Search found 62 matches
- Wed Mar 13, 2019 3:29 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: 6th edition 15.49
- Replies: 2
- Views: 358
Re: 6th edition 15.49
I think that holds true; for the steps of the rate law itself, you can include the intermediate. This allows you to substitute a factor of the later equations to find the overall rate law, which should not include the intermediate.
- Wed Mar 13, 2019 3:25 pm
- Forum: First Order Reactions
- Topic: pseudo vs 2nd order
- Replies: 4
- Views: 651
Re: pseudo vs 2nd order
Is it safe to assume that we'll always be told which reactants will be in large excess to us to be able to write the pseudo rate law?
- Wed Mar 13, 2019 3:15 pm
- Forum: Zero Order Reactions
- Topic: Concept
- Replies: 4
- Views: 764
Re: Concept
Yeah if you think about the general rate law, k[A]^n, for a zero order reaction n would equal 0. By doing that, the rate = -kt, where it can be seen that the concentration of the reactant, [A], has no effect on the rate. Thus, it can be conceptually inferred that for zero order reaction, the rate is...
- Wed Mar 06, 2019 2:25 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Integrals/derivatives
- Replies: 3
- Views: 474
Re: Integrals/derivatives
Yeah, I think we'll have to know basic integrals, especially with deriving certain equations and the rate law formulas. But for the most part, I think the one you'll really have to remember is the integral of 1/x, which turns out to be ln(x). Other than that, most of them are simple ones that you ho...
- Wed Mar 06, 2019 2:16 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Gibbs Free energy and photosynthesis
- Replies: 4
- Views: 546
Re: Gibbs Free energy and photosynthesis
Yeah at first I was thinking it'd be spontaneous because plants photosynthesize "automatically," but then it's also because of the energy of the sun that they are able to do this. Thus I would agree, that this reaction would be non spontaneous.
- Wed Mar 06, 2019 2:00 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: n in Nernst
- Replies: 4
- Views: 580
Re: n in Nernst
I also was first confused with this, but the n here is talking about the moles of electrons transferred! So yes to emphasize, you must balance the redox equations correctly and then use the moles of electrons transferred as the n.
- Wed Feb 27, 2019 3:39 pm
- Forum: Van't Hoff Equation
- Topic: van't hoff equation clarification
- Replies: 2
- Views: 525
Re: van't hoff equation clarification
Yes, you are able to rearrange the van't hoff equation to solve for K, where K = e^(-enthalpy/RT + entropy/R).
- Wed Feb 27, 2019 3:33 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.11 6th edition, part D
- Replies: 2
- Views: 327
Re: 14.11 6th edition, part D
That didn't make too much sense to me either but what Ashley is saying does make sense, as you don't include H2O(l) in cell diagrams and you still do want to separate the 2 different phases!
- Wed Feb 27, 2019 3:11 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Pt (s) [ENDORSED]
- Replies: 8
- Views: 972
Re: Pt (s) [ENDORSED]
Why do you not need to add Pt(s) to liquids?
- Tue Feb 19, 2019 1:59 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.69 in Sixth Edition
- Replies: 1
- Views: 224
Re: 9.69 in Sixth Edition
Since reactions 2 and 3 are the ones used to drive the regeneration of ATP, I would first use them with the Hess's Law-like method to figure out the ∆G of the reaction, taking into account 3 moles of NADH. Then to get it with respect to ATP, divide by the ∆G you calculated divded by the ∆G for the r...
- Tue Feb 19, 2019 1:53 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.81
- Replies: 2
- Views: 330
Re: 9.81
I would think of it more in terms of the chemical reaction first; knowing that one will be the reactant and product, and that the reactant will be oxidized with O2, you can deduce the equation and balance it accordingly. Then I would calculate ∆G using the values in Appendix 2A to then see if the fo...
- Tue Feb 19, 2019 1:49 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Cp vs Cv
- Replies: 7
- Views: 1101
Re: Cp vs Cv
I've found that most problems we deal with have to do with temperature change; you will then use Cv because it is the pressure that is changing with the system rather than the volume (even if there is volume expansion occuring but that will be taken into account with a different equation). You will ...
- Tue Feb 19, 2019 1:47 pm
- Forum: Phase Changes & Related Calculations
- Topic: Specific Heat of Ice
- Replies: 3
- Views: 2926
Re: Specific Heat of Ice
Yeah I always find it helpful to draw out the phase change diagram; if you start below 0 degrees celsius when you are still in the ice phase, then you need to use the specific heat capacity of ice. We generally use the specific heat capacity of water in our problems because we often deal with ice me...
- Tue Feb 19, 2019 1:42 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Extensive vs Intensive Properties
- Replies: 2
- Views: 302
Re: Extensive vs Intensive Properties
What makes something an extensive property is that if the path taken to achieve the final state from the initial state must be taken into account; for things like enthalpy or entropy, what we care about is the initial and final states of the reaction rather than what happens in between. But for a pr...
- Tue Feb 19, 2019 1:39 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: q=0
- Replies: 4
- Views: 530
Re: q=0
Yes, because if temperature is constant in this type of isolated system, then heat is never lost nor gained. Thus q, heat, is not changing within the system and can be set equal to 0. And because we know that ∆U = q + w, if q = 0 then that leaves ∆U = w.
- Wed Feb 06, 2019 2:22 pm
- Forum: Calculating Work of Expansion
- Topic: Degeneracy
- Replies: 6
- Views: 839
Re: Degeneracy
As others previous have said, degeneracy is the number of possible ways a system can exist in a given/specific energy state. So W, degeneracy, = (number of states) ^ number of particles.
- Wed Feb 06, 2019 2:11 pm
- Forum: Calculating Work of Expansion
- Topic: Pressure Units
- Replies: 7
- Views: 783
Re: Pressure Units
Are we always supposed to being the unit for atm in terms of calculations with pressure?
- Wed Feb 06, 2019 1:47 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Bomb Calorimeter
- Replies: 3
- Views: 387
Re: Bomb Calorimeter
Then would it be fair to say that a bomb calorimeter is an isolated system?
- Wed Jan 30, 2019 2:16 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Qsystem+Qsurr=0?
- Replies: 6
- Views: 1886
Re: Qsystem+Qsurr=0?
It's kind of like one of the laws of thermodynamics which I know we haven't covered yet but might help explain the concept of it; energy can never be created nor destroyed. With that in mind, the heat of the system + the heat of the surrounding would equal 0, because as one loses heat such that the ...
- Wed Jan 30, 2019 2:03 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Lecture Question
- Replies: 3
- Views: 325
Re: Lecture Question
Yeah just to re-iterate, the standard enthalpy of formation has to always do with getting something to its most stable form; so for diatomic molecules (like O2) it is already in its most stable form, thus its standard enthalpy of formation would be 0.
- Wed Jan 30, 2019 1:58 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Law
- Replies: 4
- Views: 645
Re: Hess's Law
I usually like to start with noticing where the products and reactants are from the final equation into the ones you're given; regardless of the coefficients it's important to first get them on the proper side. From there, often times things will begin to cancel out, but if not then I think that at ...
- Wed Jan 23, 2019 2:10 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 7th edition 5J.5 part b
- Replies: 3
- Views: 408
Re: 7th edition 5J.5 part b
Thinking similarly to Kc or Kp, we don't include solids or liquids into them. So when determining how pressure will affect equilibrium, just like we would for Kp, we would not include the solid into consideration of equilibrium shift. Thus if pressure increases, we can use the shorthand rule of favo...
- Wed Jan 23, 2019 2:08 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Endothermic vs. Exothermic
- Replies: 8
- Views: 821
Re: Endothermic vs. Exothermic
For exothermic, you can think of heat being released, so it would be reactants --> products + heat. Vice versa, for an endothermic reaction heat is required for the reaction to proceed, so it would look like reactants + heat --> products. So if temperature was increased, equilibrium would shift oppo...
- Wed Jan 23, 2019 2:06 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Ideal Gases
- Replies: 2
- Views: 369
Re: Ideal Gases
Yeah so ideal gases like Helium and nitrogen, if the reaction vessel had none of them ideal gas within the reactants or products of the reaction, then ultimately the concentrations of them do not change. We learned the quick way that if volume is decreased (pressure increased) then equilibrium would...
- Wed Jan 16, 2019 4:22 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.73 6th edition
- Replies: 2
- Views: 314
Re: 11.73 6th edition
The rule that Jane said about which side will be favored (which ever has less will be favored) is a handy trick, but just remember also that if there is the use of an inert gas to increase pressure of the system, then the system will remain at equilibrium as the gas has no real affect on neither the...
- Wed Jan 16, 2019 4:17 pm
- Forum: Ideal Gases
- Topic: Reducing Volume
- Replies: 3
- Views: 406
Re: Reducing Volume
It's easier to think in terms of concentration; although the pressure of the vessel will definitely change with the addition of the inert gas, the inert gas has no real effect to the system equation itself. This is because this inert gas will not change the concentration (or partial pressure) of nei...
- Wed Jan 16, 2019 2:45 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: question from module
- Replies: 4
- Views: 414
Re: question from module
Something that helps me with temperature shift is that when it's a negative delta h, you can think of there being a product of heat (+heat on the product side) and when delta h is positive then there is +heat on the reactant side. Thus, if temperature is decreased, then equilibrium would shift to wh...
- Wed Jan 09, 2019 3:40 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Gibbs Free Energy
- Replies: 4
- Views: 442
Re: Gibbs Free Energy
Yeah for chapter 11 the syllabus said to skip section 11.3 which slightly covers that so I think that we're meant to cover that later in the course!
- Wed Jan 09, 2019 3:38 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium constant for Gas
- Replies: 1
- Views: 199
Re: Equilibrium constant for Gas
I'd say you have to use the partial pressure notation since [] implies the concentration of something; my TA said it's okay to use the [] for Kc but use partial pressure notation for Kp
- Wed Jan 09, 2019 3:36 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Initial Concentration
- Replies: 3
- Views: 303
Re: Initial Concentration
When using ICE tables there will often be initial concentrations of 0, like for products (in forward reactions). In that example aspect, by solving for the "x" value, you are able to find the equilibrium concentration formed by the reactants for those products so I would assume you would j...
- Tue Dec 04, 2018 6:26 pm
- Forum: Dipole Moments
- Topic: net dipole
- Replies: 2
- Views: 344
Re: net dipole
The concept of being dipole to my understanding relates to polarity; if a molecule is nonpolar, like CH4, then there is no net dipole since the 4 hydrogens are equally pulling on the carbon in all 4 directions. A molecule such as water has a net dipole, as there is a net dipole from the lone pairs u...
- Tue Dec 04, 2018 6:15 pm
- Forum: Electronegativity
- Topic: Ionic Character
- Replies: 7
- Views: 1567
Re: Ionic Character
Yeah there is no "definite" trend to electronegativity like there is for trends of atomic radius (and others we learned), but in my head I like to remember that F is the most electronegative element, and then next one would be O. So I guess going down a group results in a lower electronega...
- Tue Dec 04, 2018 6:10 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: ph vs pOH
- Replies: 5
- Views: 546
Re: ph vs pOH
Just to clarify then, is the sum of pH + pOH always equal 14 then?
- Wed Nov 28, 2018 12:29 pm
- Forum: Dipole Moments
- Topic: dipole moment
- Replies: 2
- Views: 286
Re: dipole moment
The dipole moment tells us the measure of the polarity of the molecule, happening from the separation of charge. This relates to the special ionic/covalent character we learned about, how there are partial charges to make a covalent bond have ionic properties by the asymmetric distribution of electr...
- Wed Nov 28, 2018 12:26 pm
- Forum: Dipole Moments
- Topic: 4.29
- Replies: 3
- Views: 199
Re: 4.29
It is precisely because the two chlorines are positioned close together that they have the largest dipole moment, because then there is a stronger force of repulsion due to their close proximity. Figure 3 would have essentially no dipole moment, as the symmetry has the dipole moments "cancel ea...
- Wed Nov 28, 2018 12:18 pm
- Forum: Electronegativity
- Topic: Electronegativity Values
- Replies: 4
- Views: 744
Re: Electronegativity Values
Yeah I think it was mentioned in lecture that we didn't have to memorize values of electronegativity (that would also be a lot!), but rather just know the general trend of it: it increases as you go from left to right of a period where F is the most electronegative and a has somewhat trend of decrea...
- Wed Nov 21, 2018 10:47 am
- Forum: Hybridization
- Topic: unhybridized pi bonds
- Replies: 2
- Views: 339
Re: unhybridized pi bonds
I am not too sure in my knowledge about this, but I think that both have hybridized orbitals. Since hybridization is defined as the phenomenon of intermixing of the orbitals such as sp, sigma and pi bonds are just different types of covalent bonds formed depending on the way the atomic orbitals hybr...
- Wed Nov 21, 2018 10:37 am
- Forum: Hybridization
- Topic: hybridization with bonds
- Replies: 2
- Views: 252
Re: hybridization with bonds
I'm actually not sure if given only the hybridization orbital of a molecule that you can figure out the type of bonds, as the hybridization orbital doesn't really give insight into whether there are double or triple bonds present. You would mainly need the molecular formula to draw out the Lewis str...
- Wed Nov 21, 2018 10:33 am
- Forum: Hybridization
- Topic: predicting hybrids
- Replies: 3
- Views: 284
Re: predicting hybrids
For simplicity's sake, I usually do what Iris does. But you also have to take into consideration of the lone pairs, so if you had for instance 2 bonded pairs and a lone pair, that would still be sp2. This way you can also predict molecular shape, as with my example of 2 bonded and 1 lone pair, it's ...
- Wed Nov 14, 2018 12:29 pm
- Forum: Octet Exceptions
- Topic: Exceptions
- Replies: 10
- Views: 1393
Re: Exceptions
So just building onto that, the most common exceptions that we deal with are found within the 3rd period, such as Sulfur or Phosphorus, where they violate the octet rule due to having that d block available to bind to more electrons. Additionally, another exception would be for elements like Boron a...
- Wed Nov 14, 2018 12:19 pm
- Forum: Bond Lengths & Energies
- Topic: Bond Length
- Replies: 9
- Views: 1140
Re: Bond Length
It's hard to say if it would be the exact average of it as it would depend on the atoms present within the molecule and their associated pull to affect the bond length. But I would agree that it would be fairly close to the average of the two, but may lean slightly towards one of the bond lengths nu...
- Wed Nov 14, 2018 12:16 pm
- Forum: Bond Lengths & Energies
- Topic: Bond Length and Resonance
- Replies: 5
- Views: 1742
Re: Bond Length and Resonance
Just building off, the number of bonds do affect bond length in that the greater amount of bonds, the shorter the bond length (so triple bond is the shortest). This is because a triple bond is much stronger than a single bond, and thus would have much more pull between the two atoms to thus have a s...
- Wed Nov 07, 2018 11:45 am
- Forum: Properties of Light
- Topic: Exam 2 Question 4A
- Replies: 6
- Views: 886
Re: Exam 2 Question 4A
Which two light equations did you use? Because based on the wording of the question, I think only E = hv would need to be used as you are already given the energy of the photon. So you would want to convert the 3.61 x 10^-22 kJ into J (*1000), and then set that equal to hv, where h is equal to Planc...
- Wed Nov 07, 2018 11:40 am
- Forum: Einstein Equation
- Topic: Equations for Light Only
- Replies: 3
- Views: 891
Re: Equations for Light Only
You can't use E = hv (or E = hc/wavelength for that matter) for electrons as that measures the energy for photons only. De Broglie's is what you would use to figure out wavelength for electrons given it's mass and velocity (sometimes you'll have to solve for velocity using the kinetic energy formula...
- Wed Nov 07, 2018 11:26 am
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: When to multiply by 2
- Replies: 5
- Views: 971
Re: When to multiply by 2
No, if you're just given the velocity you would use that value given. You'd only multiply by 2 if given a range as stated before, like a velocity +/- .35 (for example), as that the .35 dictates the uncertainty and must multiply by 2 to account for the +/- range of it.
- Wed Oct 31, 2018 12:14 pm
- Forum: Trends in The Periodic Table
- Topic: Periodic table
- Replies: 4
- Views: 538
Re: Periodic table
Is there an explicit electronegativity trend? Or should we just know that the elements in the upper right corner (beside noble gases and He) are the most electronegative?
- Wed Oct 31, 2018 12:12 pm
- Forum: Trends in The Periodic Table
- Topic: Smallest ionic radius [ENDORSED]
- Replies: 6
- Views: 832
Re: Smallest ionic radius [ENDORSED]
In class it was discussed that anions are always larger than their parents ions (and cations are smaller). Phosphorus is the the largest anion now with the addition of 3 electrons to shield the attraction by the nucleus, whereas the chloride anion only has one extra election, providing not as much s...
- Wed Oct 31, 2018 12:09 pm
- Forum: Trends in The Periodic Table
- Topic: 1F3 Periodic Table Trends [ENDORSED]
- Replies: 2
- Views: 280
Re: 1F3 Periodic Table Trends [ENDORSED]
Just to add on, we learned that anions are always larger than their parents ions (and cations are smaller). Relating to what Brian said, phosphorus is the the largest anion now with the addition of 3 electrons to shield the attraction by the nucleus, whereas the chloride anion only has one extra ele...
- Wed Oct 31, 2018 11:59 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: 4d^10 and 5s rule
- Replies: 2
- Views: 3369
Re: 4d^10 and 5s rule
Going off of Becky, the rule is that the s orbital has higher energy than the d orbital when there is an electron present in the s orbital (s > d), but when there is no electron in the s orbital, it has lower energy than the d orbital (s < d). Thus in electron configurations, because you have to fil...
- Wed Oct 24, 2018 11:59 am
- Forum: Trends in The Periodic Table
- Topic: Ionization Energies
- Replies: 3
- Views: 470
Re: Ionization Energies
To add on, the second ionization energy will always be higher than the first ionization energy as well! The second ionization energy is the energy required to remove a second electron after one has already been removed (first ionization energy), so because of the greater positive charger from the lo...
- Wed Oct 24, 2018 11:49 am
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: How do I identify excited state through the electronic configuration
- Replies: 5
- Views: 5025
Re: How do I identify excited state through the electronic configuration
For the most part, I think you can tell an electron is in its excited state if the configuration deviates from what would be expected; in both (a) and (c) there are the abnormalities of filling Px and filling p before s respectively, thus representing an excited state. As for (b), I also want to emp...
- Wed Oct 24, 2018 11:41 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Homework Ch.2 Question 15, 6th Edition
- Replies: 6
- Views: 475
Re: Homework Ch.2 Question 15, 6th Edition
The number of nodal planes is equal to the value of l. So for 3p, l =1, thus there would only be one nodal plane.
- Tue Oct 16, 2018 11:54 am
- Forum: Einstein Equation
- Topic: Calculate Number of Photons
- Replies: 3
- Views: 251
Re: Calculate Number of Photons
I like to think of it in terms of the units; when you calculate the energy from the lamp, you get your answer in Joules. Then, when you calculate the energy corresponding to the wavelength and therefore the photon, you are left with Joules/photon. Thus, if you divide Joules by Joules/photon, you are...
- Tue Oct 16, 2018 11:42 am
- Forum: Photoelectric Effect
- Topic: Homework Ch. 1 Question 25; 6th Edition
- Replies: 4
- Views: 566
Re: Homework Ch. 1 Question 25; 6th Edition
If it helps, you can think about it in terms of units to see how it works out! The answer from part A is energy given in joules/atom: For part b, once you convert mg of Na atoms to number of Na atoms, you can multiply the energy answer from part a with the atoms you have calculated to just get overa...
- Tue Oct 16, 2018 11:12 am
- Forum: Properties of Light
- Topic: Vocab
- Replies: 5
- Views: 478
Re: Vocab
I think what JT means by discrete values are values that are of existence; it can't be something like infinity or an imaginary number like i, but rather a number that we can quantify in numerical values.
- Wed Oct 10, 2018 11:48 am
- Forum: Limiting Reactant Calculations
- Topic: How to determine the limiting reactant???? [ENDORSED]
- Replies: 8
- Views: 12609
Re: How to determine the limiting reactant???? [ENDORSED]
It also took me some time to really understand it; personally I like to just take each reactant and calculate the moles or grams of the product (depending on what the question wants for the final answer) to see which one produces the least amount of product as that would be your limiting reactant. B...
- Wed Oct 10, 2018 11:42 am
- Forum: Properties of Light
- Topic: Summary Notes Clarification
- Replies: 3
- Views: 431
Re: Summary Notes Clarification
Adding on to Seohyun's reply, this concept is due to the fact that a long wavelength has a smaller frequency than waves with a short wavelength. The energy of a photon is determined by the E = hv equation, thus not acting in accordance to what people would believe for the intensity of light to emit ...
- Wed Oct 10, 2018 11:25 am
- Forum: Properties of Electrons
- Topic: Energy per photon equation
- Replies: 3
- Views: 389
Re: Energy per photon equation
I agree with the other students who have responded; in the equation E = hv, we learned that h is Planck's constant, equal to 6.626 x 10^-34 Js. Thus if you replace that into the equation to solve for the energy of the photon, you are left with E = (6.626 x 10^-34 Js) (v), leaving frequency of the wa...
- Thu Oct 04, 2018 10:29 am
- Forum: SI Units, Unit Conversions
- Topic: Fundamentals Problem E15
- Replies: 4
- Views: 359
Re: Fundamentals Problem E15
I was also confused by the use of M and the word sulfide appearing without being in the equation; for a further clarification the M is just a variable for the "mystery element" they want you to find by subtracting the molar mass of (OH)2 from the given mass. From there, you find the mass o...
- Thu Oct 04, 2018 10:26 am
- Forum: SI Units, Unit Conversions
- Topic: Homework Question Edition 6 E1
- Replies: 4
- Views: 459
Re: Homework Question Edition 6 E1
I think the answer is given in kilometers because in class we learned that the SI fundamental unit for distance is kilometers (km), but I am also with you on generally answering in meters! I feel like most answers in the solution manual has units of meters, so I'm not sure how pressing this matter i...
- Thu Oct 04, 2018 10:06 am
- Forum: SI Units, Unit Conversions
- Topic: Naming Compounds
- Replies: 9
- Views: 850
Re: Naming Compounds
I saw that you used S, which is actually sulfur rather than sulfate. The compound of sulfate is actually SO4, so the molecular formula for magnesium sulfate heptahydrate is MgSO4(H2O)7. I personally like to keep the 7 separate to help me remember that there are 7 moles of H2O within this compound, b...
- Thu Oct 04, 2018 12:15 am
- Forum: SI Units, Unit Conversions
- Topic: formula units [ENDORSED]
- Replies: 69
- Views: 33034
Re: formula units [ENDORSED]
I think the "pm" you're referring to is the measurement unit of a picometer, which is 10^-12! So to convert that to the unit of a meter which most questions have been using, you would multiply your answer by 10^12.