Search found 61 matches
- Wed Mar 13, 2019 10:20 pm
- Forum: Zero Order Reactions
- Topic: Units of k
- Replies: 12
- Views: 2040
Re: Units of k
I would remember that zero order would be M/Sec and then divide M each time you go down an order. Here's what I mean: Zero order would be M(molarity)/second so divide by M to get units for first order. Notice how M cancels out. 1st order units: s^-1 Divide M again to get 2nd order units s^-1*M^-1 An...
- Wed Mar 13, 2019 10:14 pm
- Forum: Zero Order Reactions
- Topic: Zero order
- Replies: 10
- Views: 1607
Re: Zero order
A zero order reaction does not change if the concentration of the reactants change. It is only based on the k. Therefore, the equation would be rate=k. However, first order reactions do depend on the concentration of reactants and it's equation will be rate =k[A]. When the concentration of reactant ...
- Wed Mar 13, 2019 10:10 pm
- Forum: Zero Order Reactions
- Topic: Concept
- Replies: 4
- Views: 737
Re: Concept
Zero order means that the rate constant is independent from the concentration so therefore the concentration is to the zeroth power ie the rate law being rate= k[A]^0 which will just be rate=k.
Same applies for first and second orders.
Hope that helps!
Same applies for first and second orders.
Hope that helps!
- Wed Mar 06, 2019 7:20 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Units
- Replies: 6
- Views: 686
Re: Units
As everyone stated above, it will be M/L per second and as you increase to each order, you're constantly dividing by 1/M so for zero order, the units will be Moles per second. s^-1
1st order will be per second or s^-1M^-1
and so forth
1st order will be per second or s^-1M^-1
and so forth
- Wed Mar 06, 2019 7:15 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: how is kinetics different?
- Replies: 17
- Views: 1595
Re: how is kinetics different?
With respect to thermo, it tells us the spontaneity of a reaction and if it will occur. Kinetics allows us to determine the rate at which the reaction occurs, assuming that all reactions happen.
- Wed Mar 06, 2019 7:14 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Study Advice
- Replies: 73
- Views: 6750
Re: Study Advice
If you are in AAP or are looking to get into AAP, I recommend their PLF sessions and a lot of YouTube videos revolving the topics covered on the syllabus helps a lot. I also would recommend doing the worksheets that flow around Chem Comm. from Karen or other UAs. Best of luck!
- Wed Feb 27, 2019 7:16 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Test 3
- Replies: 14
- Views: 2274
Re: Test 3
Yes, the equations or Gibbs will also be provided to us on the constants and formulas worksheet. The test'll be on Gibbs and Electrochem but not Nernst equation. Here's his official announcement: (2/14/2019) Test 2 covers all topics on Gibbs free energy and electrochemistry up to but not including t...
- Wed Feb 27, 2019 7:13 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Value of n
- Replies: 7
- Views: 922
Re: Value of n
After you create your two half reactions, you need to balance the amount of electrons in order to combine the reaction into one. In the process of balancing out the amount of electrons, you find your number of moles of electrons. It is the number of e- you used to balance the reaction.
- Wed Feb 27, 2019 7:12 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Pt (s) [ENDORSED]
- Replies: 8
- Views: 920
Re: Pt (s) [ENDORSED]
Hi, You would add Pt(s) or C(graphite) if there is no solid acting as an anode or cathode. You still would need to add a solid (either platinum or carbon graphite) in the presence of an aqueous solution because that represents the aqueous solution that the anode/cathodes would be in. So in the absen...
- Thu Feb 21, 2019 9:49 pm
- Forum: Balancing Redox Reactions
- Topic: Helpful acronym for Redox Rxns
- Replies: 2
- Views: 2393
Re: Helpful acronym for Redox Rxns
Thank you for this creative mnemonic!
- Thu Feb 21, 2019 9:49 pm
- Forum: Balancing Redox Reactions
- Topic: balance equations
- Replies: 2
- Views: 402
Re: balance equations
Here is a helpful video for me that I hope will help you: https://www.youtube.com/watch?v=IZ1tKxsqV74
- Thu Feb 21, 2019 9:46 pm
- Forum: Balancing Redox Reactions
- Topic: OH- H+ and H2O in balancing redox reactions
- Replies: 5
- Views: 12329
Re: OH- H+ and H2O in balancing redox reactions
You would use H+ to balance the amount of hydrogens and water to balance the amount of water.
- Mon Feb 11, 2019 6:49 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Percent ionization vs percent deprotonation
- Replies: 3
- Views: 643
Re: Percent ionization vs percent deprotonation
Yes! They are used synonymously.
- Mon Feb 11, 2019 6:37 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Which does more work?
- Replies: 3
- Views: 353
Re: Which does more work?
Reversible reactions always do more work since they are infinitesimal reactions happening so it can do a maximum amount of work.
- Mon Feb 11, 2019 6:31 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: pressure effect on entropy
- Replies: 2
- Views: 392
Re: pressure effect on entropy
Right, when pressure increases, the inverse happens to volume. Entropy is the amount of possible positions that thing can have so when there is less space, the options for the different amount of positions it can hold are less so therefore, entropy is decreasing.
- Tue Feb 05, 2019 8:19 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Cup of Tea
- Replies: 17
- Views: 2433
Re: Cup of Tea
A cup of tea can interact with its surroundings, ie release heat from the tea into the universe, making the universe infitesmilly warmer. Closed systems like a hydro flask or thermos cannot do that, it is isolated/insulated.
- Tue Feb 05, 2019 8:17 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Midterm [ENDORSED]
- Replies: 49
- Views: 4693
Re: Midterm [ENDORSED]
This Wednesday Professor Lavelle is going to announce various workshops and review sessions located around campus at various times to fit your schedule. I recommend going to those as well as doing some practice problems and finding videos online that relate to the course.
- Tue Feb 05, 2019 8:15 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: positional (residual) entropy
- Replies: 2
- Views: 448
Re: positional (residual) entropy
Residual entropy is like a screenshot of the molecules in stationary form but we need to also calculate entropy for its other possible orientations since some molecules can have resonance such as COF2 and BF3. They are both trigonal planar but BF3 is non polar and doesn't have resonance; the three f...
- Wed Jan 30, 2019 7:45 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: ICE
- Replies: 19
- Views: 1424
Re: ICE
I = initial
C = change in concentration
E = equilibrium, so finding E will give you the concentration of the compound at equilibrium.
C = change in concentration
E = equilibrium, so finding E will give you the concentration of the compound at equilibrium.
- Wed Jan 30, 2019 7:44 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: When to use Quadratic equation
- Replies: 10
- Views: 6423
Re: When to use Quadratic equation
I've been told that we can use the quadratic formula when the value is greater than 10^-3. You can also use the 5% rule to double check if the value is eligible to make x equal to zero and to completely ignore it since it's infinitesimal.
- Wed Jan 30, 2019 7:42 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: pKa vs. pH
- Replies: 15
- Views: 2944
Re: pKa vs. pH
pKa is the -log of the Ka value and Ka*kB=1x10^-4 so you can find out the value of Ka/pka if you're given the value of Kb.
pH is the -log of H+ or H3O+.
The 'p' is -log-ing different values, therefore they are different. Hope this helps.
pH is the -log of H+ or H3O+.
The 'p' is -log-ing different values, therefore they are different. Hope this helps.
- Tue Jan 22, 2019 7:56 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Quiz #1 This friday
- Replies: 5
- Views: 654
Re: Quiz #1 This friday
Homework problems from the problem sets from the past two weeks often help a lot and are great practice. Also, if you can go to UAs for some worksheets for more practice, attend office hours and discussions are also a big help. I also do a lot of YouTubing videos on the topics that'll be on quiz 1 (...
- Tue Jan 22, 2019 7:54 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Problem 12.33 Sixth Edition
- Replies: 2
- Views: 318
Re: Problem 12.33 Sixth Edition
For me, I divided the final volume (.5 L) by the initial (.005 L) and multiplied it by the original concentration of hydroxide (.18) and I got 18 M which is correct according to the solutions manual.
- Tue Jan 22, 2019 7:53 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.39 6th edition
- Replies: 4
- Views: 458
Re: 11.39 6th edition
Answer: Refer to the table referenced earlier in the chapter. It tells you that K equals 377 so you would set 377 equal to your K equation (products over reactions, not including solids or liquids), and then determine K from the table again from the other elements associated with the reaction. see t...
- Mon Jan 14, 2019 8:24 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Concentration Affecting K
- Replies: 7
- Views: 961
Re: Concentration Affecting K
No, changing the concentration will not affect K over time. More specifically, at first it will try to balance out the concentrations and eventually and gradually, it will counteract the change and come back to equilibrium.
- Mon Jan 14, 2019 8:22 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Pure Liquids and Pure Solids
- Replies: 5
- Views: 1999
Re: Pure Liquids and Pure Solids
Pure solids or liquids aren't included in the equilibrium expression because their concentrations stay constant throughout the reaction so we don't include them.
- Mon Jan 14, 2019 8:20 pm
- Forum: Ideal Gases
- Topic: Concentration or Partial Pressure
- Replies: 13
- Views: 914
Re: Concentration or Partial Pressure
If the reactants are all gaseous, you can use their partial pressures to find Kp. If they are a mixture of gases and aqueous solutions, then your best bet is to use Kc. Very similar equations, just Kp uses the partial pressure of the gases while Kc uses the concentrations. Hope that helps!
- Wed Jan 09, 2019 7:13 pm
- Forum: Ideal Gases
- Topic: Memorization
- Replies: 12
- Views: 1186
Re: Memorization
I don't believe that you have to since it will be provided for you on the equations worksheet. It is posted on his website and here it is for reference: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/constants_equations.pdf Also, I think you will remember it anyways because of its...
- Wed Jan 09, 2019 7:09 pm
- Forum: Ideal Gases
- Topic: Reaction Quotient
- Replies: 4
- Views: 398
Re: Reaction Quotient
From what was gathered in lecture, the quotient reaction is the values anytime during the reaction, not necessarily at equilibrium. So not only does it determine which way the reaction goes, but if the value of Q is not equal to K then you can tell that it is not at equilibrium. If Q < K, the forwar...
- Wed Jan 09, 2019 7:06 pm
- Forum: Student Social/Study Group
- Topic: New to Lavelle
- Replies: 32
- Views: 5187
Re: New to Lavelle
Hi there, I would definitely agree with all of the above suggestions. Personally, I use his website as well as Chemistry Community a lot and the homework questions are the main source of preparation for you so I recommend doing all of them or as many as you can. I also go to the step up sessions and...
- Thu Dec 06, 2018 9:23 pm
- Forum: Photoelectric Effect
- Topic: Work Function
- Replies: 14
- Views: 2163
Re: Work Function
Work function, threshold energy, and ionization energy all mean the same thing and can be used interchangeably. The equation you'd use is ephoton - work function (or threshold or ionization energy) = Ekinetic energy, 1/2mv^2. Based on the information of the problem, you can work backwards to find on...
- Thu Dec 06, 2018 9:18 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Heisenberg and Wavelength
- Replies: 11
- Views: 1681
Re: Heisenberg and Wavelength
I wrote something along the lines of they are inversely proportional to one another and that it is impossible to know both momentum and position at the same time and got credit for it. Hope that helps!
- Thu Dec 06, 2018 9:16 pm
- Forum: Properties of Light
- Topic: Speed of light
- Replies: 13
- Views: 3840
Re: Speed of light
In the past I have used 3.0 x 10 ^ 8m/s and never saw any problems. However, a coversheet with some equations will be given to us so I'd use that if you're unsure.
- Thu Nov 29, 2018 7:04 pm
- Forum: Lewis Structures
- Topic: 7th Edition #2E.13
- Replies: 3
- Views: 2843
Re: 7th Edition #2E.13
On the periodic table, I is in the 7th column meaning that it has 7 valence electrons. We want its formal charge to be zero so we would use the equation FC (Formal charge) = valence - (Lone pair + shared pair/2). If iodine was a double bond, we'd get a formal charge of 1 (7-1) but we want a formal c...
- Thu Nov 29, 2018 6:37 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 2E.29 (7th edition)
- Replies: 2
- Views: 304
Re: 2E.29 (7th edition)
I agree that figure 1 has the largest dipole moment because the most electronegative element is Cl and it is generally going to pull towards the top right direction. The second figure is not as strong of a directional point in electronegativity as the first. And the last figure, they are pulling in ...
- Thu Nov 29, 2018 6:31 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 2E.19b (7th Edition)
- Replies: 2
- Views: 341
Re: 2E.19b (7th Edition)
Be does not need to fulfill its octet because it is the first four elements on the periodic table and the same exceptions go for Be as it does for H, He, and Li.
- Tue Nov 20, 2018 6:30 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Memorization
- Replies: 4
- Views: 400
Re: Memorization
I think that it's more reasonable to reason the shape rather than memorizing. From seeing its arrangement like tetrahedral, trigonal pyramidal, or trigonal planar, we can see if the molecule has lone pairs and whittle down from there. We would have to know both its arrangement and shape, like if it'...
- Tue Nov 20, 2018 6:26 pm
- Forum: Hybridization
- Topic: electron density in regards to # of hybrid orbitals
- Replies: 4
- Views: 429
Re: electron density in regards to # of hybrid orbitals
Adding on, your hybridization formula should equal the amount of electron density areas. For example, if your hybridization formula is sp3, then you should have four regions of electron density. 1s+ 3 p3 orbitals= 4 hybrid orbitals = 4 regions of electron density.
- Tue Nov 20, 2018 6:23 pm
- Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
- Topic: AXE
- Replies: 7
- Views: 686
Re: AXE
When writing the VSEPR formula, you begin with A which is the central atom. And then E and the subscript number will depend on the amount of "things" it's attached to, regardless of bonds. Then you'd finish with E and the subscribe number will depend on the amount of LONE PAIRS. We count e...
- Wed Nov 14, 2018 5:21 pm
- Forum: Bond Lengths & Energies
- Topic: Bond length question
- Replies: 4
- Views: 537
Re: Bond length question
Professor Lavelle said that we won't need to draw the structures in a 3D form with shading and wedges but we will be asked to name the shape. And regarding resonance, he said that we can use either resonance structure since they do average out to the be the same bond length. Hope that helped! :)
- Wed Nov 14, 2018 5:13 pm
- Forum: Properties of Electrons
- Topic: Polar vs Non polar
- Replies: 15
- Views: 4559
Re: Polar vs Non polar
When you determine that the atom is polar or non polar (from the previous comments), the direction of polarity points to the atom that is most electromagnetic! :)
- Wed Nov 14, 2018 5:08 pm
- Forum: Hybridization
- Topic: Bond Angels
- Replies: 5
- Views: 645
Re: Bond Angels
Bond angles are the angles that are created by electrons. Since they are the same negative charges, they repel and that creates a bond. For example, a central atom bonded by two other atoms creates a linear bond. Electrons are so repulsed by one another, they want to remain as far apart from each ot...
- Tue Nov 13, 2018 7:51 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Why do the orbitals of an atom only hold a certain amount of electrons?
- Replies: 7
- Views: 991
Re: Why do the orbitals of an atom only hold a certain amount of electrons?
At 0, 0=s At 1, 1=p At 2, 2=d At 3, 3=f S has to have one orbital and one orbital can contain 2 electrons. P has three orbitals (the orbitals increase in increments of 2), containing 6 electrons. D has five orbitals, containing 10 electrons. F has seven orbitals which can hold 14 electrons. The s or...
- Mon Nov 05, 2018 10:14 pm
- Forum: Balancing Chemical Reactions
- Topic: Combustion Chemical Reactions
- Replies: 2
- Views: 495
Re: Combustion Chemical Reactions
I agree with what Tony said! A combustion reaction will always be (some compound) + O2 --> CO2 + H2O And according to that compound, that will help you balance the equation. Sometimes, there will be more things produced but the question will have to tell you the extra molecules that were produced.
- Mon Nov 05, 2018 10:12 pm
- Forum: Limiting Reactant Calculations
- Topic: Midterm Review Q2
- Replies: 8
- Views: 4311
Re: Midterm Review Q2
Start off with what you have and convert it to grams of NO using the chemical equation. 21.1 g NH3x(1 mol NH3/ 17.031 g NH3)x(4 mol NO/4 mol NH3) [because we are trying to get grams of NO, use the chemical equation ratio] x(30.006 g NO/1mol NO) That will give you 37.2 grams of NO produced using 21.1...
- Mon Nov 05, 2018 10:05 pm
- Forum: Significant Figures
- Topic: Exam
- Replies: 6
- Views: 1149
Re: Exam
From my understanding, we will use the smallest amount of significant figures that is used in the problem for example, if the question gives us a velocity of 3.00x10^3 and energy of 1.2x10^-10, the smallest amount of sig figs would be 2 because 1.2 has two significant figures. Therefore, our answer ...
- Mon Nov 05, 2018 10:04 pm
- Forum: Empirical & Molecular Formulas
- Topic: Garlic Bread #1
- Replies: 2
- Views: 748
Re: Garlic Bread #1
First, convert all that was given in the equation to moles. .561 g CO2*(1mol Co2/44.01 g CO2) Then, use the mole ratio of how many carbons there are in CO2. So your entire table should look something like: .561 g CO2*(1mol Co2/44.01 g CO2)*(1 mol C/1 mol CO2) CO2 is on the Botton to cancel out, leav...
- Tue Oct 30, 2018 8:14 pm
- Forum: Ionic & Covalent Bonds
- Topic: Covalent Bonds
- Replies: 16
- Views: 1549
Re: Covalent Bonds
Covalent bonds tend to share electrons to complete the octet rule which makes them anions.
- Tue Oct 30, 2018 8:13 pm
- Forum: Ionic & Covalent Bonds
- Topic: Octet Rule
- Replies: 7
- Views: 646
Re: Octet Rule
The first 4 elements are the ones that are an exception (along with others). The elements H, He, Li, and Be are exceptions to the octet rule because they are in the 2s orbital, meaning they can only have most 2 electrons. Here's more info on them in case you are confused: https://lavelle.chem.ucla.e...
- Tue Oct 30, 2018 8:07 pm
- Forum: Ionic & Covalent Bonds
- Topic: 2B.7
- Replies: 3
- Views: 402
Re: 2B.7
Since in D orbitals l=2, that means that l can be 1 and 2 so it breaks the octet rule
- Wed Oct 24, 2018 10:42 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Advice from a Medical Student [ENDORSED]
- Replies: 297
- Views: 382381
Re: Advice from a Medical Student [ENDORSED]
This is what I needed to read! Thanks so much for sharing. During the gap year, aside from studying for the MCAT what do you recommend we do in that free time?
- Mon Oct 22, 2018 8:15 pm
- Forum: DeBroglie Equation
- Topic: Given the energy, calculate the wavelength of y-rays
- Replies: 12
- Views: 3254
Re: Given the energy, calculate the wavelength of y-rays
Hi, First I converted the given keV to eV. I did that by multiplying 140.511 by 1000 and got 140511 eV. Then I used that and multiplied it by 1.602*10^-19 J because that is how many joules per eV. That gave me 2.251x10^-14 J for my energy. Then I set that equal to the (E=hc/lambda) equation to find ...
- Mon Oct 22, 2018 7:16 pm
- Forum: Properties of Light
- Topic: Photons of light
- Replies: 5
- Views: 7725
Re: Photons of light
First, I converted my given wavelength of 420 nm to m (420 x 10^-9) and multiplied the given 32 W to joules (32 J/s). Since the problem gave us a time, I multiplied the time with the Joules/sec to give me 64 total Joules. That is what I would call the TOTAL Energy, not the energy for one photon. So ...
- Wed Oct 17, 2018 3:19 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Planck's Constant
- Replies: 4
- Views: 333
Re: Planck's Constant
In peer learning sessions, we use 6.626*10^-34.
- Wed Oct 17, 2018 3:17 pm
- Forum: Photoelectric Effect
- Topic: Threshold energy [ENDORSED]
- Replies: 17
- Views: 2075
Re: Threshold energy [ENDORSED]
An electron will only be launched/ejected if the threshold energy is greater than or equal to the amount of work needed to do so.
- Wed Oct 17, 2018 3:15 pm
- Forum: *Black Body Radiation
- Topic: Question 1.21
- Replies: 5
- Views: 2313
Re: Question 1.21
The photoelectric effect should be the answer because during the experiment, light (which is also electromagnetic radiation) was shined on a metal in attempt to launch an electron. In trying to do so, they tried amping up the intensity but that still did not eject the electron. Upon more attempts, t...
- Wed Oct 17, 2018 3:08 pm
- Forum: DeBroglie Equation
- Topic: 1B.9
- Replies: 3
- Views: 259
Re: 1B.9
A watt is also J/seconds so you can multiply 32 by 2 to get the total energy, 64 J. Then, use E=(hc)/lambda which will give you the energy of only 1 photon. I believe you then multiply the energy of 1 photon. You can then proceed to use 6.022*10^23 (Avogadro's) to find the number of moles.
- Wed Oct 10, 2018 6:53 pm
- Forum: Empirical & Molecular Formulas
- Topic: F.13
- Replies: 6
- Views: 4273
Re: F.13
I began the question by turning the given 4.14 g P into moles. (4.14 g P)/(30.97g P) = .13367 moles P. (I got 30.97 g from its molar mass on the Periodic Table). Then I went back to the question and subtracted the total mass of what the equation produced, 27.8 g, and subtracted it with the original ...
- Tue Oct 02, 2018 9:23 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Molar Mass
- Replies: 5
- Views: 818
Re: Molar Mass
You would look at the element first and then see its atomic mass on the periodic table. Ex: O=~16g/mol. And since you're asking about O2, we'd multiply that by 2 because we have two of them so it would be 32g/mol.
- Tue Oct 02, 2018 9:16 pm
- Forum: SI Units, Unit Conversions
- Topic: Audio-Visual Focus-Topics, Assessments & Surveys
- Replies: 16
- Views: 1510
Re: Audio-Visual Focus-Topics, Assessments & Surveys
Hi! Even though they aren't required and it is no way identifiable, I found them extremely useful and great as a tool to refresh what happened in the lecture. There are also a lot of good example problems that might help for future quizzes and a great way to review for the final. I plan on doing the...
- Tue Oct 02, 2018 9:12 pm
- Forum: Significant Figures
- Topic: F.15
- Replies: 4
- Views: 356
Re: F.15
When you are adding or dividing or multiplying or subtracting with significant figures, your answer will have the least amount of sig figs because your answer can not be more precise than the least precise measurement. In this case, yes, I also agree that the answer would be 4.56 because 3 is less t...