Search found 64 matches
- Mon Mar 11, 2019 6:58 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Pre equilibrium approach
- Replies: 2
- Views: 281
Re: Pre equilibrium approach
We can substitute the [N2O2] with [NO]^2 because we can use the equilibrium concept that K = [N2O2]/[NO]^2 because the second step in the reaction is slow and creates a bottleneck. So you just have to solve for [N2O2], which is K[NO]^2.
- Tue Mar 05, 2019 7:07 pm
- Forum: First Order Reactions
- Topic: How to calculate [A]t in 7B.3 7th ed
- Replies: 1
- Views: 265
Re: How to calculate [A]t in 7B.3 7th ed
What you are doing is finding the leftover [A] because if you just do (2molA/1molB)(0.034molB/L), it gives you the change in [A] based on how much the [B] changed. That's why you have to subtract the change from the initial to get the [A]t.
- Tue Mar 05, 2019 7:01 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Finding Order for Each Reactant
- Replies: 4
- Views: 503
Re: Finding Order for Each Reactant
It doesn't matter what the ratio is but you should logically get a whole number as 1/4 doesn't make sense when dealing with rates and n value. In one of the cases you will have to flip the ratio after you do the calculations.
- Tue Mar 05, 2019 6:59 pm
- Forum: General Rate Laws
- Topic: does k change if rxn is multiplied?
- Replies: 6
- Views: 1470
Re: does k change if rxn is multiplied?
The Kr shouldn't really change as when you multiply the rxn, both the reactants and products will be affected and should cancel the multiplication.
- Tue Mar 05, 2019 6:47 pm
- Forum: First Order Reactions
- Topic: First Order Integrated Law
- Replies: 3
- Views: 331
Re: First Order Integrated Law
The difference is probably due to what the [A] value is. The - is probably associated with reactants while the equation with the - is associated with products.
- Tue Feb 26, 2019 12:07 am
- Forum: Balancing Redox Reactions
- Topic: Oxidizing vs Reducing Agent
- Replies: 3
- Views: 326
Re: Oxidizing vs Reducing Agent
In this reaction, the element Cr is oxidized while Mn is reduced. Thus, Cr3+ will be the reducing agent and MnO2 will be the oxidizing agent.
- Tue Feb 26, 2019 12:02 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: ln(k2/k1)=(-delta H/R)(1/T2 - 1/T1)
- Replies: 3
- Views: 6394
Re: ln(k2/k1)=(-delta H/R)(1/T2 - 1/T1)
This is a derived equation from the formula of delta G (standard) = -RTlnK where you solve for K for two different T value. You have to combine the two different T equations to find the equation ln(k2/k1)=(-delta H/R)(1/T2 - 1/T1). So technically it is on the equation sheet, but you can memorize it ...
- Mon Feb 25, 2019 11:57 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Pt(s) in cell diagram
- Replies: 4
- Views: 497
Re: Pt(s) in cell diagram
You would add Pt to the Ce4+ side of the diagram as it has no solids to conduct the charges. We use Pt when there are no solids in the half reaction.
- Mon Feb 25, 2019 4:50 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Addition of Platinum to a Cell
- Replies: 1
- Views: 202
Re: Addition of Platinum to a Cell
Platinum is added as it is an inert metal that is able to conduct the current but doesn't interact with the solutions itself. Platinum can be added to either, cathode and anode, or both. Its usually used when there is no solid to conduct the charge in the cell.
- Mon Feb 25, 2019 4:42 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Homework question
- Replies: 3
- Views: 389
Re: Homework question
After balancing the two half-reactions, the number of electrons will be the value of n for the equation.
- Mon Feb 25, 2019 4:40 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: reversing reaction [ENDORSED]
- Replies: 3
- Views: 357
Re: reversing reaction [ENDORSED]
One of the half reactions has to be reversed in order to balance the number of electrons on both sides of the overall reaction.
- Mon Feb 25, 2019 4:37 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Calculations [ENDORSED]
- Replies: 3
- Views: 374
Re: Calculations [ENDORSED]
When referencing entropy, the temperature should be 0 degrees Kelvin as that's when a compound will not have any vibrational or rotational entropy.
- Mon Feb 25, 2019 4:14 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Standard Cell Potential [ENDORSED]
- Replies: 1
- Views: 221
Re: Standard Cell Potential [ENDORSED]
When calculating the standard cell potential, we must to look at appropriate half reactions and combine them to get the appropriate cell potential. Most likely if there is no half-reaction for a compound, then it probably can't be used to find the standard cell potential.
- Tue Feb 12, 2019 6:29 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: 6th edition 9.75
- Replies: 1
- Views: 259
Re: 6th edition 9.75
You will have to draw an octahedral shape with two particles of X and 4 particles of Y. For cis, you have to look at the arrangements were the X particles can be close to each other. This would result in 12 different orientations while for tran you have to make sure the X particles are opposite to e...
- Tue Feb 12, 2019 6:04 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Problem 9.13 6th edition
- Replies: 1
- Views: 328
Re: Problem 9.13 6th edition
The key wording of the problem is "assume ideal behavior", this basically states that assume all the unknown values are that of an ideal gas. In this case that would be that the n value is 1 mole.
- Tue Feb 12, 2019 6:01 pm
- Forum: Calculating Work of Expansion
- Topic: moles and entropy
- Replies: 3
- Views: 402
Re: moles and entropy
1 mol should be used in the equations whenever the problem states that the gas is an Ideal Gas and thus has ideal gas properties. When dealing with ideal gas, you assume 1 mole if not explicitly stated.
- Sat Feb 09, 2019 11:18 am
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: Calculating W
- Replies: 1
- Views: 260
Re: Calculating W
Yes, you are correct that the degeneracy of BF3 would be 1 as you can't distinguish between the positions the F atoms occupy. For BF2Cl the degeneracy value would be 3 as the Cl could occupy any of the three bonds in the triangular planar shape.
- Sat Feb 09, 2019 11:15 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Nonspontaneous Changes
- Replies: 1
- Views: 207
Re: Nonspontaneous Changes
Any form of heat transfer into the system should satisfy the need for nonspontaneous changes as long as the energy is sufficient enough to cause the change.
- Sat Feb 09, 2019 11:12 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Units for Ccal?
- Replies: 1
- Views: 487
Re: Units for Ccal?
I doubt they would mark you off for switching the signs from K to C but most equations use the value of K for a proper answer and if you use the C value you would get an incorrect answer. So it better to just stick to K.
- Sat Feb 09, 2019 11:10 am
- Forum: Calculating Work of Expansion
- Topic: Reversible and constant pressure
- Replies: 2
- Views: 300
Re: Reversible and constant pressure
When the process is reversible and the pressure is held constant, there are changes in the temperature that causes the gas to expand or contract, resulting in the work done on or by the system.
- Fri Feb 01, 2019 8:35 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Enthalpy
- Replies: 3
- Views: 395
Re: Enthalpy
Enthalpy is equal to the internal energy plus the pressure times the volume. Thus, it can be calculated for any reactions. For reactions with a fixed volume, the change in v is 0, therefore, the enthalpy value equals the internal energy.
- Fri Feb 01, 2019 8:19 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Heating Curve
- Replies: 2
- Views: 392
Re: Heating Curve
Melting and vaporizing do have to do with breaking small intermolecular bonds, however, I think that the majority of the heat is transferred to the molecules themselves and is translated into kinetic energy. I infer this as the molecules in the liquid state move around at greater speeds than those a...
- Fri Feb 01, 2019 2:44 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Analogy of Boulder
- Replies: 5
- Views: 996
Re: Analogy of Boulder
This analogy was related to the concept of a reaction occurring spontaneously. I think that Dr. Lavelle just wanted to convey the idea that reactions will have a certain outcome, say the rock rolling down, if not influenced by the surrounding. The rock is not likely to go up the hill, similarly, a r...
- Fri Jan 25, 2019 12:34 am
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: determining x for weak acids and bases
- Replies: 6
- Views: 748
Re: determining x for weak acids and bases
Conjugate acid is the molecule formed on the product side when a base reacts with water and gains hydrogen. Similarly, a conjugate base is the molecule formed when the acid loses hydrogen during a reaction. Thus, each reaction has a conjugate acid/base pair.
- Fri Jan 25, 2019 12:30 am
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Ka
- Replies: 6
- Views: 650
Re: Ka
Since it is a strong acid it doesn't have a Ka value as it dissociates completely in the reaction.
- Fri Jan 25, 2019 12:29 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: K and temp changes
- Replies: 2
- Views: 195
Re: K and temp changes
The simplest way to look at is that for an exothermic reaction, Kc decreases with the increase in temperature, while for an endothermic reaction the Kc increases with increase in temperature. The amount that the Kc value changes by is probably dependent on the reaction.
- Fri Jan 18, 2019 10:39 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Example in class
- Replies: 5
- Views: 555
Re: Example in class
I agree Q is used when the reaction is not at equilibrium. The Q value then will allow us to predict the direction the reaction will proceed when compared to Kc.
- Fri Jan 18, 2019 3:24 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: When is X negligible.
- Replies: 6
- Views: 513
Re: When is X negligible.
Usually, it depends on the problem. You can ignore the X when the K value is smaller than 10^-3. However, for the numerator, it would be best to keep the x as usually, that's the x value that is being solved for.
- Fri Jan 18, 2019 3:21 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Kp or Kc
- Replies: 3
- Views: 268
Re: Kp or Kc
I agree, typically use Kc for aqueous solutions while Kp for gases. You can also convert Kp to Kc using the equation Kp = Kc [RT]^ng, where ng is the sum of the constants of products and reactants.
- Fri Jan 18, 2019 3:14 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE table
- Replies: 2
- Views: 256
Re: ICE table
When the is a coefficient in front of a product, you must use it while constructing the equation for K when solving for x. The coefficient is used as a power value in the equation. Ex. 2NO would be [NO]^2 in a K equation.
- Thu Jan 10, 2019 6:01 pm
- Forum: Ideal Gases
- Topic: Temperature Units
- Replies: 4
- Views: 149
Re: Temperature Units
The temperature will be unique for each of the problems as reactions can take place at different temperatures and the Kc value is also affected by the temperature thus each reaction will have a unique value of K (Kelvin).
- Thu Jan 10, 2019 5:59 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Partial Pressures Definition
- Replies: 2
- Views: 226
Re: Partial Pressures Definition
Partial Pressure is basically the pressure exerted by one gas in a mixture of multiple in a specific volume. You don't have to use partial pressure for gases per se, but gases are often measured by their pressure instead of their concentration. Also, you can't measure pressure for other states thus ...
- Thu Jan 10, 2019 5:55 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE Table Units
- Replies: 4
- Views: 546
Re: ICE Table Units
I agree you should use molarity during the ICE table problems as you have to deal with the value of K, and K requires you to solve for the molarity. So it's more convenient to use the molarity.
- Sat Dec 08, 2018 8:19 pm
- Forum: Hybridization
- Topic: hybridization of the central atom
- Replies: 1
- Views: 375
Re: hybridization of the central atom
So far in this class, we only focus on the central atoms. That and because the atoms usually are not hybridized and thus we don't have to account for that.
- Sat Dec 08, 2018 8:17 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: polarity
- Replies: 1
- Views: 353
Re: polarity
I don't think there is a specific method for determining the polarity of a single atom. When talking about polarity, it is a molecule with two or more atoms. To find the polarity of a molecule, you just have to look at the electronegativity and then treat them as vectors. If the vectors cancel then ...
- Sat Dec 01, 2018 5:46 pm
- Forum: Naming
- Topic: kE in textbook
- Replies: 1
- Views: 192
Re: kE in textbook
I am not completely sure if we need to know this or not. But as for how do we know which is the connecting atom, the atom after the k is the atom that connects to the metal. In the first case N bonds with Fe from NCS while in the second one S bonds with Fe.
- Tue Nov 27, 2018 8:21 am
- Forum: Hybridization
- Topic: hybridization and radicals
- Replies: 2
- Views: 371
Re: hybridization and radicals
The radical should be included in the hybridization as it would still influence the bonds by repelling them away from itself.
- Tue Nov 27, 2018 8:15 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Double Bonds as Regions of Electron Density
- Replies: 5
- Views: 431
Re: Double Bonds as Regions of Electron Density
Double bonds still count as a single region of electron density. This is a crucial idea for the VSEPR model and the shapes of the molecules.
- Tue Nov 27, 2018 8:13 am
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: Drawing a Dipole
- Replies: 6
- Views: 621
Re: Drawing a Dipole
I think Professor Lavelle said that in this chem the arrow should point to the negative charge.
- Wed Nov 21, 2018 10:45 pm
- Forum: Dipole Moments
- Topic: Repulsion
- Replies: 2
- Views: 291
Re: Repulsion
To me, it seems that the electrons would be the cause of the repulsion as the elections with similar spins would repel each other, causing repulsion between the atoms itself.
- Wed Nov 21, 2018 10:24 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lone Pairs on Axial vs. Equatorial Positions
- Replies: 2
- Views: 296
Re: Lone Pairs on Axial vs. Equatorial Positions
I think Dr. Lavelle said that this happens as the molecule wants to be in a state where the lone pairs are not in 90 degrees from the least number of bonds. So if the lone pairs occupy the axial position, they will be 90 degrees to 3 bonds while if they don't then it would be only 2 bonds.
- Wed Nov 21, 2018 10:06 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: HW CH4 4.75 6TH EDITION
- Replies: 1
- Views: 104
Re: HW CH4 4.75 6TH EDITION
a) The elemental composition would give us the empirical formula CH4O, which agrees with the molar mass. This would give us a Lewis structure of the compound methanol. The Lewis structure would have a single bond between C and O with three H atoms bonded to O with single bonds and the fourth H bonde...
- Sat Nov 17, 2018 9:38 pm
- Forum: Sigma & Pi Bonds
- Topic: Sigma and Pi Bonds affect on structure
- Replies: 4
- Views: 413
Re: Sigma and Pi Bonds affect on structure
The number of sigma and pi bonds don't influence the shape or structure of a molecule. This is because when determining the molecular shape, instead of the number of bonds we look at the electron regions as a single charged region.
- Sat Nov 17, 2018 9:23 pm
- Forum: Sigma & Pi Bonds
- Topic: Sigma and pi bond
- Replies: 1
- Views: 136
Re: Sigma and pi bond
The first bond to form is always the sigma bond. There can only be one sigma bond between two atoms and any bonds formed after are the pi bonds due to the way the electrons of each atom interact with each year. Thus, a double bond has one sigma bond and pi bond, while triple bonds are made of one si...
- Sat Nov 17, 2018 9:19 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Molecular Shape
- Replies: 2
- Views: 222
Re: Molecular Shape
Molecule shapes can be predicted based on the Lewis dot structures using the VSEPR theory. It states that electron pairs in the valence shell of an atom repel each other, resulting in the molecular geometry and shape.
- Sat Nov 10, 2018 8:48 pm
- Forum: Bond Lengths & Energies
- Topic: Energy of Intermolecular forces
- Replies: 2
- Views: 319
Re: Energy of Intermolecular forces
The negative values that we discussed represent the fact that -250KJ/mol of energy is needed to break a mole of the bonds. It is negative as the surrounding loses energy in the process of breaking the bonds.
- Sat Nov 10, 2018 9:46 am
- Forum: General Science Questions
- Topic: The 4th Dimension
- Replies: 1
- Views: 188
Re: The 4th Dimension
This question is quite interesting and really got me thinking. Personally, I don't think chemical structures can be expressed in 4-D. The reasoning for my thinking is because all these chemical structures exist in our world and thus must be bound by the 3-D model of our world. Even if there were 4-D...
- Sat Nov 10, 2018 9:35 am
- Forum: Dipole Moments
- Topic: Fluctuation
- Replies: 3
- Views: 359
Re: Fluctuation
Based on what we have learned so far, to me it seems that we can never be certain of electrons location in the orbitals. However, when a significant amount of the electrons are supposedly at one side of the atom, they can create a dipole moment that then induces a dipole moment in the adjacent molec...
- Sat Nov 10, 2018 9:30 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Repulsion Strength
- Replies: 2
- Views: 211
Re: Repulsion Strength
That list compares the bond strength of the three different interactions that occur around the central atom involving electrons. The list just states that the repulsion between two pairs of lone electrons is greater than the repulsion between a lone pair and a bonded pair. Finally, the interaction b...
- Sat Nov 03, 2018 9:50 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Valence shells
- Replies: 1
- Views: 206
Re: Valence shells
Yes, when there are elections in the 4s and 4p shells, 3d shell doesn't contribute to the number of valence electrons due to the fact that valence electrons are calculated from the highest energy levels. The d orbitals only contribute to the valence electrons for any transition metals as the d orbit...
- Sat Nov 03, 2018 9:41 am
- Forum: Ionic & Covalent Bonds
- Topic: Delta Negative
- Replies: 2
- Views: 177
Re: Delta Negative
To my understanding, the Delta negative region is referencing the part of a molecule where the difference in the electronegativity between the atoms of a molecule forms a negative charge. This is also the concept of the dipoles, where one part of the molecule become negative and the becomes positive...
- Sat Nov 03, 2018 9:36 am
- Forum: Lewis Structures
- Topic: How many valence electrons?
- Replies: 1
- Views: 1768
Re: How many valence electrons?
There are 5 valence electrons in [Ar] 3d10 4s2 4p3 (Arsenic), and 7 valence electrons in [Ar] 3d5 4s2 (Manganese). The transition metals can use the electrons in their d orbitals to bond while the other elements can't. Thus, Arsenic only has valence electrons in the s and p shells, while Manganese h...
- Fri Oct 26, 2018 10:08 pm
- Forum: Octet Exceptions
- Topic: Octet Exceptions
- Replies: 3
- Views: 513
Re: Octet Exceptions
H and He are exceptions to the Octet Rule because for them a full shell is only 2 electrons (the 1s shell). Li and Be are metals, and according to the periodic trends they are likely to lose electrons and become cations (Li+, Be2+). When that happens, they would have an electron configuration of a &...
- Fri Oct 26, 2018 9:57 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 2.43 6th edition
- Replies: 1
- Views: 232
Re: 2.43 6th edition
You are supposed to always write the lowest energy levels based on their n number. Initially, it is easier to fill the 6s orbital but when the electrons begin to fill the 5d orbital, the overall energy of the 5d orbital is lower than the 6s orbital. Thus, we have to write the configuration from the ...
- Fri Oct 26, 2018 9:44 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Writing electron configurations [ENDORSED]
- Replies: 5
- Views: 270
Re: Writing electron configurations [ENDORSED]
I don't think it matters if its positive or negative, but we usually keep it positive as to not get confused while working with more than one electrons in the p, d, and f orbitals. This is because all the electrons must be of the same charge before they start pairing up. But there is probably a much...
- Sat Oct 20, 2018 3:23 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Homework Question 1E.1
- Replies: 1
- Views: 243
Re: Homework Question 1E.1
When the electron undergoes the transition from 1s-orbital to a 2p-orbital all of the choices are correct. This is because, when it jumps to a higher value of n (n=2) it gains energy as the n increased from 1 to 2. Also, the value of l increases as the p subshell represents l = 1 while the s subshel...
- Sat Oct 20, 2018 3:18 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Homework Question 1.D13
- Replies: 2
- Views: 224
Re: Homework Question 1.D13
There will be 7 values for l when n=7 but the highest value of l would be a 6. This is because l includes 0 as one of the values and to find the highest value of l we can use the equation (n-1).
- Sat Oct 20, 2018 3:15 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: angular node
- Replies: 1
- Views: 86
Re: angular node
Yes, they are the same thing. As far as I know, they represent the same concept that no electron can be found within the region they represent. However, the plane might represent a plane shaped like a square, but the angular node could be other shapes such as the cone in the dz2 orbital.
- Thu Oct 11, 2018 10:44 pm
- Forum: Photoelectric Effect
- Topic: Kinetic Energy of Electrons
- Replies: 2
- Views: 202
Re: Kinetic Energy of Electrons
This is because the work function, or the energy required to cause an electron ejection is fixed. So, based on the equation E = hv, as the frequency increases, the energy of the photon increases. ANd because the work function is a fixed value, the remaining energy is transferred to the ejected elect...
- Thu Oct 11, 2018 10:39 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Uncertainty in Kinetic energy of electron [ENDORSED]
- Replies: 2
- Views: 445
Re: Uncertainty in Kinetic energy of electron [ENDORSED]
In this case yes!! Just use the equation E = 1/2 m v^2. Using the uncertainty in the velocity in the equation will give you the uncertainty in the kinetic energy of the electron.
- Thu Oct 11, 2018 10:32 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: H-Atom
- Replies: 1
- Views: 144
Re: H-Atom
When the hydrogen gets excited, the electron doesn't completely leave the hydrogen. The electron just absorbs the energy and gains energy levels, meaning it is in an excited state but still influenced by the pull from the hydrogen atom's nucleus. When the electron goes back to the ground state(where...
- Wed Oct 03, 2018 10:24 pm
- Forum: Empirical & Molecular Formulas
- Topic: Molar Mass: E21
- Replies: 3
- Views: 339
Re: Molar Mass: E21
Yes, first convert all the values to their SI Unit equivalent. Then, find the molar mass of the compound in question using the periodic table. After getting the molar mass, fin the moles of the substance by dividing the amount present by the molar mass of that substance. Finally, to find the number ...
- Wed Oct 03, 2018 10:04 pm
- Forum: Significant Figures
- Topic: E23C Sig Figs
- Replies: 2
- Views: 220
Re: E23C Sig Figs
The number of sig figs that you should use should be based on the measured value in the question, in this case, 25.2 kg. Thus, you should have 3 sig figs for your answer.
- Wed Oct 03, 2018 9:57 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: V=n/c
- Replies: 7
- Views: 8880
Re: V=n/c
That equation is derived from the equation c = n/V where c represents the molarity of a solution, n the moles of the substance being dissolved and V represents the volume of the solution.