Search found 62 matches
- Sat Mar 17, 2018 11:14 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Strength of Oxidation/Reducing Agents
- Replies: 2
- Views: 512
Re: Strength of Oxidation/Reducing Agents
The basic idea is that most positive potential is more oxidizing and the most negative are more reducing. page 579 offers a great chart to see how oxidizing power changes with cell potential.
- Sat Mar 17, 2018 11:13 pm
- Forum: Balancing Redox Reactions
- Topic: Acidic/ basic conditions
- Replies: 4
- Views: 689
Re: Acidic/ basic conditions
For basic equations, they say balance the O's by adding H2O to the side that needs it, then add to the other side the an H2O for every H atom present. And add to the same side an OH for every H present.
See page 567 for more details.
See page 567 for more details.
- Sat Mar 17, 2018 11:08 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Galvanic/Voltaic Cells
- Replies: 2
- Views: 464
Re: Galvanic/Voltaic Cells
Galvanic cells consist of an anode and a cathode that would produce a spontaneous reaction between them. In order for there to be a spontaneous reaction, the gibbs free energy must be negative, so that means that the standard cell potential must be positive as G=-nFE. Knowing this, this means that o...
- Sat Mar 17, 2018 6:58 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Finding Standard Cell Potential
- Replies: 1
- Views: 260
Re: Finding Standard Cell Potential
Because in this case, you're not finding the Standard Cell Potential where the electrons cancel everything out evenly. You're combining two half-reactions to form a half-reaction, which is why we have to do a different method of finding the Gibbs Free Energy, because Gibbs Free energy is a state fun...
- Fri Mar 16, 2018 11:15 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.61 units
- Replies: 1
- Views: 289
Re: 15.61 units
I supposed that it doesn't, but for simplicity sakes, I think it would be easier to convert your final answer into kJ to make it easier to read.
- Fri Mar 16, 2018 10:53 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagram
- Replies: 3
- Views: 449
Re: Cell Diagram
For each part (reduction and oxidation), the reactants go on the left, and the products go on the right and are separated by a |. Thus, it looks like this (reactant of anode)|(product of anoode)||(reactant of cathode)|(product of cathode) If the reactants and products are in the same phase (ie aqueo...
- Fri Mar 16, 2018 9:26 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.63
- Replies: 2
- Views: 364
Re: 15.63
The lnA's cancel each other out. It was just an error in the solutions manual. The minus sign is supposed to be an equal sign. Its .59 is equal to -(activation energy)/(gas constant)*(1/t1)-(1/t2)
or
or
- Fri Mar 16, 2018 9:21 pm
- Forum: Balancing Redox Reactions
- Topic: determining cathode from anode
- Replies: 11
- Views: 1836
Re: determining cathode from anode
I am pretty sure that the only way to determine the cathode from the anode is based on their standard cell potential. Cathodes are the ones that are reduced and the the Anodes are the ones that are being oxidized. And we know our Galvanic cell needs to be a positive number so whichever combination o...
- Fri Mar 16, 2018 9:12 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.61
- Replies: 2
- Views: 341
Re: 15.61
The change in T can be derived as the two ln equations subtracting from each other.
Which means that the lnA cancel each other out when the two equations subtract from one another.
Which means that the lnA cancel each other out when the two equations subtract from one another.
- Fri Mar 16, 2018 9:11 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Change in Entropy Equations
- Replies: 2
- Views: 475
Re: Change in Entropy Equations
Alright, so the first equation is typically the base for all the equations. It is useful for all scenarios because you can modify it. It's based on a constant temperature too. The only instance where I found the second equation is when it talked about phase change. So, when there's a phase change, y...
- Wed Feb 21, 2018 3:11 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.31c
- Replies: 1
- Views: 291
Re: 14.31c
Pb^4+ + 2e- ---> Pb^2+ should be the anode where Pb^4+ will be on the right side of the half-reaction. And this is mainly because of the fact that in the final equation, we can see hat Pb^4+ should be on the right side. Thus it is the anode. 2 Pb^2(aq)--->Pb(s)+ Pb^4+(aq) Now for Pb(s)--->Pb^2+ + 2e...
- Mon Feb 19, 2018 5:03 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: cell diagram
- Replies: 1
- Views: 246
Re: cell diagram
I think they said that we have to put Pt when the reactant and product are of the same phases.
- Mon Feb 19, 2018 5:02 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: how to write a cell diagram [ENDORSED]
- Replies: 7
- Views: 1981
Re: how to write a cell diagram [ENDORSED]
So the order is typically Inert Metal (if applicable) | (Anode reactant) | (anode product) || (Cathode reactant) | (cathode product)| (Inert Metal if applicable).
- Tue Feb 13, 2018 10:41 pm
- Forum: Van't Hoff Equation
- Topic: Van't Hoff equation
- Replies: 4
- Views: 944
Re: Van't Hoff equation
I would imagine so. Isn't the Van't Hoff equation broken down into those two equations that you mentioned? So I feel like it would be ok.
- Tue Feb 13, 2018 10:40 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Endothermic Favorable?
- Replies: 2
- Views: 1952
Re: Endothermic Favorable?
If you have an endothermic reaction, that means Delta H is positive. Delta S depends. If Delta S was positive, then by Delta G = Delta H - T*Delta S, the reaction would be favorable at high temperatures. If Delta S was negative, then the reaction would not be favorable at any temperature.
- Tue Feb 13, 2018 10:22 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.15 confusion
- Replies: 2
- Views: 421
Re: 9.15 confusion
Since we're calculating phase change you can find the value of 43.5 on page 284, where it lists all the values of the standard enthalpies of phase changes.
- Tue Feb 13, 2018 10:19 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Heat Transfer with no temperature given
- Replies: 2
- Views: 364
Re: Heat Transfer with no temperature given
Since it gives pressure, I am assuming that it involves gas. That simply means you would need to use the PV=nrt formula to figure out temperature, solving for T.
- Tue Feb 13, 2018 10:17 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.57
- Replies: 1
- Views: 249
Re: 9.57
In regards to that, essentially, since we know that at standard temperature, it would 298 K. So you would just need to look up the Gibbs Free Energy from the appendix and calculate it from that, not from delta G = Delta H - T * Delta S.
- Tue Feb 13, 2018 9:54 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Standard Molar Entropy Question
- Replies: 3
- Views: 660
Re: Standard Molar Entropy Question
Remember that entropy also depends on the complexity of the molecule itself, so that also means the elements that it consists of. So in this case, F is more complex than H in terms of its atomic structure. It has more particles in its atom than H and each of those particles has more orientations pos...
- Tue Feb 13, 2018 9:51 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Calculating Degeneracy
- Replies: 3
- Views: 624
Re: Calculating Degeneracy
You would use the formula to calculate the entropy. W representing degeneracy is simply the amount of orientations possible for the molecule raised to Avogadro's number if given only the fact that you have to fine the degeneracy of a just a molecular formula. Otherwise I agree with the above stateme...
- Tue Feb 13, 2018 9:48 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Change in the Entropy of the Universe
- Replies: 1
- Views: 258
Re: Change in the Entropy of the Universe
So the total entropy change 0 for reversible processes is 0 because essentially it's like the entropy of the system with the entropy of its surroundings. When the system increases or decreases in entropy, then its surroundings must do the opposite in a reversible process. If it can go back to its or...
- Tue Feb 13, 2018 9:41 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.115
- Replies: 1
- Views: 296
Re: 11.115
Basically yes. So this goes back to about our ideas of equilibrium constants. When you compress a gas (decrease the volume), the reaction will shift to the side with less moles of gas, but since there's the same amount of moles of gas on either side, there's no effect. For the water, we only take in...
- Sun Feb 11, 2018 9:40 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Comparing Molar Entropies
- Replies: 2
- Views: 1422
Re: Comparing Molar Entropies
Mainly due to higher molar mass and complexity of the atom/molecule. Typically, the more complex an object is, the more entropy it has due to more "disorder" or orientations that it could potentially have. Lead has more entropy than carbon due to having more particles in its atom than carb...
- Sun Feb 11, 2018 9:37 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Sublimation
- Replies: 3
- Views: 499
Re: Sublimation
Sublimation is the change of a solid to a gas. So in order to find the enthalpy for it, simply do the enthalpy of fusion + enthalpy of vaporization.
- Sun Feb 04, 2018 3:37 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Using dS vs change in S
- Replies: 3
- Views: 634
Re: Using dS vs change in S
Most likely never. Using dS involves using calculus and integrals, which isn't mainly the whole focus of the class. I see that there's no real reason in order to derive the answer from dS. In fact, most of the time you can relate dS to change in S. After all, dS can be thought of as the change in S.
- Sun Feb 04, 2018 3:35 pm
- Forum: Calculating Work of Expansion
- Topic: 8.9
- Replies: 2
- Views: 398
Re: 8.9
Since we're trying to find the Internal Energy, our final answer should have energy units. By using these two R constants, our units would all cancel out to get joules as a final unit. That's why we use the two R constants.
- Sun Feb 04, 2018 2:33 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 2764624
Re: Post All Chemistry Jokes Here
A line I used with my boyfriend:
"You know, I usually hate chemistry, but I love having chemistry with you."
"You know, I usually hate chemistry, but I love having chemistry with you."
- Sun Feb 04, 2018 2:31 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Heating Capacity
- Replies: 3
- Views: 385
Re: Heating Capacity
In Problem 9.43, we use the molar heat capacity of water in a liquid form because we have that information to do so. We know the heating capacity of water, and we needed to find the temperature in K (which is why we used Molar heat capacity). In regards to your other questions, the R is the ideal ga...
- Sun Feb 04, 2018 2:26 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Change in Entropy of Universe
- Replies: 3
- Views: 1049
Re: Change in Entropy of Universe
Delta S of the universe is equal to 0, because it's essentially Delta S of the system + Delta S of the surroundings. The system experiences an increase in entropy in a reversible process, but since its a reversible process, the Delta S of the surroundings experience a decrease in entropy as well. Th...
- Sun Feb 04, 2018 2:19 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Using Q
- Replies: 2
- Views: 314
Re: Using Q
If the reaction is not at equilibrium, then we can use Q and that equation to solve for Gibbs Free Energy. If the reaction comes to equilibrium, then we use K.
- Sat Feb 03, 2018 10:11 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Isolated
- Replies: 9
- Views: 1052
Re: Isolated
So if we go back to the drawn system or like lets say a bomb calorimeter right? The whole system itself is isolated, but the things thats inside the bomb calorimeter and the isolated system are not isolated. So that's why there can be a change in temperature within the individual system thats within...
- Sat Feb 03, 2018 10:06 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: entropy vs. Degeneracy
- Replies: 4
- Views: 544
Re: entropy vs. Degeneracy
Degeneracy is the number of ways of achieving a given energy state, so basically how many different orientations (of lets say a molecule) that will give the same energy level. Entropy on the other hand is disorder, or if we do not want to use the term disorder, it is the proper of the system needed ...
- Sat Feb 03, 2018 5:05 pm
- Forum: Calculating Work of Expansion
- Topic: Reversible vs Irreversible [ENDORSED]
- Replies: 3
- Views: 639
Re: Reversible vs Irreversible [ENDORSED]
A reversible pathway has more work done than an irreversible pathway mainly due to the fact that in a reversible process, it is pushing against that external pressure and tries to achieve the maximum expansion work possible. In an irreversible pathway some of the potential of the system to do work i...
- Sat Feb 03, 2018 4:57 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.52
- Replies: 3
- Views: 449
Re: 9.52
An endothermic process is usually not spontaneous due to the fact that by putting energy into the system, there is an external influence now acting on the system. However, the only way that an endothermic process can be spontaneous is if the increase in entropy will be greater than the increase in e...
- Sat Dec 09, 2017 4:18 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration of Br-
- Replies: 6
- Views: 18518
Re: Electron Configuration of Br-
Yes it would be correct to just write as [Kr] if they ask for the shorthand of it.
- Sat Dec 09, 2017 4:17 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Confused about What Lavelle Mentioned in class
- Replies: 1
- Views: 248
Re: Confused about What Lavelle Mentioned in class
Since the concentration of hydronium atoms will be so small, that they will not make that huge of an impact on the pH level of the solution. This only works if the concentration is way smaller than 10 -7 , ie like 10 -10 , in which subtracting that from 10 -7 , will not impact that severely leading ...
- Sat Dec 09, 2017 4:09 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Using electronegativity to determine acidity
- Replies: 2
- Views: 644
Re: Using electronegativity to determine acidity
As for the reading from the textbook, on page 469 in Chapter 12. "Metals typically form basic oxides and nonmetals typically form acidic oxides." And then they talk about the examples of amphoteric oxides. Acidic strength, in one part, is determined by how weak the bonds are. The weaker th...
- Sat Dec 09, 2017 4:07 pm
- Forum: Conjugate Acids & Bases
- Topic: 12.61
- Replies: 1
- Views: 413
Re: 12.61
To find Ka, you have to go to the Tables 12.1 and 12.2 on pages 477 and 478. They give you the Ka and Kb of various weak acids and bases.
- Sat Dec 09, 2017 4:05 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration for ions
- Replies: 2
- Views: 497
Re: Electron Configuration for ions
Fe2+ and other metals within the d-block are much harder to write the electron configuration for due to their nature. Some of them are transition metals and can have different oxidation numbers too. For the most part, focus on the s-block and p-block in order to write electron configuration and crea...
- Sat Dec 09, 2017 4:03 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: SigFigs
- Replies: 1
- Views: 229
Re: SigFigs
The TAs told us (well I wasn't at the review session) not to worry about sig figs. Of course this doesn't mean write out all the numbers of the answer, but do it as many as required. And try not to round before coming to your final answer.
- Sat Dec 09, 2017 3:59 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Combining fractional roots in K expression?
- Replies: 1
- Views: 570
Re: Combining fractional roots in K expression?
The problem you're referring to comes from Chapter 11. From pages 439 to 440, the reading will tell you the general properties of the Equilibrium constant. The general basis is that because we divided our balanced equation by a coefficient of 1/2, our equilibrium constant will be raised to a power o...
- Sat Dec 09, 2017 3:55 pm
- Forum: Bronsted Acids & Bases
- Topic: 12.9 (c)
- Replies: 1
- Views: 395
Re: 12.9 (c)
Proton transfers are usually represented by the transfer of a hydrogen atom. Whether the proton of a water molecule went to a base, or when the proton of an acid went to a water molecule. In this case, there was no proton transfer, but rather an OH bond was transferred from and replaced with NH 2 . ...
- Sat Dec 09, 2017 3:53 pm
- Forum: Electronegativity
- Topic: Kr
- Replies: 3
- Views: 451
Re: Kr
Krypton is a noble gas and has a complete valence shell. For it to be ionized, or to achieve ionization, it would take an insane amount of energy to remove an electron from a complete shell. It is easier for Mg to remove an electron than Krypton.
- Sun Nov 26, 2017 12:12 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.1 part a.
- Replies: 1
- Views: 228
Re: 11.1 part a.
When a reaction is in equilibrium, the reaction still happens. The idea is that rate of formation (from reactants to products) and the rate of decomposition (from products to reactants) are happening at the same rate so the concentration remains constant. It's as if you're assembling a toy from a pa...
- Sat Nov 25, 2017 11:59 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: KC/KP/QC/QP
- Replies: 2
- Views: 7496
Re: KC/KP/QC/QP
K c is the equilibrium constant, referring to when the reactants and products are in terms of molarity. K p is the equilibrium constant when the products and reactants are given in terms of atm (usually when they're gases) and so it's know as the equilibrium constant of partial pressures. Q c and Q ...
- Sun Nov 19, 2017 1:44 pm
- Forum: Hybridization
- Topic: Energy of Hybrid Orbitals
- Replies: 4
- Views: 1177
Re: Energy of Hybrid Orbitals
Hybrid orbitals occur because we want to essentially form more bonds. For instance, a Carbon atom, by theory, says we can form 4 bonds because we have 4 valence electrons. However, when we take a look at the electron configuration, we can see that 2 of its electrons (in the s-orbital are already pai...
- Sun Nov 19, 2017 1:41 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: All bond angles
- Replies: 3
- Views: 413
Re: All bond angles
We can't be certain of the exact bond angles of every shape of the molecules, because sometimes lone pairs affect it. So, I simply would recommend memorizing the angles of molecules where they're symmetrical or stated in the book, and memorize how lone pairs affect the angles.
- Sun Nov 19, 2017 12:48 pm
- Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
- Topic: Polar or non-polar molecule
- Replies: 2
- Views: 390
Re: Polar or non-polar molecule
It depends on the shape of the molecule. Since the electron arrangement is 4 areas of electron density, and then 4 paired atoms, then the shape of the atom is a tetrahedral. And for a tetrahedral atom, unless all the paired atoms are the same like CH 4 , then it'll most likely be polar due to the we...
- Sun Nov 19, 2017 12:43 pm
- Forum: Hybridization
- Topic: Forming the hybridization
- Replies: 1
- Views: 259
Re: Forming the hybridization
We do not just put the superscript after the p. We sometimes put it after the d too. It depends on the atom itself. In SF 6 , the S atom forms hybrid orbitals of sp 3 d 2 . The amount of orbitals used in the structure responds to the hybridization. Since we used 6 orbitals, we have to go to s (1 orb...
- Sat Nov 11, 2017 6:01 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Beryllium Octet Rule
- Replies: 6
- Views: 10673
Re: Beryllium Octet Rule
Beryllium doesn't fulfill the octet rules because of its valence electrons and the orbitals they occupy. Just like H, He, Li, and Be, they are all exceptions to the octet rules mainly because they don't need 8 to feel complete. Their s-orbitals do not need 8 electrons to complete the shell to feel f...
- Sat Nov 11, 2017 5:51 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Polar vs nonpolar vs ionic
- Replies: 11
- Views: 27829
Re: Polar vs nonpolar vs ionic
The polar and nonpolar is a little bit confusing to understand. Essentially, it is based on the partial charges of the compound. Based on the reading of Chapter 3, we understand that all molecules can be seen as resonance hybrids between their purely covalent bonds and purely ionic bonds. You can ea...
- Sun Nov 05, 2017 7:46 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: 3.49
- Replies: 3
- Views: 625
Re: 3.49
For 49, it's NO+, so in total there should be ten electrons. That means the formal charge for Nitrogen would be 5-(6/2 + 2 electrons [1 lone pair]) = 0. For oxygen, it would be 6- (6/2 +2 electrons [1 lone pairs]) = +1.
Therefore the charge for the ion is +1.
Therefore the charge for the ion is +1.
- Fri Nov 03, 2017 1:51 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Midterm
- Replies: 8
- Views: 884
Re: Midterm
I don’t believe so. The midterm covers the fundamentals and chapter 1-3. We finished chapter 3 this Monday too I believe. VSEPR is a part of chapter 4.
- Sun Oct 29, 2017 10:34 pm
- Forum: Trends in The Periodic Table
- Topic: Electron affinity versus ionization energy [ENDORSED]
- Replies: 3
- Views: 585
Re: Electron affinity versus ionization energy [ENDORSED]
Electron Affinity is the energy that is released when an electron is added to a gas phase atom, while ionization energy is the energy needed to remove an electron from an atom. The further away an electron is from a nucleus, the easier it becomes to remove the electron. Thus, for ionization energy, ...
- Sun Oct 29, 2017 10:26 pm
- Forum: Trends in The Periodic Table
- Topic: Electron Affinity [ENDORSED]
- Replies: 3
- Views: 593
Re: Electron Affinity [ENDORSED]
I believe that if you take a look at the electron configuration you can also figure out whether or not the electron affinity will be greater than 0, less than 0, or if it becomes more negative. For instance, the EA becomes more negative moving down the Group 11 elements because n increases. Because ...
- Sun Oct 22, 2017 1:29 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Quantum Numbers [ENDORSED]
- Replies: 1
- Views: 363
Re: Quantum Numbers [ENDORSED]
You have to find the principle quantum number first based on the row of the elements (n=1,2,3, etc) and then you can figure out the rest of the quantum numbers.
- Sun Oct 22, 2017 1:22 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration for silver [ENDORSED]
- Replies: 1
- Views: 672
Re: Electron Configuration for silver [ENDORSED]
So I believe that since 3d has a lower energy level than 4s, we fill up the 3d orbital before the 4s orbital. Thus for Silver, we start to fill up the 5s orbital. However, on page 47 in the textbook, it states that the "half-complete subshell configuration d 5 and complete subshell configuratio...
- Sun Oct 15, 2017 5:35 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Upcoming Test
- Replies: 5
- Views: 530
Re: Upcoming Test
In my Discussion, my TA told me that we will be tested on material up to Chapter 1, Section 5. She said that it was unlikely that we will be tested on the Heisenberg Uncertainty Principle. She also said to make sure that we understand the experiments discussed in the book.
- Sun Oct 15, 2017 5:19 pm
- Forum: Properties of Light
- Topic: Question about speed of light rounding
- Replies: 4
- Views: 607
Re: Question about speed of light rounding
I believe is that in problem 15, they had a number of 102.6, which has 4 significant figures so they round the speed of light to 4 sig figs. Meanwhile, in problem 17, they didn't specify a number so they used the actual number of the speed of light.
- Fri Oct 06, 2017 10:27 pm
- Forum: Balancing Chemical Reactions
- Topic: Net Moles Produced [ENDORSED]
- Replies: 5
- Views: 858
Re: Net Moles Produced [ENDORSED]
It just basically asks for how much the equation produced. If it's using the number of moles of equation, then it's how many gets produced in the equation (ie 1H 2 O -> 2 moles of H and 1 mole of O, giving 3 moles produced). If it's using an amount of moles, then you have to figure out the ratios fi...
- Fri Oct 06, 2017 10:24 pm
- Forum: Empirical & Molecular Formulas
- Topic: Friday Oct 6 Test
- Replies: 7
- Views: 1108
Re: Friday Oct 6 Test
For the problem, here's what I did. I converted all grams of the products into moles, and then used the moles of those to figure out the moles of the individual elements of Carbon and Hydrogen, because we can't figure out Oxygen due to it coming from the two sources of compound X and the Oxygen used...
- Fri Oct 06, 2017 10:02 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Rounding Rule for 5
- Replies: 5
- Views: 1084
Re: Rounding Rule for 5
The reason why we round to the nearest even number is because many different people have different ideas about rounding for numbers ending in 5. For some people, it means rounding down, for others it means rounding up. So to clear away the confusion, we just round to the nearest even number to make ...