Search found 60 matches

by Heesu_Kim_1F
Sat Mar 09, 2019 1:04 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: zero order
Replies: 2
Views: 50

Re: zero order

The molecularity of a reaction cannot be zero. Most likely, a zero order reaction is not an elementary reaction, a reaction which occurs in a single step, but rather is a part of complex reactions that occur in two or more steps. Hope this helps!
by Heesu_Kim_1F
Sat Mar 09, 2019 12:43 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: reaction rate
Replies: 2
Views: 58

Re: reaction rate

The slowest rate determines the reaction rate because the reaction cannot be any faster than the slowest step. In addition, the high activation energy for the slow step limits the reaction rate. Hope this helps!
by Heesu_Kim_1F
Sat Mar 09, 2019 12:28 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: catalyst/enzyme
Replies: 4
Views: 72

Re: catalyst/enzyme

^I agree. A catalyst/enzyme does change/increases the rate of reaction by lowering the activation energy. However, a catalyst/enzyme does not change the equilibrium constant for a reaction. Also, a catalyst or enzyme (itself) is not changed during a reaction.
Hope this helps!
by Heesu_Kim_1F
Tue Mar 05, 2019 4:20 pm
Forum: First Order Reactions
Topic: First order graph
Replies: 7
Views: 123

Re: First order graph

The x-intercept (meaning y=0) of the first order graph would represent the time (usually in seconds) when the concentration of the reactant reaches 0, or essentially when there is no more reactant concentration left.
Hope this helps!
by Heesu_Kim_1F
Tue Mar 05, 2019 4:09 pm
Forum: General Rate Laws
Topic: overall reaction rate
Replies: 2
Views: 75

Re: overall reaction rate

The reaction rate for a given chemical reaction is a measure of the change in concentration of the reactants or products per unit time. Essentially, it tells us the speed, how fast or slow, of the reaction.
by Heesu_Kim_1F
Tue Mar 05, 2019 2:33 pm
Forum: General Rate Laws
Topic: 6th edition 15.9
Replies: 4
Views: 86

Re: 6th edition 15.9

I agree, the units for rate constant k depend on the order of reaction.
For zero order reactions, the unit for k is (mol x L^-1 x s^-1).
For first order reactions, the unit for k is (s^-1).
For second order reactions, the unit for k is (L x mol^-1 x s^-1).
Hope this helps!
by Heesu_Kim_1F
Mon Feb 25, 2019 11:59 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: order of cell notation
Replies: 1
Views: 45

Re: order of cell notation

I think the order of cell notation matters. The anode components are written on the left. The cathode components are written on the right. Usually, the outside is the metal and the inside is the ion/aqueous solution. Thus, a generalized cell notation is anode metal | anode ion || cathode ion | catho...
by Heesu_Kim_1F
Mon Feb 25, 2019 11:33 am
Forum: Calculating Work of Expansion
Topic: -DeltanRT
Replies: 2
Views: 212

Re: -DeltanRT

Yes, you can use -delta n RT as -P delta V because of the ideal gas law, which states PV = nRT (P=pressure, V=volume, n=# of moles, R=gas constant, T=temperature). Hope this helps!
by Heesu_Kim_1F
Mon Feb 25, 2019 11:30 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Standard Gibbs free energy
Replies: 1
Views: 42

Re: Standard Gibbs free energy

The standard Gibbs free energy of formation of a compound is the change of Gibbs free energy that accompanies the formation of 1 mole of that substance from its component elements, at their standard states (25 Celsius and 100kPa). Hope this helps!
by Heesu_Kim_1F
Wed Feb 20, 2019 1:04 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 9.61
Replies: 1
Views: 49

Re: 9.61

I would find the delta G for each product using their enthalpy and entropy values and then add the delta G for all the products. I would then do the same for the delta G of all reactants and then do sum of delta G products - sum of delta G reactants to find the delta G for the reaction. Hope this he...
by Heesu_Kim_1F
Wed Feb 20, 2019 12:52 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Relationship between K and G
Replies: 3
Views: 88

Re: Relationship between K and G

I believe you're right.
If (Q/K) <1 or Q < K, then delta G will be negative and thus spontaneous reaction.
If (Q/K) >1 or Q > K, then delta G will be positive and thus non-spontaneous reaction.
Hope this helps!
by Heesu_Kim_1F
Wed Feb 20, 2019 12:43 pm
Forum: Administrative Questions and Class Announcements
Topic: Midterm Question 1A
Replies: 3
Views: 89

Re: Midterm Question 1A

I would think it is because the given Kp value, 3 x 10^4, has 1 sig fig along with the given temperature value, 700K. Hope this helps!
by Heesu_Kim_1F
Thu Feb 14, 2019 5:47 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: gibbs free energy
Replies: 10
Views: 151

Re: gibbs free energy

If delta G is negative, the reaction is spontaneous.
If delta G is positive, the reaction is not spontaneous.
Hope this helps!
by Heesu_Kim_1F
Thu Feb 14, 2019 5:42 pm
Forum: Phase Changes & Related Calculations
Topic: Phase Change Diagram of Water
Replies: 6
Views: 129

Re: Phase Change Diagram of Water

Boiling water contains only a specific amount of heat energy required for it to boil. On the other hand, as steam is formed from boiling water, it contains the heat energy of boiling water, along with the heat of vaporization due to phase change. Thus, as steam has more heat energy, it can cause mor...
by Heesu_Kim_1F
Thu Feb 14, 2019 5:27 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Delta G
Replies: 2
Views: 52

Re: Delta G

The Gibbs free energy equation is (delta G) = (delta H) - (temperature)(delta S). When (delta S) is positive and (delta H) is positive, then the reaction will be spontaneous at high temperature. When (delta S) is positive and (delta H) is negative OR when (delta S) is negative and (delta H) is posit...
by Heesu_Kim_1F
Mon Feb 04, 2019 11:03 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: enthalpy of combustion vs formation
Replies: 3
Views: 86

Re: enthalpy of combustion vs formation

I believe the enthalpy of combustion is a type of enthalpy of formation that involves a specific reaction where one mole of a compound, such as methane, completely combines with oxygen gas. Hope this helps!
by Heesu_Kim_1F
Mon Feb 04, 2019 10:49 pm
Forum: Ideal Gases
Topic: Ideal gas constant R
Replies: 9
Views: 366

Re: Ideal gas constant R

I think it just depends on which units you are working with: R = 0.08206 L x atm x K^-1 x mol ^-1 and R = 8.314 J x K^-1 x mol^-1.
by Heesu_Kim_1F
Mon Feb 04, 2019 10:36 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 8.19
Replies: 3
Views: 81

Re: 8.19

You would use q=m x Cp x delta T, in which m = mass, Cp = specific heat capacity, and delta T= change in temperature, as your equation. Hope this helps!
by Heesu_Kim_1F
Wed Jan 30, 2019 1:21 am
Forum: Phase Changes & Related Calculations
Topic: 6th Edition 8.31
Replies: 1
Views: 72

Re: 6th Edition 8.31

(b) The molar heat capacity of a monatomic ideal gas at constant volume is Cv,m = 3R/2. with R = 8.314 J x mol^-1 x C^-1
Thus, q = (5.025g/83.80g x mol^-1) x (25.0C - 97.6 C) x (12.5J x mol^-1 x C^-1) = =54.4J. Hope this helps!
by Heesu_Kim_1F
Wed Jan 30, 2019 12:33 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 6th edition 8.31
Replies: 2
Views: 60

Re: 6th edition 8.31

R=8.314 J x C^-1 x mol^-1 is the gas constant.
(a) The molar heat capacity of a monatomic ideal gas at constant pressure is Cp,m = 5R/2 = (20.8 J x C^-1 x mol ^-1).
(b) The molar heat capacity of a monatomic ideal gas at constant volume is Cv, m = 3R/2 = (12.5 J x C^-1 x mol ^-1).
by Heesu_Kim_1F
Wed Jan 30, 2019 12:17 am
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: 6th Edition Problem 8.21
Replies: 8
Views: 165

Re: 6th Edition Problem 8.21

Hi, this might help.
The idea comes from Qsystem+Qsurr=0.
Qsystem = -Qsurr, in which copper = system and water = surrounding.
Thus, heat lost by copper = -heat gained by water.
Then use the calorimetry equation q = m x Cp x delta T to solve.
by Heesu_Kim_1F
Wed Jan 23, 2019 5:02 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy and q
Replies: 3
Views: 55

Enthalpy and q

From the lecture, I got that enthalpy is denoted by q = delta H and that it is a state property, but also confusingly, I got from the powerpoint that heat is denoted by q and that it is NOT a state property. Can someone clarify? Thank you in advance!
by Heesu_Kim_1F
Wed Jan 23, 2019 4:53 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: ICE Table Clarification
Replies: 4
Views: 70

Re: ICE Table Clarification

I think the assumption was that x is negligible when k is equal to or less than 10^-3. To check, if x is less than 5% of initial concentration, then the approximation is valid. Hope this helps!
by Heesu_Kim_1F
Wed Jan 23, 2019 4:38 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Endothermic vs. Exothermic
Replies: 8
Views: 155

Re: Endothermic vs. Exothermic

Exothermic reaction gives a net release of heat and thus would result in delta H to be a negative value.
Endothermic reaction requires heat and thus would result in delta H to be a positive value.
by Heesu_Kim_1F
Wed Jan 16, 2019 5:29 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Catalysts
Replies: 7
Views: 146

Re: Catalysts

A catalyst speeds up the forward and back reaction to the same extent. The relative rates of the two reactions won't be affected and thus equilibrium composition will not be changed. Hope this helps!
by Heesu_Kim_1F
Wed Jan 16, 2019 5:19 pm
Forum: Conjugate Acids & Bases
Topic: Cojugate Acids and Bases
Replies: 10
Views: 303

Re: Cojugate Acids and Bases

In a chemical equation, the acid and the base on the reactant side would be known as the "regular" acid and base while the acid and the base on the product side would be known as the conjugate acid and conjugate base. Also they are related in a way that professor Lavelle called "the c...
by Heesu_Kim_1F
Wed Jan 16, 2019 5:05 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Percentage Ionization
Replies: 4
Views: 80

Re: Percentage Ionization

Percent ionization refers to the amount of a weak acid that exists as ions at a particular concentration. It can be calculated with the formula: % ionization = (acid ionized/initial acid) x 100%. Hope this helps!
by Heesu_Kim_1F
Wed Jan 09, 2019 9:04 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Spectator Ions and Equilibrium Constants
Replies: 2
Views: 128

Re: Spectator Ions and Equilibrium Constants

A spectator ion is an ion that appears in the same form on both the reactant and product sides of a chemical reaction. Even in the case that you include it in the equilibrium constant, they will reduce to 1 in the end and have no effect. Hope this helps!
by Heesu_Kim_1F
Wed Jan 09, 2019 7:12 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Brackets vs P For Equilibrium Constants
Replies: 4
Views: 48

Re: Brackets vs P For Equilibrium Constants

Although I think we can write the answers either way, my TA mentioned today that if the given equation is all in gas form, use partial pressure for K, while if the given equation is all in aqueous form, use concentration for K. He also mentioned that for heterogeneous equilibrium, use concentration ...
by Heesu_Kim_1F
Wed Jan 09, 2019 7:07 pm
Forum: Ideal Gases
Topic: Reaction Quotient
Replies: 4
Views: 80

Re: Reaction Quotient

The reaction quotient (Q) measures the amount/concentration of products and reactants at a given time, as opposed to K, which is equal to the value of Q only at equilibrium. One solves Q in the same way as one does K. Thus, for problem 11.13a, Q = 1/((P BCl3)^2). This is because Hg is liquid and B2C...
by Heesu_Kim_1F
Tue Dec 04, 2018 5:31 pm
Forum: Naming
Topic: bis-,tris-,tetrakis-
Replies: 2
Views: 76

Re: bis-,tris-,tetrakis-

You would use bis-, tris-, and tetrakis- if the ligand already contains a prefix. So, for example, Pd[P(C6H5)3]4 is palladium(0) tetrakis(triphenylphosphine). Tetrakis- is used instead of tetra- since tri- is already used in the ligand's name. Hope this helps!
by Heesu_Kim_1F
Tue Dec 04, 2018 5:26 pm
Forum: Naming
Topic: Difference between cis and trans
Replies: 4
Views: 85

Re: Difference between cis and trans

Ligands next to one another/on the same side are cis, while ligands across from one another/on opposite sides are trans. Hope this helps!
by Heesu_Kim_1F
Tue Dec 04, 2018 5:07 pm
Forum: Naming
Topic: Naming using Prefixes
Replies: 1
Views: 47

Re: Naming using Prefixes

It seems that you would use di-, tri-, and tetra- for the compounds first to denote the number of each ligand type and use bis-, tris-, and tetrakis- if the ligand already contains a prefix. Hope this helps!
by Heesu_Kim_1F
Tue Nov 27, 2018 3:47 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Determining Bond Angle
Replies: 3
Views: 37

Re: Determining Bond Angle

I would consider both atom arrangement and its actual shape. The reason the bond angle for SOCl2 is less than 109.5 is because its actual shape as a trigonal pyramidal has one lone pair of electrons, which affects the bond angles to be slightly less than 109.5 rather than 109.5 exactly. Also, Dr. La...
by Heesu_Kim_1F
Tue Nov 27, 2018 2:16 pm
Forum: Hybridization
Topic: Lone Electrons and Hybridization
Replies: 3
Views: 104

Re: Lone Electrons and Hybridization

I believe lone pairs of electrons count as an area of electron density and thus contributes to hybridization. For problem 4.35, if you draw the Lewis structure for SeF3+, you can see that Se (the central atom) has four areas of electron density. This is because Se has 6 electrons, Fluorine has 7 ele...
by Heesu_Kim_1F
Tue Nov 27, 2018 2:09 pm
Forum: Ionic & Covalent Bonds
Topic: Ion-Ion and Ionic Bonds
Replies: 1
Views: 61

Re: Ion-Ion and Ionic Bonds

I believe so, because ion-ion interactions refer to the attractive forces between two ions, which is essentially an ionic bond. These intermolecular forces are what hold together ionic compounds. Hope this helps!
by Heesu_Kim_1F
Mon Nov 19, 2018 9:45 pm
Forum: Dipole Moments
Topic: Dipole-Dipole Forces
Replies: 2
Views: 70

Re: Dipole-Dipole Forces

Dipole-dipole interactions happen between polar molecules. Even for our previous tests, we were never asked to calculate the electronegativity difference between atoms, but we had to know the general idea of the difference between polar and nonpolar molecules. Thus, if you based off that knowledge a...
by Heesu_Kim_1F
Mon Nov 19, 2018 9:26 pm
Forum: Bond Lengths & Energies
Topic: double bonds vs. single bonds
Replies: 5
Views: 145

Re: double bonds vs. single bonds

I understand that double bonds are more stable because of the fact that they have both sigma and pi bonds. Can another reason be that because double bonds have shorter distances than single bonds, thus making the doubles bonds stronger and more stable?
by Heesu_Kim_1F
Mon Nov 19, 2018 9:02 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: Difference
Replies: 1
Views: 66

Re: Difference

According to the textbook, "a diamagnetic substance is one that tends to move out of a magnetic field, and a paramagnetic substance is one that tends to move into a magnetic field." Essentially, I think an easy way to think about is that if a certain molecule has unpaired electrons, it is ...
by Heesu_Kim_1F
Wed Nov 14, 2018 12:02 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Trigonal planar and bent shape
Replies: 2
Views: 65

Re: Trigonal planar and bent shape

The answer actually is that the shape is angular, which also means bent. However, it goes further to say that it is in a trigonal planar arrangement. I think what it's trying to explain is that its basic geometry is trigonal planar because the central atom consists of 3 regions of electron density. ...
by Heesu_Kim_1F
Wed Nov 14, 2018 11:54 am
Forum: Lewis Structures
Topic: HW #3.49 (6th Ed)
Replies: 1
Views: 76

Re: HW #3.49 (6th Ed)

The +1 charge on the Oxygen is referring to the formal charge. For part (a), the total number of electrons is 5 (N) + 6 (O) - 1 (overall +1 charge) = 10 electrons. For part (c), the total number of electrons is C (4) +6 (O) = 10 electrons. For both parts, the only way to fulfill 10 electrons without...
by Heesu_Kim_1F
Wed Nov 14, 2018 12:03 am
Forum: Determining Molecular Shape (VSEPR)
Topic: Shape
Replies: 5
Views: 94

Re: Shape

The two terms "angular" and "bent" are interchangeable. Both refer to molecules with three groups of electrons and one lone pair.
Hope this helps!
by Heesu_Kim_1F
Wed Nov 07, 2018 10:11 pm
Forum: Lewis Structures
Topic: Radicals 3.55
Replies: 3
Views: 132

Re: Radicals 3.55

Only (b) and (c) would be radicals, which means that they are species with an unpaired electron, because when you draw out the Lewis structures, you see that they have an odd number of electrons compared to (a) and (d), both of which have even number of electrons. Hope this helps!!
by Heesu_Kim_1F
Wed Nov 07, 2018 10:08 pm
Forum: Lewis Structures
Topic: Electron Pairs 3.61
Replies: 2
Views: 53

Re: Electron Pairs 3.61

(a) 2 bonding pairs and 2 lone pairs

(b) 4 bonding pairs and 2 lone pairs

(c) 3 bonding pairs and 2 lone pairs

(d) 5 bonding pairs and 1 lone pair

Hope this helps!!
by Heesu_Kim_1F
Wed Nov 07, 2018 10:05 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Electron Configuration 3.27
Replies: 1
Views: 94

Re: Electron Configuration 3.27

(a) TlCl3

(b) Al2S3

(c) MnO2

Hope this helps!!
by Heesu_Kim_1F
Tue Oct 30, 2018 10:45 pm
Forum: Hybridization
Topic: Question 2.43
Replies: 4
Views: 162

Re: Question 2.43

Professor Lavelle said in class that we can write it in both ways. I think the solution manual writes out from the lowest quantum number to the highest quantum number. He also mentioned that if he prefers a certain way, then he will make sure to let us know. Hope this helps!
by Heesu_Kim_1F
Tue Oct 30, 2018 10:41 pm
Forum: Ionic & Covalent Bonds
Topic: 3.61 6TH EDITION HW
Replies: 1
Views: 113

Re: 3.61 6TH EDITION HW

(a) 2 bonding pairs and 2 lone pairs

(b) 4 bonding pairs and 2 lone pairs

(c) 3 bonding pairs and 2 lone pairs

(d) 5 bonding pairs and 1 lone pair

Hope this helps!!
by Heesu_Kim_1F
Tue Oct 30, 2018 10:37 pm
Forum: Ionic & Covalent Bonds
Topic: Electron Configuration
Replies: 1
Views: 58

Re: Electron Configuration

(a) [Ar] 3d^10

(b) [Xe] 4f^14 5d^10 6s^2

(c) [Ar] 3d^10

(d) [Xe] 4f^14 5d^10

Hope this helps!!
by Heesu_Kim_1F
Mon Oct 22, 2018 1:48 am
Forum: Photoelectric Effect
Topic: module question 28 a
Replies: 1
Views: 59

Re: module question 28 a

Problem 28A is only asking for you to solve for the kinetic energy. Thus, the only equation to use for this problem is (1/2)mv^2. Velocity is given in the problem: 6.61 x 10^5 m/s, and the mass of electron is a number/constant that will be given on the equation sheet: 9.11 x 10^-31 kg. Plug those nu...
by Heesu_Kim_1F
Mon Oct 22, 2018 1:38 am
Forum: Photoelectric Effect
Topic: Module question 30 c
Replies: 4
Views: 170

Re: Module question 30 c

Problem 30C is a problem that uses the answers that you get for 28A and 29B. You can get the value of E by using the photoelectric equation: E(hv)=work function+Ek (kinetic energy). The kinetic energy value is the answer for 28A and the work function energy value is the answer for 29B. Thus, in orde...
by Heesu_Kim_1F
Mon Oct 22, 2018 1:16 am
Forum: Properties of Light
Topic: kg or g [ENDORSED]
Replies: 11
Views: 311

Re: kg or g [ENDORSED]

I would assume that most quantum equations, if not all, use kg because kg is the SI unit for mass. Looking at the equation sheet from our last test, it seems that all the equations will involve kg. If the given is in grams, then you would change it to kg before solving the problem. Hope this helps!
by Heesu_Kim_1F
Mon Oct 15, 2018 7:36 pm
Forum: DeBroglie Equation
Topic: HW 1.41
Replies: 6
Views: 70

Re: HW 1.41

The v in the c=λv equation represents frequency, not velocity, so it won't help you solve the problem. On the other hand, the v in the De Broglie's Equation λ=h/mv represents velocity, which is given in the problem.
Hope this helps!
by Heesu_Kim_1F
Mon Oct 15, 2018 6:05 pm
Forum: DeBroglie Equation
Topic: Question 1.37 sixth edition
Replies: 1
Views: 58

Question 1.37 sixth edition

1.37) Protons and neutrons have nearly the same mass. How different are their wavelengths? Calculate the wavelength of each particle when traveling at 2.75 x 10^5 ms^-1 in a particle accelerator and report the difference as a percentage of the wavelength of the neutron. I was able to solve the wavel...
by Heesu_Kim_1F
Mon Oct 15, 2018 5:10 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Question 1.15 sixth edition
Replies: 10
Views: 170

Question 1.15 sixth edition

1.15) In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line. I know that problem 15 is just the opposite of problem 13, which I s...
by Heesu_Kim_1F
Mon Oct 08, 2018 5:15 pm
Forum: Properties of Light
Topic: Photoelectric Effect Module #28
Replies: 1
Views: 45

Photoelectric Effect Module #28

28) Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 10^5 m s^-1. The work function for sodium is 150.6 kJ mol^-1. A. What is the kinetic energy of the ejected electron? I know that in order to find the kinetic energy, I need to use the (1/2) mv^2 equation. Veloci...
by Heesu_Kim_1F
Mon Oct 08, 2018 4:26 pm
Forum: Empirical & Molecular Formulas
Topic: F.7
Replies: 5
Views: 409

Re: F.7

(a) If we assume the compound to be 100g, then 88.8 g is M and 11.2 g would be O. First, find the molar mass of the whole compound through stoichiometry: (100 g M2O/11.2 g O) x (16.00 g O/1 mol O) x (1 mol O/1 mol M2O) will result in 143 g/mol M2O. Since O's molar mass is 16.00 g/mol, the molar mass...
by Heesu_Kim_1F
Mon Oct 08, 2018 2:12 pm
Forum: Limiting Reactant Calculations
Topic: M1 Question
Replies: 1
Views: 39

Re: M1 Question

I think you either did the calculation wrong or the rounding because you are supposed to get around 32.9 g N2H4 for the theoretical yield. Check your math again maybe. To set up the equation: 35.0 g NH3 x 1 mol NH3/17.031 g NH3 x 1 mol N2H4/2 mols NH3 x 32.046 g N2H4/1 mol N2H4 = 32.9 g N2H4 Then th...
by Heesu_Kim_1F
Mon Oct 01, 2018 11:35 pm
Forum: Empirical & Molecular Formulas
Topic: Question M19 (6th edition)
Replies: 3
Views: 317

Re: Question M19 (6th edition)

You can know that caffeine includes O because the mass of C, H, and N do not add up to the mass of caffeine. Thus, in order to follow the law of conservation of mass, you can infer that it also includes O as well. My explanation might have been confusing, so I'll do the first part of the problem. Wh...
by Heesu_Kim_1F
Mon Oct 01, 2018 9:31 pm
Forum: Molarity, Solutions, Dilutions
Topic: Molarity
Replies: 3
Views: 234

Re: Molarity

The concept that is confusing (even for me at first) and needs to be understood is that in 1 mole of MgSO4+7H2O (epsom salt), there are 7 moles of H2O (water molecules). Thus, you need to first convert the given information 5.15g MgSO4+7H2O into moles of MgSO4+7H2O. Its molar mass is 246.466 g/mol. ...
by Heesu_Kim_1F
Mon Oct 01, 2018 9:08 pm
Forum: Limiting Reactant Calculations
Topic: Question M.15 Part B & C (Sixth Edition)
Replies: 2
Views: 107

Re: Question M.15 Part B & C (Sixth Edition)

Since you know that the limiting reactant is Al, you use the given information about aluminum and do stoichiometry to find the mass of aluminum chloride. First, convert 255g of Al to moles using its molar mass. Then use the balanced equation to convert moles of Al to moles of AlCl3. Then use the mol...

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