

In order for one H3O+ to be formed, one A- must also be formed (in an HA acid), because the proton (H+) transfers from the acid to the H2O. It is a one to one mole ratio. Therefore, they must be equivalent.
by ThomasLai1D
Tue Jan 15, 2019 9:36 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Kw
Replies: 4
Views: 79

Re: Kw

Because the conversion of H2O,in pure water, into H3O+ and OH- is an autoprolytic reaction, this necessitates that the concentrations of OH- and H3O remain the same. In other words, for H3O+ to be formed, a proton must dissociate from an H2O molecule and bind to another H2O molecule, thus creating O...
by ThomasLai1D
Thu Jan 10, 2019 9:28 am
Forum: Ideal Gases
Topic: Algebra
Replies: 5
Views: 78

Re: Algebra

By dividing P=(conc)RT by RT, the resulting equation would be P/RT=conc. I do not think P is necessarily restricted to partial pressures, just that it represents pressure in the formula.
by ThomasLai1D
Thu Jan 10, 2019 9:22 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 7th edition 5H.1
Replies: 1
Views: 18

Re: 7th edition 5H.1

Recall that when you stoichiometrically multiply the reaction by a magnitude of two, you square the Keq of the original reaction. Similarly, if you divide the reaction by a magnitude of two (1/2N2+3/2H2---->NH3), you would take the square root of the original Keq (Keq^1/2).
by ThomasLai1D
Thu Jan 10, 2019 9:19 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: PArtial pressure= concentration?
Replies: 2
Views: 35

Re: PArtial pressure= concentration?

While not equivalent, both partial pressures and concentration are used in calculating K. The equation P=(conc)(RT) allows you to convert between the two. With respect to solving for the partial pressure, you would use the same algebra to calculate the pressures and concentrations (For partial press...
by ThomasLai1D
Wed Dec 05, 2018 7:08 pm
Forum: Naming
Topic: so4 2- ligand name
Replies: 1
Views: 123

Re: so4 2- ligand name

Due to the lone pairs on oxygen, oxygen will always bond with the metal in SO4 2- to create a ligand. This is the same in H2O, which binds to the metal with O in a ligand, and is written as OH2.
by ThomasLai1D
Wed Dec 05, 2018 6:59 pm
Forum: Conjugate Acids & Bases
Topic: Strong acids and bases to memorize
Replies: 9
Views: 425

Re: Strong acids and bases to memorize

There are only 7 common strong acids:
HCl
HNO3
H2SO4
HBr
HI
HClO4
HClO3

Strong Bases tend to be composed of group 1 and 2 metals.
by ThomasLai1D
Tue Dec 04, 2018 8:33 pm
Forum: Trends in The Periodic Table
Topic: Group 5 Transition Metals
Replies: 2
Views: 201

Re: Group 5 Transition Metals

Group 5 transition metals refers to the column with vanadium.
by ThomasLai1D
Tue Nov 27, 2018 4:49 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: hydrogen bonding
Replies: 1
Views: 37

Re: hydrogen bonding

N, O, and F are the most electronegative, causing the greatest polarization within the molecule. This polarization is so strong that the positive dipole in the H atom will interact with the lone pairs of N, O, or F atom of another molecule.
by ThomasLai1D
Tue Nov 27, 2018 4:45 pm
Forum: Hybridization
Topic: Lone pairs
Replies: 2
Views: 70

Re: Lone pairs

The hybridization of lone pairs is the equivalent to the hybridization of the same central atom with the lone pair substituted for the bonding pair.
by ThomasLai1D
Tue Nov 27, 2018 4:41 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Polar vs. Nonpolar? (Q25, C4, 6th edition)
Replies: 1
Views: 46

Re: Polar vs. Nonpolar? (Q25, C4, 6th edition)

The direction the dipole moments point to are towards the most electronegative atom in the molecule Because Cl is the most electronegative atom, the dipole moment is more negative towards the Cl (Cl attracts the electrons), and more positive towards the H atoms. Therefore, the molecule is polar.
by ThomasLai1D
Tue Nov 27, 2018 4:37 pm
Forum: Sigma & Pi Bonds
Topic: Pi bond concepts
Replies: 4
Views: 135

Re: Pi bond concepts

The shape of a single p-orbital has the shape of a dumbbell. In a pi bond, one "head" of the dumbbell extends upwards, and the other extends downwards. The upward head interacts with the upward head of the other atom and the downward head corresponds with the downward head of the other ato...
by ThomasLai1D
Tue Nov 27, 2018 4:33 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Sigma and Pi bonds
Replies: 2
Views: 40

Re: Sigma and Pi bonds

Sigma bonds are overlapping s-orbitals of two neighboring atoms that create a tubelike shape. Pi bonds consist of two unbonded p-orbitals of neighboring atoms. Atoms tend to bond first with sigma bonds, so single bonds will always be sigma bonds. A double bond consists of a sigma and a pi bond, and ...
by ThomasLai1D
Fri Nov 16, 2018 9:19 am
Forum: Lewis Structures
Topic: Electron configuration
Replies: 3
Views: 93

Re: Electron configuration

I think you are confusing cations with anions. Cations are ions that lose electrons (yielding a + charge), whereas anions are ions that gain electrons (yielding a - charge). Therefore, germanium +1 would have an electron configuration of [Ar] 3d10 4s2 4p1, not [Ar] 3d10 4s2 4p3.
by ThomasLai1D
Fri Nov 16, 2018 9:09 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Sulfite Ion and its Bond Angle
Replies: 3
Views: 54

Re: Sulfite Ion and its Bond Angle

I think Dr. Lavelle said that for the specific sulfite ion, the bond angle is 106 degrees, however it is less important that you know the specific angle, but rather that you know the bond angle is slightly less than 109.5 degrees. The reason the sulfite ion has a slightly smaller bond angle than 109...
by ThomasLai1D
Fri Nov 16, 2018 9:01 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Bond strength
Replies: 2
Views: 53

Re: Bond strength

Lone pairs are held closer to the nucleus of an atom than a bonding pair. Therefore, because the area in which they exist is the smaller area closer to the nucleus, lone-lone pairs will be closer to each other, causing greater repulsions. The same reasoning applies to lone-bonding pair, however, the...
by ThomasLai1D
Thu Nov 08, 2018 6:40 pm
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Polarizability
Replies: 3
Views: 59

Re: Polarizability

The greater the number of electrons, the less control the positive nuclear charge exerts on the electrons due to shielding, thus causing the larger electron cloud to distort more and increasing polarizability.
by ThomasLai1D
Sun Nov 04, 2018 9:05 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Heisenberg and Wavelength
Replies: 11
Views: 522

Heisenberg and Wavelength

Hi, I am just going over the last test, and one of the conceptual questions asked "How does the change in uncertainty in momentum affect the uncertainty in its wavelength?" I answered that there would be no impact because the Heisenberg principle does not relate momentum and wavelength, bu...
by ThomasLai1D
Wed Oct 31, 2018 9:40 pm
Forum: Lewis Structures
Topic: Double bonds [ENDORSED]
Replies: 3
Views: 103

Re: Double bonds [ENDORSED]

To determine the kind of bond, you must first determine the amount of valence electrons available for use in bonding. Once you have determined that, you need to find a bonding configuration that utilizes all available electrons. If necessary, you need to use the formal charge (LC=V-(L+B/2)) to deter...
by ThomasLai1D
Wed Oct 31, 2018 9:36 pm
Forum: Bond Lengths & Energies
Topic: Triple Bonds Shorter than Double Bonds
Replies: 4
Views: 191

Re: Triple Bonds Shorter than Double Bonds

The additional electrons involved in a triple bond exert greater attractive forces on the nuclei, thereby shortening the length of the bond.
by ThomasLai1D
Wed Oct 31, 2018 3:56 pm
Forum: Octet Exceptions
Topic: Valence Electrons for Exceptions to Octets
Replies: 2
Views: 99

Re: Valence Electrons for Exceptions to Octets

I think that's one of the exceptions to the octet rule. He was referring to that fact that Boron is stable with only 6 valence electrons, and therefore only shares three bonds with fluoride in BF3 and has no lone pairs (a lone pair would complete the octet rule, if it applied to Boron).
by ThomasLai1D
Tue Oct 23, 2018 2:13 pm
Forum: Quantum Numbers and The H-Atom
Topic: Quantum Numbers
Replies: 3
Views: 62

Re: Quantum Numbers

Hi, First, you need to determine the energy level, or n. Say the orbital is on the 3rd energy level, so n=3. If n=3, you can determine l, or the angular momentum function (the name is largely irrelevant for finding the number). l is determined by n-1, so the maximum value of l can be n=2, with the p...
by ThomasLai1D
Sun Oct 21, 2018 5:54 pm
Forum: Properties of Light
Topic: Atomic Spectrum Module #27
Replies: 1
Views: 68

Re: Atomic Spectrum Module #27

I believe what's given is the number of wavelengths per meter (1,650,763.73/m). So if you take the inverse of the number, you should be able to find the wavelength (1/1,650,763.73*m^-1 = 6.05x10^-7 m)
by ThomasLai1D
Sun Oct 21, 2018 5:48 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Quantum numbers
Replies: 5
Views: 100

Re: Quantum numbers

The electron is located in 4f.
by ThomasLai1D
Sun Oct 21, 2018 5:45 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Energy levels of orbitals?
Replies: 10
Views: 144

Re: Energy levels of orbitals?

The energy level goes s<p<d<f. This is because s-orbitals are closer to the nucleus and therefore experience greater pull from the positive nucleus. Thought about in a different way, f-orbitals need to be of higher energy to remain farther outside the nucleus and resist the inward pull of the nucleus.
by ThomasLai1D
Fri Oct 19, 2018 3:37 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: energy in shells closer to the nucleus
Replies: 4
Views: 96

Re: energy in shells closer to the nucleus

I think of it this way: An negatively charged electron farther from the nucleus must be of higher energy in order to resist the pull of the positively charged nucleus. Therefore, an electron closer to the nucleus is of lower energy.
by ThomasLai1D
Mon Oct 15, 2018 8:16 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Building Up Principle
Replies: 1
Views: 40

Building Up Principle

Hi,

Would someone explain how to do the building up principle for the d-block elements? I'm really confused and I can't seem to recognize a pattern.

Thanks
by ThomasLai1D
Mon Oct 15, 2018 8:01 pm
Forum: DeBroglie Equation
Topic: Question 1.37 sixth edition
Replies: 1
Views: 57

Re: Question 1.37 sixth edition

Hi, Take the absolute the difference of the two wavelengths, so subtract the wavelength of the neutron from the wavelength of the proton, and divide that number by the wavelength of the neutron. If I recall, the difference between the two wavelengths was insignificant, so the percentage should be ze...
by ThomasLai1D
Mon Oct 15, 2018 7:57 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Textbook pages
Replies: 3
Views: 113

Re: Textbook pages

In the 7th edition, pages 19-21.
by ThomasLai1D
Mon Oct 08, 2018 11:54 am
Forum: Properties of Light
Topic: Intensity of Light and Wavelength
Replies: 3
Views: 41

Re: Intensity of Light and Wavelength

In the wave model of light, increased intensity relates to increased amplitude of the wave. However, in the photoelectric model of light, increased intensity of light means an increase in the number of photons. So yes, increasing/decreasing amplitude of wavelength results in the increase/decrease of...
by ThomasLai1D
Mon Oct 08, 2018 11:49 am
Forum: Properties of Light
Topic: Problem 1b 7b
Replies: 1
Views: 50

Problem 1b 7b

Hi,
I'm having difficulty converting mass and wavelength into energy. The problem states that a sodium vapor lamp emits yellow light at a wavelength 589 nm. How much energy is emitted by 5.00 mg of sodium atoms emitting light at this wavelength?

Thanks
by ThomasLai1D
Sun Oct 07, 2018 3:39 pm
Forum: Properties of Light
Topic: Problem #9 6th Edition
Replies: 1
Views: 36

Re: Problem #9 6th Edition

You can determine the wave's placement on the electromagnetic spectrum based on the size of the wavelength. For example, a wavelength of 3.4x10-7m, therefore corresponding to a wavelength in the ultraviolet range (420-100 nm), or a sunburn.
by ThomasLai1D
Tue Oct 02, 2018 8:48 pm
Forum: Balancing Chemical Reactions
Topic: Problem H. 11
Replies: 3
Views: 71

Problem H. 11

Hi,

I am having difficulty balancing the equation Fe3O4 + CO ----> Fe + CO2 . No matter what I do, I cannot seem to balance both the Carbons or the Oxygens in the equation.

Thanks
by ThomasLai1D
Tue Oct 02, 2018 8:38 pm
Forum: Empirical & Molecular Formulas
Topic: Fundamental L.39
Replies: 6
Views: 117

Re: Fundamental L.39

Hi, The roman numerals following the metals name are an indication the number of positive cations of that metal. Some transition metals may have multiple cation forms, therefore the number of positive cations of the ionic compound are notated with roman numerals. So copper (II) oxide would have two ...

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