Search found 61 matches
- Sun Mar 10, 2019 7:00 pm
- Forum: First Order Reactions
- Topic: first order reactions examples
- Replies: 3
- Views: 407
Re: first order reactions examples
SO2Cl2(g) → SO2(g) + Cl2(g) is an example of a first order reaction
- Sun Mar 10, 2019 6:51 pm
- Forum: Second Order Reactions
- Topic: Second order reactions
- Replies: 2
- Views: 331
Re: Second order reactions
An example of a 2nd order reaction is
O + O3 → O2+ O2
O + O3 → O2+ O2
- Sun Mar 10, 2019 6:30 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 7A.15
- Replies: 4
- Views: 452
Re: 7A.15
Technically, yes, because when you find the order of A and B by dividing the rate laws, the exponent (z) of [C] would be 0 because it is a zero-order reaction, which would make the value of [C]^z = 1 and thus irrelevant to finding the orders of A and B.
- Tue Mar 05, 2019 7:41 pm
- Forum: Zero Order Reactions
- Topic: 7A 15
- Replies: 4
- Views: 526
Re: 7A 15
You can tell that molecule C is a zero order reaction because when you divide the equations (you select the 2 equations where the concentration of A and B are the same and only C is different, which would be in experiment 1 and 4), you get 2/2 = (7/4)^z Since 2/2 is 1, Z has to be 0 in order for (7/...
- Tue Mar 05, 2019 7:34 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Finding Order for Each Reactant
- Replies: 4
- Views: 549
Re: Finding Order for Each Reactant
It doesn't matter what order you go in, you just have to make sure the ratio of the rates is in the same 'order' as the ratio of the concentrations. For example, rate 1/rate 2 = k [A]1 / k[A]2.
- Tue Mar 05, 2019 7:27 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Which experimented do you use to find rate constant?
- Replies: 2
- Views: 408
Re: Which experimented do you use to find rate constant?
For 7A.17, they just happened to choose the first experiment and plugged in the values to get the rate constant k. All the experiments should get you approximately the same k value.
- Sat Mar 02, 2019 11:23 pm
- Forum: Zero Order Reactions
- Topic: Units
- Replies: 3
- Views: 369
Re: Units
IsabelLight2H wrote:Are there units for K?
For a zero-order reaction, the units for k will be units of concentration/time (M/s).
- Sat Mar 02, 2019 10:56 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: E vs E naught
- Replies: 5
- Views: 712
Re: E vs E naught
E naught is under standard conditions so it is a constant for a reaction, while E can vary under different conditions.
- Sat Mar 02, 2019 10:53 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Reaction order
- Replies: 3
- Views: 452
Re: Reaction order
The reaction order is the relationship between the concentrations of species and the rate of a reaction, so a zero-order reaction would be one where the rate is independent of the concentration of the reactants, while a first-order reaction is one where the rate of reaction is directly proportional ...
- Thu Feb 21, 2019 11:10 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy vs Entropy
- Replies: 4
- Views: 800
Re: Gibbs Free Energy vs Entropy
Gibbs free energy (G) is the measure of the total amount of energy available to do work while entropy (S) is the measure of the disorder of the system. Gibbs free energy is calculated using enthalpy (H), temperature (T), and entropy (S) with the equation delta G = delta H - T(delta S).
- Tue Feb 19, 2019 8:08 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Indications
- Replies: 3
- Views: 390
Re: Indications
Since the equation to calculate for delta G is delta G = delta H - T(delta S), a small/negative delta H (enthalpy), large/positive T (temperature), and a large/positive S (entropy) would make for a very small/negative delta G.
The opposites would make delta G larger (more nonspontaneous).
The opposites would make delta G larger (more nonspontaneous).
- Tue Feb 19, 2019 8:06 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Factors
- Replies: 2
- Views: 272
Re: Factors
Since delta G is calculated for by the equation delta G = delta H - T(delta S), the factors that affect delta G are the enthalpy of reaction, the temperature that the reaction takes place at, and the entropy. A negative enthalpy, high temperature, and high entropy makes for a spontaneous reaction (n...
- Thu Feb 14, 2019 6:16 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Reversible vs. Irreversible
- Replies: 5
- Views: 1051
Re: Reversible vs. Irreversible
A reversible reaction would also have more work done and is isothermal (temperature remains constant.)
An irreversible reaction would have less work done and is isobaric (pressure remains constant).
An irreversible reaction would have less work done and is isobaric (pressure remains constant).
- Thu Feb 14, 2019 6:14 pm
- Forum: Phase Changes & Related Calculations
- Topic: Phase Change Diagram of Water
- Replies: 6
- Views: 648
Re: Phase Change Diagram of Water
The heat of vaporization is a lot larger than the heat of fusion (40 kJ vs 334 J) so when steam condenses back into water, it releases all 40 kJ into the skin, which creates more severe burns.
- Thu Feb 14, 2019 6:12 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Stability
- Replies: 3
- Views: 367
Re: Stability
If the reaction is exothermic, heat is a product and increasing the temperature would cause the reaction to shift toward the reactant side, making reactants more stable and products less stable. If the reaction is endothermic, heat is a reactant and increasing the temperature would cause the reactio...
- Tue Feb 05, 2019 2:53 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: 4I.3
- Replies: 5
- Views: 10595
Re: 4I.3
In part (a), the equation you use to calculate the standard enthalpy of vaporization is delta S (vap) = delta H (vap) / T(boiling point in K). So you get delta H (vap) = (85 J/K)(80 + 273 K) = 30,005 J or 30.0 kJ. For part (b), the standard entropy change of the system is calculated by multiplying t...
- Tue Feb 05, 2019 2:27 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: ΔH and ΔU
- Replies: 6
- Views: 880
Re: ΔH and ΔU
The degree symbol means that the reaction is happening in the standard state, which is at 1atm and 298.15K (or 25°C).
- Tue Feb 05, 2019 11:43 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: U
- Replies: 7
- Views: 655
Re: U
U is a state property/function because it does not depend on the path taken to get to its present state and also describes the equilibrium state of the system. That is why you can add or subtract separate internal energies to calculate the total internal energy.
- Mon Jan 28, 2019 9:39 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Delta U=0 for isothermal compression?
- Replies: 1
- Views: 242
Re: Delta U=0 for isothermal compression?
delta U is 0 for it to be isothermal, but that doesn't mean work also has to be 0. Since delta U = q + w, that just means that q + w has to equal 0. Work has to cancel out the heat.
- Mon Jan 28, 2019 9:22 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond Enthalpies
- Replies: 7
- Views: 690
Re: Bond Enthalpies
Bond enthalpies are the least accurate way of measuring reaction enthalpies because they are made of multiple averages and thus an overall average.
- Mon Jan 28, 2019 9:20 pm
- Forum: Calculating Work of Expansion
- Topic: value of R?
- Replies: 4
- Views: 500
Re: value of R?
R is 0.08206 in the PV=nRT formula because its units are L atm K-1 mol-1 in that case.
R is 8.314 in the w=-nRTlnV1/V2 formula because its units are J K-1 mol-1 in that case.
R is 8.314 in the w=-nRTlnV1/V2 formula because its units are J K-1 mol-1 in that case.
- Sun Jan 20, 2019 11:00 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: The Conjugate Seesaw
- Replies: 13
- Views: 998
Re: The Conjugate Seesaw
An example of the conjugate seesaw could be NaOH.
NaOH is a strong base, but its conjugate acid, Na+, is so weak that it's basically a neutral cation and won't affect the pH of the solution.
NaOH is a strong base, but its conjugate acid, Na+, is so weak that it's basically a neutral cation and won't affect the pH of the solution.
- Sun Jan 20, 2019 10:57 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Acids and Bases
- Replies: 3
- Views: 374
Re: Acids and Bases
Acids generally have an H+ in their chemical formula (e.g. HCl, HNO3, HClO4, NH4+ etc.)
Bases usually have an OH- in their chemical formula (e.g. NaOH, KOH etc.)
Bases usually have an OH- in their chemical formula (e.g. NaOH, KOH etc.)
- Sun Jan 20, 2019 10:49 pm
- Forum: Calculating the pH of Salt Solutions
- Topic: Identifying which part of the salt does not affect PH
- Replies: 1
- Views: 382
Re: Identifying which part of the salt does not affect PH
The anions of strong acids, which include Cl-, Br-, I-, NO3-, ClO4-, are such weak bases that they have no significant effect on the pH of a solution so they are basically neutral. The cations of Group ½ elements (also those with charge +1 from other groups) are such weak acids that the hydrated ion...
- Tue Jan 15, 2019 9:46 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE table
- Replies: 3
- Views: 444
Re: ICE table
Solids and liquids are excluded from K calculations because their concentrations/densities remain constant (1) throughout the reaction. On the other hand, the concentrations/densities of gases and aqueous solutions are changeable and thus we need the ICE chart to solve for concentrations at equilibr...
- Tue Jan 15, 2019 9:40 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th edition Question 11.45
- Replies: 2
- Views: 173
Re: 6th edition Question 11.45
We can deduce that Cl2 is more thermodynamically stable than F2 because there is more Cl2 at equilibrium than F2 with the same pre-conditions.
- Tue Jan 15, 2019 9:32 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Temperature Relation
- Replies: 3
- Views: 329
Re: Temperature Relation
Temperature changes the equilibrium constant because temperature permanently affects the rate constants of the forward and backward reactions, which thus, in turn, affects Kc. On the other hand, if pressure or concentration of reactants/products are changed, the reaction will shift towards equilibri...
- Wed Jan 09, 2019 10:40 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: "Shifting"
- Replies: 15
- Views: 1779
Re: "Shifting"
I think he meant to differentiate between "sitting" and "shifting". When equilibrium 'sits' to the right, reactants are favored, and when it 'sits' on the left, products are favored. On the other hand, "shifting" means that the reaction is moving towards equilibrium (wh...
- Wed Jan 09, 2019 10:34 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium Expression K
- Replies: 3
- Views: 197
Re: Equilibrium Expression K
The equation definitely has to be balanced before you plug things in to solve K because the stoichiometric coefficients matter as the exponents.
- Wed Jan 09, 2019 10:33 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Chemical Equilibrium Part 2 Question 24
- Replies: 2
- Views: 243
Re: Chemical Equilibrium Part 2 Question 24
You can use the equation Q = [(pC)^c(pD)^d] / [(pA)^a(pB)^b] and plug in the partial pressures into A, B, C, D and the stoichiometric coefficients into a, b, c, d.
- Sun Dec 02, 2018 10:48 pm
- Forum: Properties & Structures of Inorganic & Organic Bases
- Topic: Bases for final
- Replies: 1
- Views: 287
Re: Bases for final
The 8 strong bases would probably be good to know:
1. LiOH
2. NaOH
3. KOH
4. RbOH
5. CsOH
6. Ca(OH)2
7. Sr(OH)2
8. Ba(OH)2
1. LiOH
2. NaOH
3. KOH
4. RbOH
5. CsOH
6. Ca(OH)2
7. Sr(OH)2
8. Ba(OH)2
- Sun Dec 02, 2018 10:44 pm
- Forum: Bronsted Acids & Bases
- Topic: Strong Acids
- Replies: 4
- Views: 344
Re: Strong Acids
The definition of a strong acid is that they completely dissociate in solutions and thus, they give off more protons. Weak acids may not completely dissociate and/or dissociate less than strong acids so they won't give off as many protons.
- Sun Dec 02, 2018 10:42 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Acids for final
- Replies: 2
- Views: 218
Re: Acids for final
We should probably also know the general 8 strong bases and 7 strong acids:
Bases
1. LiOH
2. NaOH
3. KOH
4. RbOH
5. CsOH
6. Ca(OH)2
7. Sr(OH)2
8. Ba(OH)2
Acids
1. HI
2. HBr
3. HCl
4. HClO3
5. HClO4
6. H2SO4
7. HNO3
Bases
1. LiOH
2. NaOH
3. KOH
4. RbOH
5. CsOH
6. Ca(OH)2
7. Sr(OH)2
8. Ba(OH)2
Acids
1. HI
2. HBr
3. HCl
4. HClO3
5. HClO4
6. H2SO4
7. HNO3
- Sun Nov 25, 2018 6:35 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Sigma vs pi bonds
- Replies: 7
- Views: 776
Re: Sigma vs pi bonds
Sigma bonds are also stronger than pi bonds, so a double bond (sigma + pi) would overall still be weaker than 2 sigma bonds.
- Sun Nov 25, 2018 4:51 pm
- Forum: Dipole Moments
- Topic: Polarizability
- Replies: 3
- Views: 326
Re: Polarizability
Polarizability means that the electron clouds of an atom readily undergo a large distortion. Anions have high polarizability because they are larger and have a bigger electron cloud. The nucleus exerts weak control over valence electrons because there is a lower effective nuclear charge and the elec...
- Sun Nov 25, 2018 4:49 pm
- Forum: Bond Lengths & Energies
- Topic: Dissociation Energy
- Replies: 1
- Views: 333
Re: Dissociation Energy
The graph shows that: If the bond is weaker, the dissociation energy is smaller because it is easier to break apart the bond. If the bond is stronger, the dissociation energy is higher because it is harder to break apart the bond. As the internuclear distance increases, the dissociation energy decre...
- Sat Nov 24, 2018 10:15 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 4.13 6th Edition: VSEPR formula for I3-
- Replies: 1
- Views: 86
Re: 4.13 6th Edition: VSEPR formula for I3-
The molecular structure for I3- is linear and not bent since the formula is AX2E3 (2 attached atoms, 3 electron pairs), the 3 electron pairs will settle on the axial plane and the other two iodine atoms will settle on the equatorial plane, which will form a linear shape.
- Sat Nov 24, 2018 10:11 pm
- Forum: Sigma & Pi Bonds
- Topic: Sigma Bonds
- Replies: 2
- Views: 281
Re: Sigma Bonds
It's impossible to have more than one sigma bond (with one sigma bond, any more bonds would have to be pi bonds). This is because sigma bonds are when orbitals are overlapping head-to-head, and it's impossible to have 2+ lobes pointing at each other and overlapping each other's orbitals in the head-...
- Sat Nov 24, 2018 10:08 pm
- Forum: Sigma & Pi Bonds
- Topic: Pi bond and e- density
- Replies: 1
- Views: 99
Re: Pi bond and e- density
This means that the electrons in the p-orbitals pair with the electrons of the p-orbital on the other atom and the orbitals overlap in a side-by-side arrangement both on the top and bottom (on each side of the internuclear axis). These pi bonds basically lock the molecule into place so that it is un...
- Sun Nov 18, 2018 4:42 pm
- Forum: Hybridization
- Topic: Sigma and Pi Bonds [ENDORSED]
- Replies: 12
- Views: 1097
Re: Sigma and Pi Bonds [ENDORSED]
Something to add on is that all single bonds are sigma bonds, all double bonds have one sigma and one pi bond, and all triple bonds have one sigma and two pi bonds and so on (all have one sigma, the rest are pi bonds). The sigma bonds allow the molecule to rotate, but the pi bonds are rigid and lock...
- Sun Nov 18, 2018 4:37 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: AX2E3 Shape [ENDORSED]
- Replies: 2
- Views: 258
Re: AX2E3 Shape [ENDORSED]
AX2E3 is linear because all three lone pairs are equatorial, leaving one upper axial and one lower axial atom that forms a linear shape.
AX2E2 is bent because the two lone pairs repel the two atoms (e.g. water molecule).
AX2E2 is bent because the two lone pairs repel the two atoms (e.g. water molecule).
- Sun Nov 18, 2018 4:32 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Water Molecule
- Replies: 3
- Views: 372
Re: Water Molecule
The two lone pairs look like they are next to each other (and therefore closer) when you draw them in 2D, but because it is a tetrahedral shape, all the angles/positions are technically the same distance from each other and it doesn't matter where the two lone pairs are located.
- Thu Nov 08, 2018 10:45 pm
- Forum: Empirical & Molecular Formulas
- Topic: Empirical Formula Problem Solving
- Replies: 2
- Views: 589
Re: Empirical Formula Problem Solving
I would divide all the moles by the smallest value, and set that small value as x. The other values would then be multiples (not necessarily integer) of x. I would then try to multiply all these multiples of x by a number to get whole numbers. There's really no easy way besides trial and error for t...
- Thu Nov 08, 2018 5:43 pm
- Forum: Dipole Moments
- Topic: Permanent Dipole Moments
- Replies: 4
- Views: 4001
Re: Permanent Dipole Moments
A permanent dipole moment is when the molecule is polar. The two or more atoms within the molecule must have substantially different electronegativities (one must attract electrons more than the other and becomes more negative while the other becomes positive).
- Thu Nov 08, 2018 5:40 pm
- Forum: Dipole Moments
- Topic: Induced Dipole
- Replies: 4
- Views: 350
Re: Induced Dipole
Induced dipole implies that before a heavily charged ion came near the original atom, the atom was nonpolar. Even though the atom was nonpolar, it has a cloud of electrons surrounding it. Putting a charged ion near the nonpolar atom disturbs the arrangement of the electrons so that the originally no...
- Thu Nov 01, 2018 7:36 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Formal Charge
- Replies: 8
- Views: 759
Re: Formal Charge
V is the # of valence electrons the atom has (look at the periodic table, count over groups from left to right)
L is how many lone electrons that element has on its Lewis structure
S is the number of shared electrons between multiple atoms' Lewis structures
L is how many lone electrons that element has on its Lewis structure
S is the number of shared electrons between multiple atoms' Lewis structures
- Thu Nov 01, 2018 6:03 pm
- Forum: Properties of Light
- Topic: Midterm Question 5a
- Replies: 4
- Views: 494
Re: Midterm Question 5a
Not any wavelength of light can be emitted from the atoms in the lamp because there needs to be a specific amount (quanta) of energy in order to eject an electron from a metal surface. Thus, the wavelength (because it is related to the frequency and thus energy of the atom) also has to be a specific...
- Thu Nov 01, 2018 4:58 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: 5s Quantum Numbers
- Replies: 1
- Views: 1037
Re: 5s Quantum Numbers
The general formula for quantum numbers is (n, l, ml, ms).
Since it's the 5s orbital, n=5.
Since it's the s orbital, l=0.
ml would also be 0, and ms would be either +1/2 or -1/2.
(5, 0, 0, +1/2 or -1/2)
Since it's the 5s orbital, n=5.
Since it's the s orbital, l=0.
ml would also be 0, and ms would be either +1/2 or -1/2.
(5, 0, 0, +1/2 or -1/2)
- Fri Oct 26, 2018 12:01 am
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration and Unfilled Orbitals
- Replies: 2
- Views: 305
Electron Configuration and Unfilled Orbitals
Given that an Au atom (gold) has an electron configuration of [Xe] 5d9, 6s2, does that mean the 6s subshell would start to be filled before the 5d subshell was completely filled? Is that possible and why?
- Thu Oct 25, 2018 10:50 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Units
- Replies: 8
- Views: 723
Re: Units
deltap x deltax >= h/4pi
p (momentum) - kg x m/s
x (position) - m
p (momentum) - kg x m/s
x (position) - m
- Thu Oct 25, 2018 10:48 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: nodal planes for d-orbitals
- Replies: 3
- Views: 392
Re: nodal planes for d-orbitals
P-orbitals do have one plane!
And d-orbitals have 2 nodes no matter the orientation. Four of the d-orbitals have 2 nodal planes, and one of them has 2 nodes that are actually two cones that point at the origin.
And d-orbitals have 2 nodes no matter the orientation. Four of the d-orbitals have 2 nodal planes, and one of them has 2 nodes that are actually two cones that point at the origin.
- Fri Oct 19, 2018 2:46 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: Quantum number M [ENDORSED]
- Replies: 7
- Views: 853
Re: Quantum number M [ENDORSED]
You can determine how many values of Ml there are based on the number of L through the equation M = (2L + 1) For example, if L = 1, there are (2(1)+1) = 3 values of M and they are -1, 0, +1 if L = 2, there are (2(2)+1) = 5 values of M and they are -2, -1, 0, +1, +2 and so on.
- Fri Oct 19, 2018 12:53 am
- Forum: DeBroglie Equation
- Topic: Rest Mass
- Replies: 7
- Views: 713
Re: Rest Mass
Actually, you can use DeBroglie's equation on photons. Even though photons have no rest mass, they do have a nonzero momentum. Thus, the wavelength relationship still works for photons, you just have to use their momentum and the equation λ=h/p for photons.
- Fri Oct 19, 2018 12:45 am
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Modules
- Replies: 2
- Views: 501
Re: Modules
I asked Prof. Lavelle last lecture and he said the Heisenberg module is the last one he will be posting. No more posts but he did say that Sapling Learning (something that comes with the textbook) has more resources like videos.
- Thu Oct 11, 2018 4:47 pm
- Forum: Properties of Light
- Topic: Threshold Energy
- Replies: 11
- Views: 830
Re: Threshold Energy
Threshold energy is the minimum amount of energy needed for the photon to remove an electron from the metal surface. It takes energy for the photon to remove the electron, and any less energy than needed (the threshold) would not eject an electron.
- Thu Oct 11, 2018 4:44 pm
- Forum: Photoelectric Effect
- Topic: Electrons
- Replies: 2
- Views: 178
Re: Electrons
In the experiment setup, the detector is connected to a polarized electric circuit, meaning there is a positive end and a negative end. Even though the velocity of the electron is 0, it would float up towards the positively charged part of the circuit near the detector (because opposite charges attr...
- Thu Oct 11, 2018 4:40 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect
- Replies: 1
- Views: 245
Re: Photoelectric Effect
Usually, if you were given the equation E=1/2MV^2, you would be asked to find V. The M in this equation is the mass of a single electron (not the sum mass of all the electrons), which is the same for all elements no matter the type of element/metal. So yes, it would have to be given and it is 9.11x1...
- Thu Oct 11, 2018 4:34 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Audio-Visual Assignments
- Replies: 4
- Views: 277
Re: Audio-Visual Assignments
The audio-visual assignments are not part of the homework grade (they are anonymous and the TAs/prof can't keep track of whether you complete them or not, but they are probably helpful in terms of helping you understand the content and doing better on tests!
- Tue Oct 02, 2018 3:17 pm
- Forum: Significant Figures
- Topic: Significant Figures
- Replies: 6
- Views: 814
Re: Significant Figures
The answer should have as many sigfigs as the LEAST # of sigfigs in the given problem.
For example, if you were told to subtract 1.702 (4 sigfigs) from 5.0 (2 sigfigs), your answer should only have 2 sigfigs and is 3.3.
For example, if you were told to subtract 1.702 (4 sigfigs) from 5.0 (2 sigfigs), your answer should only have 2 sigfigs and is 3.3.
- Tue Oct 02, 2018 3:07 pm
- Forum: SI Units, Unit Conversions
- Topic: State Symbols
- Replies: 2
- Views: 304
Re: State Symbols
I think Prof.Lavelle did say in a recent lecture that we should include state symbols when writing chemical equations in the homework as well as tests! I'm pretty sure he also said that there won't be a super random compound that we'll have know the state of--it'll generally be something we're famil...
- Tue Oct 02, 2018 3:01 pm
- Forum: Student Social/Study Group
- Topic: Anyone do G15 A?
- Replies: 2
- Views: 471
Re: Anyone do G15 A?
Hey! So you can first list out the basic formulas needed here: M(molarity) = mol/volume(L) MiVi=MfVf (because the # of moles are constant; i means initial, f means final) You can then list out the given information: Mi = 0.778 M Vi = ? Mf = 0.0234 M Vf = 150.0 mL You just have to plug all the given ...