Search found 60 matches

by Kelsey Warren 1I
Mon Mar 11, 2019 7:48 pm
Forum: General Rate Laws
Topic: 7A15 how to find order of reactant (no similar C values)
Replies: 2
Views: 37

Re: 7A15 how to find order of reactant (no similar C values)

First you have to notice that when A and B are constant and C changes, the rate doesn't change so C must be 0 order. Then you can ignore C and just compare A and B when B and A respectively are constant to find their orders and the overall order.
by Kelsey Warren 1I
Mon Mar 11, 2019 7:47 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Y-Axis Units
Replies: 2
Views: 129

Re: Y-Axis Units

For 0 order reactions, the y-axis is [A] to make a straight line graph. For 1st order, it's ln[A], and for 2nd order it's 1/[A]. You can figure out the units by always making the rate have units mol/Ls (or whichever concentration/time units you're given).
by Kelsey Warren 1I
Mon Mar 11, 2019 7:43 pm
Forum: General Rate Laws
Topic: Unique Rate Law
Replies: 4
Views: 63

Re: Unique Rate Law

We learned the unique rate law as unique rate=-(1/a)(d[A]/dt), and you can set this equal to the same relationship for products if you make the 1/coefficient term positive instead of negative.
by Kelsey Warren 1I
Tue Mar 05, 2019 8:12 pm
Forum: First Order Reactions
Topic: First Order Reactions Graph
Replies: 3
Views: 54

Re: First Order Reactions Graph

Both graphs really mean the same thing and can be about the exact same reaction, you're just plotting [A] vs time and then ln[A] vs time (which produces a linear graph from the exponential decay).
by Kelsey Warren 1I
Tue Mar 05, 2019 8:11 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Rate Constant
Replies: 3
Views: 49

Re: Rate Constant

Something else to remember is that you can calculate k if the reaction rate, concentration of reactants, and the order of the reaction are known. Also, k is always positive and is the slope of the first order graph of ln[A] vs time.
by Kelsey Warren 1I
Tue Mar 05, 2019 8:06 pm
Forum: First Order Reactions
Topic: First Order Reaction
Replies: 3
Views: 41

Re: First Order Reaction

[A] is just the concentration of the reactant A at any time t throughout the reaction, whereas [A]o is the concentration exactly at the beginning of the reaction.
by Kelsey Warren 1I
Wed Feb 27, 2019 5:09 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Thermodynamic Stability
Replies: 2
Views: 73

Re: Thermodynamic Stability

If you're given a reaction and told that the products are thermodynamically stable, then the forward reaction will be favored. In contrast, if the reactants are thermodynamically stable, then the reverse reaction would be favored. (And vice versa for if you're told the products/reactants are thermod...
by Kelsey Warren 1I
Wed Feb 27, 2019 5:05 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.15 6th Edition
Replies: 1
Views: 31

Re: 14.15 6th Edition

I don't have the 6th edition textbook so I don't know which problem you're talking about, but usually the cell diagram will include KOH (aq) if the reaction is taking place in a basic solution. So it basically just means that OH- is present in the solution and is part of the electron transfer or is ...
by Kelsey Warren 1I
Wed Feb 27, 2019 5:03 pm
Forum: Balancing Redox Reactions
Topic: oxidation numbers (hw 6k.1 7th ed)
Replies: 2
Views: 37

Re: oxidation numbers (hw 6k.1 7th ed)

I think I understand what you're asking (?), and I'm pretty sure that what they give you depends on the compounds involved in the problem and what they're asking you to calculate in the problem. I'm sure that Dr. Lavelle will give us the appropriate oxidation numbers on any test/final problem.
by Kelsey Warren 1I
Mon Feb 18, 2019 4:13 pm
Forum: General Science Questions
Topic: Delta G equaling w
Replies: 2
Views: 169

Re: Delta G equaling w

I could definitely be wrong, but I think that delta G always equals the maximum energy available to do work because delta G takes into account entropy/temp changes/heat changes already.
by Kelsey Warren 1I
Mon Feb 18, 2019 4:10 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 6th edition 9.65
Replies: 2
Views: 51

Re: 6th edition 9.65

The book uses the delta S values of the elements that make up the compound because the question is asking about their stability "with respect to the elements." So just use those delta S values to see which becomes less stable as temp is raised.
by Kelsey Warren 1I
Mon Feb 18, 2019 4:06 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Compound stability with respect to decomposition
Replies: 3
Views: 41

Re: Compound stability with respect to decomposition

If delta G is negative, then the forward reaction is favored, which means that the reactants are unstable and will decompose into the products spontaneously.
by Kelsey Warren 1I
Tue Feb 12, 2019 5:12 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: units
Replies: 2
Views: 51

Re: units

q and U will always be in J or kJ, but delta H can be in J/mol or J/g or just J or kJ, etc., depending on what type of delta H it is in that problem.
by Kelsey Warren 1I
Tue Feb 12, 2019 5:10 pm
Forum: Phase Changes & Related Calculations
Topic: HOTDOG #3B
Replies: 1
Views: 53

Re: HOTDOG #3B

Basically, you calculate the heat that was used to increase the temp of the 125g from -2.8C to 0C. Then, subtract that amount from the total energy and you'll have the amount of heat that was used to convert half the ice cream (62.5g) into liquid, so you divide that leftover heat by 62.5g to get the...
by Kelsey Warren 1I
Tue Feb 12, 2019 5:08 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: 8.45
Replies: 2
Views: 51

Re: 8.45

Yep, delta H is assumed to be in kJ/mol if it's not specified, and also it's assumed that if delta H is given, then the reaction is at a constant pressure.
by Kelsey Warren 1I
Wed Feb 06, 2019 8:14 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Constant Pressure/Volume
Replies: 6
Views: 200

Re: Constant Pressure/Volume

Also, the pressure has to be constant in an open beaker because the pressure in the beaker would equal the atmospheric pressure, which is constant.
by Kelsey Warren 1I
Wed Feb 06, 2019 8:12 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: kj v j in calculations?
Replies: 2
Views: 41

Re: kj v j in calculations?

If the heat/energy value is really large (>10^3), then use kJ so that your numerical answer won't be so huge. If your heat/energy value is smaller, you can use J. It's really just dependent on the size of the numbers you're dealing with.
by Kelsey Warren 1I
Wed Feb 06, 2019 8:10 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Entropy Change Due to Temperature Change
Replies: 3
Views: 60

Re: Entropy Change Due to Temperature Change

I'm not positive either, but I agree with what she said. Cv is for constant volume and Cp is for constant pressure.
by Kelsey Warren 1I
Tue Jan 29, 2019 7:52 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Calorimeter
Replies: 4
Views: 53

Re: Calorimeter

Basically a calorimeter is a reaction vessel, so it acts as a container for a reaction to take place. It allows you to make measurements about temperature changes, volume, etc. that you can use to calculate enthalpy changes/heat capacities, etc.
by Kelsey Warren 1I
Tue Jan 29, 2019 7:49 pm
Forum: Phase Changes & Related Calculations
Topic: Delta Hº versus delta H
Replies: 7
Views: 131

Re: Delta Hº versus delta H

Delta H naught is just delta H in standard/ideal conditions. You use them the same but one is for standard conditions.
by Kelsey Warren 1I
Tue Jan 29, 2019 7:47 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat capacity vs. enthalpy
Replies: 4
Views: 70

Re: Heat capacity vs. enthalpy

Enthalpy is basically the heat possessed within a system, but we usually talk about enthalpy changes, which are the changes in heat within a system. Heat capacity is the heat required to raise the temperature of a substance by 1 degree C. So heat capacity has to do with heat entering/leaving a syste...
by Kelsey Warren 1I
Mon Jan 21, 2019 7:26 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Ionized vs Deprotonated
Replies: 2
Views: 66

Re: Ionized vs Deprotonated

Ionized and deprotonated are basically the same thing when talking about acids and bases; these words just mean the amount of acid/base that has reacted divided by the amount of acid/base you initially started with (times 100).
by Kelsey Warren 1I
Mon Jan 21, 2019 7:20 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Percent Protonation
Replies: 6
Views: 82

Re: Percent Protonation

Percent protonation is calculated the second way you mentioned, which is % protonation=concentration of conjugate acid / initial concentration of weak base x 100 (or the opposite if you start with a weak acid).
by Kelsey Warren 1I
Mon Jan 21, 2019 7:16 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Salt Solution
Replies: 4
Views: 159

Re: Salt Solution

Yes, if they're just asking you to calculate the pH and there's an ion that does not take part in the reaction, then you always ignore that ion because it won't affect the pH.
by Kelsey Warren 1I
Tue Jan 15, 2019 8:12 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Reaction Quotient
Replies: 8
Views: 155

Re: Reaction Quotient

Just to add on a little, the reaction quotient can be used to determine if the reaction will proceed forward or backward to reach equilibrium. For example, if the reaction quotient is smaller than the equilibrium constant, then the reaction will proceed forward to make more products to increase the ...
by Kelsey Warren 1I
Tue Jan 15, 2019 8:08 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Same/different @ equilibrium
Replies: 2
Views: 41

Re: Same/different @ equilibrium

The equilibrium constant stays the same at equilibrium because it's a constant: it's a calculated factual value for that exact reaction at the given temperature. The ratio with the exponents is the equilibrium constant, so it will stay the same. And like Vincent said, the other ratio you mentioned i...
by Kelsey Warren 1I
Tue Jan 15, 2019 8:03 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: HW problem 5I.17
Replies: 2
Views: 45

Re: HW problem 5I.17

As far as the ICE table, sometimes they're helpful because they make a nice visual representation of what's happening in the reaction, but if you are given enough information to just use the K equation, that will work too (and it's faster).
by Kelsey Warren 1I
Wed Jan 09, 2019 10:01 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Difference between brackets and parentheses?
Replies: 4
Views: 137

Re: Difference between brackets and parentheses?

Another way to think about it is that brackets are used when you're referring to a concentration and parentheses are used when you substitute in the actual numerical value of the concentration into the K equation. For example, if one of the compounds in your chemical reaction is NH3 and its concentr...
by Kelsey Warren 1I
Wed Jan 09, 2019 9:42 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Compostition vs Concentration for Equilibrium
Replies: 2
Views: 53

Re: Compostition vs Concentration for Equilibrium

I think the book uses the word composition just to encompass both concentrations and partial pressures into its definition of equilibrium. So it's basically correct to say that that the compositions and concentrations in the mixture stay constant at equilibrium. You could also say the partial pressu...
by Kelsey Warren 1I
Wed Jan 09, 2019 9:37 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: K
Replies: 4
Views: 51

Re: K

I'm pretty sure that when you're dealing with gases, you use partial pressures to calculate K, so you'd use Kp. When you're using aqueous compounds, etc., so you're plugging in concentrations into your calculation for K, you'd use Kc.
by Kelsey Warren 1I
Mon Dec 03, 2018 7:55 pm
Forum: Bronsted Acids & Bases
Topic: Fundamentals J.17 (7th)
Replies: 2
Views: 103

Re: Fundamentals J.17 (7th)

NH4 can't be a base because it can't accept a proton or donate an electron. It can't accept a proton because the N already has 4 bonds each with one H, so neither the N nor the H's can make another bond. NH4 can't donate a base because there are no lone pairs of electrons and all single bonds. Becau...
by Kelsey Warren 1I
Mon Dec 03, 2018 7:52 pm
Forum: Calculating the pH of Salt Solutions
Topic: pH 1-14
Replies: 2
Views: 61

Re: pH 1-14

Only a super acid or a super base would make the H+ concentration exceed .1M or less than 10^-14M, which correspond to ph 1 and 14. Because super acids and bases aren't as common as regular ones, values outside the regular pH range are rare.
by Kelsey Warren 1I
Mon Dec 03, 2018 7:46 pm
Forum: Naming
Topic: 6th Edition 29c
Replies: 3
Views: 47

Re: 6th Edition 29c

I was a little confused about this too, but the aqua comes from the water ligand that is part of the complex (the OH2). Because the aqua comes before the cyano in alphabetical order, the aqua is the start of the name.
by Kelsey Warren 1I
Mon Nov 26, 2018 8:30 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: Intermolecular forces and melting/boiling points
Replies: 4
Views: 97

Re: Intermolecular forces and melting/boiling points

Another easy way to think about this is that increasing the strength of intermolecular attractions is like a bunch of molecules holding on tighter and tighter to each other. If they're holding on really tightly, it becomes very difficult for one to escape the others, therefore it takes more energy t...
by Kelsey Warren 1I
Mon Nov 26, 2018 8:27 pm
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Polarizability
Replies: 5
Views: 103

Re: Polarizability

Polarizability increases as electronegativity decreases and as atomic radius increases. Therefore, polarizability increases right to left and top to bottom.
by Kelsey Warren 1I
Mon Nov 26, 2018 8:26 pm
Forum: Hybridization
Topic: The Number Before Hybridization
Replies: 3
Views: 63

Re: The Number Before Hybridization

Yes, the number is the period the element is in just because that's the period that the valence electrons (the ones actually doing the bonding) are in. Also, writing out the electron configuration is very helpful as they said above!
by Kelsey Warren 1I
Mon Nov 26, 2018 8:23 pm
Forum: Hybridization
Topic: hybridization
Replies: 3
Views: 65

Re: hybridization

I think all we really need to know about the cause of hybridization is that it makes the molecule formed more stable and that a hybrid orbital interacting with the other atom's orbital makes the bond stronger/more balanced, whichever makes the most sense to you.
by Kelsey Warren 1I
Mon Nov 26, 2018 8:18 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Can an octahedral be polar?
Replies: 4
Views: 78

Re: Can an octahedral be polar?

If a central atom has 6 regions of electron density surrounding it and they're all other atoms, then it will be nonpolar, but if there's one or more lone pair on the central atom then the shape is no longer symmetrical and could be polar.
by Kelsey Warren 1I
Mon Nov 26, 2018 8:15 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: Dipole-dipole and H bonding
Replies: 1
Views: 44

Re: Dipole-dipole and H bonding

H bonding is a form of dipole-dipole because the H has a slight positive charge and the O, N, or F has a slight negative charge, so yes they can.
by Kelsey Warren 1I
Wed Nov 14, 2018 9:54 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Bond Strengths
Replies: 5
Views: 92

Re: Bond Strengths

An ionic bond is considered intramolecular within a molecule and intermolecular within an ionic solid. Since the metal completely gives one or more electrons to the nonmetal, the bond is much stronger than a H bond which is completely intermolecular and is only caused by the attraction of positive a...
by Kelsey Warren 1I
Wed Nov 14, 2018 9:50 pm
Forum: Ionic & Covalent Bonds
Topic: Hydrogen Bond
Replies: 7
Views: 126

Re: Hydrogen Bond

Since a H bond is intermolecular and a covalent bond is intramolecular, a molecule can have a covalent bond and then share a H bond with another molecule of the same species. I guess the effect of that would be that the molecule with the covalent bond (like H2O) would be attracted to another H2O via...
by Kelsey Warren 1I
Wed Nov 14, 2018 9:44 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: 3D Structure
Replies: 2
Views: 26

Re: 3D Structure

I'm pretty sure you just draw one wedge because the wedge just represents the direction of the bond rather than the bond itself.
by Kelsey Warren 1I
Wed Nov 07, 2018 6:41 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: Greater polarizability of larger molecules
Replies: 6
Views: 90

Re: Greater polarizability of larger molecules

I was under the impression that the larger the molecule, there's more electrons and more electron shielding. The outer electrons of a large atom are less tightly held than for a smaller atom so they're able to be pulled by an attractive force more easily than for a smaller atom.
by Kelsey Warren 1I
Wed Nov 07, 2018 6:38 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: Why are larger hydrocarbons more solid?
Replies: 2
Views: 53

Re: Why are larger hydrocarbons more solid?

Polarizability is increased with increased size of atoms, number of electrons, and amount of negative charge (if you're talking about ions). Because larger hydrocarbons are larger and have more electrons, they are more polarizable. Their greater number of electrons experience more shielding and can ...
by Kelsey Warren 1I
Wed Nov 07, 2018 6:33 pm
Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
Topic: Similar Names of Interactions between Ions and Molecules
Replies: 2
Views: 41

Re: Similar Names of Interactions between Ions and Molecules

Yep, induced dipole-induced dipole is probably the most descriptive name for the intermolecular interaction but you can also use dispersion, VDW's, and London forces interchangeably.
by Kelsey Warren 1I
Tue Oct 30, 2018 8:41 pm
Forum: Resonance Structures
Topic: Resonance merging
Replies: 3
Views: 99

Re: Resonance merging

I don't know how many details I can explain but I'll try! Basically what he means is that in real molecules, instead of a molecule clearly having two single bonds and one double bond (as shown by observed bond lengths), the molecule would have three bonds that were each "somewhere in between&qu...
by Kelsey Warren 1I
Tue Oct 30, 2018 8:29 pm
Forum: Lewis Structures
Topic: bond length
Replies: 2
Views: 47

Re: bond length

I think he cares more about us knowing what each bond represents and the possible resonance for each molecule than drawing the bond lengths correctly. Also, the bond length differences for triple/double/single/etc. bonds are relatively small and would be extremely difficult to draw accurately on a t...
by Kelsey Warren 1I
Tue Oct 30, 2018 8:25 pm
Forum: Trends in The Periodic Table
Topic: Electron Affinity
Replies: 2
Views: 62

Re: Electron Affinity

In its ground state, C has one electron in the px orbital and one electron in the py orbital and an unoccupied pz orbital. When an e- is added, it occupies the previously empty p orbital, which releases more energy than an additional e- in N pairing up with another in the px orbital.
by Kelsey Warren 1I
Wed Oct 24, 2018 5:10 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Energy of Electron in many-electron atom
Replies: 2
Views: 45

Re: Energy of Electron in many-electron atom

Yes, electron shielding occurs in multi-electron atoms, so an electron far from the nucleus will experience less of the positive charge of the nucleus. This results in what is called an "effective nuclear charge." I don't know if we have to know that term but probably should be familiar wi...
by Kelsey Warren 1I
Wed Oct 24, 2018 5:07 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Electron Configurations for Chromium and Copper
Replies: 2
Views: 44

Re: Electron Configurations for Chromium and Copper

The Cu and Cr are more stable when their electron configurations have a full (or half full) d shell. I think we just need to know these exceptions and what they look like rather than understand exactly why they occur.
by Kelsey Warren 1I
Wed Oct 24, 2018 5:04 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 2.53 Predict how many unpaired electrons...
Replies: 2
Views: 34

Re: 2.53 Predict how many unpaired electrons...

Unpaired electrons are any electrons that don't have a another electron (with opposite spin) in their orbital. Draw out the Aufbau diagrams for each element using the rules from lecture and you'll be able to see which electrons don't have a buddy!
by Kelsey Warren 1I
Tue Oct 16, 2018 1:21 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Formula
Replies: 3
Views: 82

Re: Formula

The equation is (usually?) used for an electron losing energy and consequently emitting a photon. Because the electron loses energy, its energy is "negative" while the energy for the photon is positive. Dr. Lavelle also said in lecture that the negative sign means that the bound electron h...
by Kelsey Warren 1I
Tue Oct 16, 2018 1:14 pm
Forum: Properties of Light
Topic: Mass of Light
Replies: 5
Views: 84

Re: Mass of Light

Like they said above, light doesn't have mass. A photon acts like a particle rather than a wave, but that "particle" is only used for visualizing discrete amounts of energy, 1:1 photon:electron interactions, etc. A photon has a velocity as well.
by Kelsey Warren 1I
Tue Oct 16, 2018 1:11 pm
Forum: Einstein Equation
Topic: 7th edition 1B.5
Replies: 3
Views: 68

Re: 7th edition 1B.5

I agree with everything stated above I just wanted to clarify that 1 eV=1.6x10^-19 J, not the other way around. Hope this helps!
by Kelsey Warren 1I
Wed Oct 10, 2018 5:17 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Balmer and Lyman series
Replies: 3
Views: 78

Re: Balmer and Lyman series

I'm not 100% sure, but it's derived from the equation in lecture today, E=-hR/(n^2). If you substitute E=hv, then you can divide both sides by h to get rid of Planck's constant, as in the equation you posted. Besides that, I'd guess that you would use it to find the frequency of a photon emitted fro...
by Kelsey Warren 1I
Wed Oct 10, 2018 5:12 pm
Forum: Properties of Electrons
Topic: Behavior of Electrons and Energy Levels
Replies: 3
Views: 67

Re: Behavior of Electrons and Energy Levels

I agree with the points they made above about the Bohr model vs. VESPR model and electrons jumping from ground states to excited states. I'm also wondering if when an electron absorbs such high frequency/high energy radiation that it approaches n=infinity and leaves the atom, does it leave the atom ...
by Kelsey Warren 1I
Wed Oct 10, 2018 5:06 pm
Forum: Properties of Electrons
Topic: Kinetic energy
Replies: 4
Views: 102

Re: Kinetic energy

You can't really measure "how fast" the kinetic energy of anything is. Kinetic energy is directly proportional to the velocity of the particle squared, so I guess you could say a physical sign of the kinetic energy of an ejected electron is the square root of its velocity.
by Kelsey Warren 1I
Mon Oct 01, 2018 4:46 pm
Forum: Balancing Chemical Reactions
Topic: States of Matter
Replies: 11
Views: 331

Re: States of Matter

Addressing the second part of your question, including the state of matter when writing and balancing equations has been very helpful for me in that I'm able to picture conducting the reaction in a lab, which we obviously aren't dealing with yet but we will in the future! It's also helpful when iden...
by Kelsey Warren 1I
Mon Oct 01, 2018 4:37 pm
Forum: Limiting Reactant Calculations
Topic: Limiting Reactant Calculation Question [ENDORSED]
Replies: 2
Views: 102

Re: Limiting Reactant Calculation Question [ENDORSED]

Another way to approach this is using dimensional analysis:

You have 1.56 mol reactant and you want to convert to mols of product, so multiply 1.56 mol reactant times (5 mol product/4 mol reactant). I think the last answer just got the 4:5 ratio flip flopped.
by Kelsey Warren 1I
Mon Oct 01, 2018 4:27 pm
Forum: Limiting Reactant Calculations
Topic: Limiting Reactants Post- Module
Replies: 2
Views: 71

Re: Limiting Reactants Post- Module

Hi Hannah,

I'm under the impression that the given pressure and temperature are not needed for calculating the amount of moles of carbon dioxide produced. I think that those come in handy when dealing with Ideal Gas Law problems, etc.

Hopefully it's true that we didn't need to use that information!

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