## Search found 62 matches

Wed Mar 13, 2019 3:35 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: 6th edition 15.67
Replies: 1
Views: 60

### 6th edition 15.67

The presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kJ per mol to 75 kJ per mol. a) By what factor does the rate of reaction increase at 298K, all other factors being equal? What is the relationship between Ea, catalyzed and...
Tue Mar 12, 2019 12:56 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6th ed 14.97
Replies: 1
Views: 57

### Re: 6th ed 14.97

They tell you that one of the reactions is F2 (g) + 2H+ (aq) + 2e- -> 2HF (aq) and since you are finding the Ka of HF, HF must be the reactant (the denominator) when you calculate Ka. So you would want to flip the given half-reaction 2HF (aq) -> F2 (g) + 2H+ (aq) + 2e- and find the other half reacti...
Mon Mar 11, 2019 1:54 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Concentration cells
Replies: 2
Views: 67

### Concentration cells

What is a concentration cell? How is this different from a galvanic cell and an electrolytic cell?
Tue Mar 05, 2019 3:22 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6th edition 14.97
Replies: 1
Views: 69

### 6th edition 14.97

14.97 Use the data in Appendix 2B and the fact that for the half-reaction F2(g) + H+ (aq) + 2e- -> 2 HF (aq), E=+3.03 V, to calculate the value for Ka for HF. The log K equation is the one shown in our notes, but the solution manual uses ln K. I tried to use the log K equation but it resulted in a d...
Mon Mar 04, 2019 8:10 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6th edition 14.47
Replies: 4
Views: 121

### 6th edition 14.47

Calculate the reaction quotient, Q, for the following cell reactions, given the measured values of the cell potential. Balance the chemical equations by using the smallest whole number coefficients: a) Pt(s) | Sn4+(aq),Sn2+(aq) || Pb4+(aq),Pb2+(aq) | C(gr), E=+1.33 V The answer I got for this proble...
Mon Mar 04, 2019 4:54 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 6th Edition 14.41
Replies: 3
Views: 85

### 6th Edition 14.41

14.41 Calculate E for each of the following concentration cells:
b) Pt(s) I H2 (g, 1 bar) I H+ (aq, pH = 4.0) II H+ (aq, pH = 3.0) I H2 (g, 1 bar) I Pt(s)

In the solution manual, it says that n=1. Why is that?
Fri Mar 01, 2019 7:50 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: In the 6th Edition in Problem 14.23, Part C
Replies: 1
Views: 51

### Re: In the 6th Edition in Problem 14.23, Part C

Part C of 14.23 is supposed to be spontaneous. To determine spontaneity you would have to solve for delta G using delta G = -nFE. If delta G is negative, then the reaction is spontaneous as written (favors forward reaction). If delta G is positive, then the reaction is unfavorable and not spontaneou...
Thu Feb 28, 2019 8:23 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 6th edition 14.25
Replies: 1
Views: 57

### 6th edition 14.25

14.25 Arrange the following metals in order of increasing strength as reducing agents for species in aqueous solution: (a) Cu, Zn, Cr, Fe; (b) Li, Na, K, Mg; (c) U, V, Ti, Al; (d) Ni, Sn, Au, Ag I understand how to do (a) because they all share a common reaction in which 2 electrons are lost. But fo...
Wed Feb 27, 2019 5:49 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 9.65 6th edition
Replies: 1
Views: 53

### Re: 9.65 6th edition

The coefficients are fractions because you trying to balance the reactants such that you would only get 1 mole of product. The questions does not state how many moles, but you can assume 1 mole since you are solving for standard entropy of formation
Wed Feb 20, 2019 10:06 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 6th edition 9.67
Replies: 3
Views: 90

### 6th edition 9.67

Assume that the standard change in enthalpy and entropy are independent of temperature and calculate the change in the standard Gibbs Free Energy for each of the following at 80 C. Over what temperature range will each reaction be spontaneous under standard conditions? a) B2O3 + 6HF -> 2BF3 + 3H2O W...
Wed Feb 20, 2019 8:51 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Determining stability of a compound/reaction when there's an increase in temperature
Replies: 2
Views: 69

### Re: Determining stability of a compound/reaction when there's an increase in temperature

You can calculate the standard entropies of formation (delta S)
If delta S is negative, then the compound will be less stable at a higher temperature.
If delta S is positive, the compound will be more stable at a higher temperature
Wed Feb 20, 2019 8:48 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: stoichiometric coefficients for calculating Gibbs free energy
Replies: 2
Views: 67

### Re: stoichiometric coefficients for calculating Gibbs free energy

Unless stated otherwise in the problem, I think you are to assume the problem is asking for the change in Gibbs Free Energy for the creation of one mole of product hence why it would be 1/2N2 + 3/2H2 --> NH3 and not N2 + 3H2 --> 2NH3.
Mon Feb 11, 2019 11:20 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 7th Edit 4D21
Replies: 1
Views: 57

### Re: 7th Edit 4D21

Enthalpy of formation is specifically referring to ∆H during the formation of 1 mole of the substance. In the appendix near the top for each column, it should be labeled with the units and does say kJ/mol for enthalpy of formation
Mon Feb 11, 2019 11:16 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Replies: 1
Views: 76

Are there any situations in which we would need to apply adiabatic and diathermic? or are these just concepts we should know?
Mon Feb 11, 2019 11:12 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Ideal Gas
Replies: 2
Views: 63

### Re: Ideal Gas

An ideal gas is a gas whose molecules don't have volume and don't interact with each other. It's a theoretical concept used to make calculations easier.
Tue Feb 05, 2019 10:39 pm
Forum: Calculating Work of Expansion
Topic: Self-Test 8.6B
Replies: 1
Views: 75

### Self-Test 8.6B

Suppose that 1.00 kJ of energy is transferred as heat to oxygen in a cylinder fitted with a piston; the external pressure is 2.00 atm. The oxygen expands from 1.00 L to 3.00 L against this constant pressure. Calculate w and U for the entire process by treating the O2 as an ideal gas. I got an answer...
Mon Feb 04, 2019 6:35 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Homework Problem 8.21 in 6th edition: Q=mcat
Replies: 1
Views: 58

### Re: Homework Problem 8.21 in 6th edition: Q=mcat

I think the textbook used the equations: -q=qcal and Ccal= qcal/change in T Ccal is the heat capacity of the calorimeter q is the change in internal energy. So you can determine qcal is 22.5 kJ then you can plug this into Ccal=qcal/change in T change in temp= T2 - T1 = 23.97 C - 22.45 C = 1.52 C Cca...
Mon Feb 04, 2019 6:26 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Self-Test 8.4B
Replies: 1
Views: 44

### Self-Test 8.4B

A calorimeter was calibrated by mixing two aqueous solutions together, each of volume 0.100L. The heat output of the reaction that took place was known to be 4.16kJ, and the temperature of the calorimeter rose by 3.24 degrees C. Calculate the heat capacity of this calorimeter when it contains 0.200 ...
Wed Jan 30, 2019 10:46 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: 4B.3
Replies: 1
Views: 64

### Re: 4B.3

Your answer should be right. It should be +490 J or +4.90 x 10^2 J (3 sig figs).

The same question in the 6th edition has the answer listed as +4.90 x 10^2 J.
Wed Jan 30, 2019 10:38 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Work Question
Replies: 4
Views: 90

### Re: Work Question

Work and energy are different. Work refers to a way that energy can be transferred by a system to its surroundings. Energy is the capacity of a system to do work.
So if a system can do a lot of work, it possesses a lot of energy.
Wed Jan 30, 2019 1:57 pm
Forum: Calculating Work of Expansion
Topic: Units for Work
Replies: 6
Views: 114

### Units for Work

The units for work is just joules right? Or is there more?

When calculating work, what units should we use for pressure and volume?
Wed Jan 23, 2019 3:58 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: pKa, Ka, and relative acid strength
Replies: 3
Views: 244

### Re: pKa, Ka, and relative acid strength

Yes, you are right.

The larger the Ka, the stronger the acid. The smaller the pKa, the stronger the acid
And the same applies for bases, the larger the Kb, the stronger the base. The smaller the pKb, the stronger the base
Wed Jan 23, 2019 3:54 pm
Forum: Ideal Gases
Topic: Ideal Gas Law
Replies: 7
Views: 180

### Re: Ideal Gas Law

The ideal gas law (PV=nRT) is used to convert between molar concentration and partial pressure (Kc and Kp).
Tue Jan 22, 2019 6:12 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6th edition 12.29
Replies: 1
Views: 53

### Re: 6th edition 12.29

When dilution is involved you would use this equation: M1*V1=M2*V2 where M1 = the initial concentration in molarity, V1 = the initial volume before dilution, M2 = the concentration in molarity of the dilute solution, V2 = the volume after dilution So like in (d) you have 2.0 mL of 0.175 M KOH(aq) af...
Fri Jan 18, 2019 10:17 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Le Chatelier's Principle
Replies: 3
Views: 60

### Re: Le Chatelier's Principle

Le Chatelier’s principle states that "when a stress is applied to a system in dynamic equilibrium, the equilibrium tends to adjust to minimize the effect of the stress" (textbook). So if any changes were to be made to a reaction like increasing or decreasing the concentration, pressure, et...
Thu Jan 17, 2019 10:20 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Question 11.89
Replies: 1
Views: 62

### Re: Question 11.89

Based on the plot you know the initial kPa and the kPa at equilibrium. So you can draw out an ICE box. A B C I 28 0 0 C (-10) (+5) (+10) E 18 5 10 Hence knowing the initial and equilibrium you can determine the change, which would be 10, 5, and 10. You can simplify this ratio to 2:1:2, thus you get ...
Thu Jan 17, 2019 3:46 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: How temperature affects K
Replies: 4
Views: 103

### Re: How temperature affects K

Temperature permanently affects the rate constant of the forward and reverse reactions. Increasing/decreasing the temperature would add/remove heat changing the rate of collision of molecules of the whole reaction. But for changes in pressure, heat, or concentration, the system eventually will retur...
Wed Jan 09, 2019 9:18 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Calculating the Reaction Quotient
Replies: 3
Views: 53

### Re: Calculating the Reaction Quotient

Yes, you would use the QC equation. Since the problem only gives you moles and the equilibrium constant equation uses concentration (i.e. molarity which is mol/L), you would need to divide the given moles by 3.00 L in order to be able to calculate the QC. And you would not include the solid as a par...
Wed Jan 09, 2019 9:13 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: #26 on post-assessment
Replies: 1
Views: 46

### Re: #26 on post-assessment

You can use the I.C.E. box to help visualize/solve this. Initial molar concentration (I) Change in molar concentration (C) Equilibrium molar concentration (E) N2 H2 NH3 I 0.250 0.500 0 C -0.075 -0.225 +0.15 E 0.175 0.275 0.15 Given the change in the concentration of NH3, you can find the change in N...
Wed Jan 09, 2019 8:03 pm
Forum: Ideal Gases
Topic: Value of K
Replies: 6
Views: 120

### Re: Value of K

When K is larger than 1, there are more products at equilibrium, so equilibrium favors products ("equilibrium sits to the right"). When K is smaller than 1, there are more reactants at equilibrium, so equilibrium favors reactants ("equilibrium sits to the left"). This is because ...
Wed Jan 09, 2019 7:38 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6th edition, 11.23
Replies: 4
Views: 119

### Re: 6th edition, 11.23

One of the sig fig rules is your answer should have no more significant figures than the least accurately known number. So, 0.031 which has 2 sig figs is the "least accurately known number" and is the number of sig figs you should base your answer on.
Thu Dec 06, 2018 7:35 pm
Forum: Conjugate Acids & Bases
Topic: 12.3 6th edition
Replies: 1
Views: 81

### Re: 12.3 6th edition

Yes, you are correct.

acid: C6H5NH3+ because it donates a H+ to H2O
base: H2O because it accepts a H+ from C6H5NH3+
conjugate acid: H3O+ because it gains H+
conjugate base: C6H5NH2 because it loses H+
Thu Dec 06, 2018 7:30 pm
Forum: Lewis Structures
Topic: Lewis structure of SO2
Replies: 1
Views: 63

### Lewis structure of SO2

Can sulfur form more than an octet? I have seen SO2 drawn with two double bonds, but the textbook draws sulfur with a single bond and a double bond. Would the SO2 structure with two double bonds between O and S be more stable? Which would be the correct lewis structure to draw for SO2 to be the most...
Thu Dec 06, 2018 3:48 pm
Forum: Naming
Topic: [Co(P(CH3)3)4]2(SO4)3
Replies: 1
Views: 461

### Re: [Co(P(CH3)3)4]2(SO4)3

tetrakis(trimethylphosphine)cobalt(III) sulfate

If the ligand is polydentate (able to attach at more than one binding site), then for 4 ligands you would say tetrakis- instead of tetra-
you can figure out the charge for cobalt by solving for the oxidation state.
Tue Nov 27, 2018 9:35 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: 7th edition 2E.13b
Replies: 2
Views: 43

### Re: 7th edition 2E.13b

The central atom is the atom with the lower ionization energy. Ionization energy increases across a period, so N would have a lower ionization energy then O. Also if you were to calculate the formal charges for the structure with N as the central atom and with O as the central atom, the one with N a...
Mon Nov 26, 2018 6:56 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Problem 2E # 1 7th edition
Replies: 2
Views: 67

### Re: Problem 2E # 1 7th edition

In instances where you have structures with VSEPR formulas of AX2E3 or AX2E4, the molecular shape will be linear.; there are 2 bonded atoms and lone pairs present. However, the lone pair electrons are arranged in such a way that their repulsions cancel out to make the shape linear.
Mon Nov 26, 2018 5:24 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: 4.19 6th Edition
Replies: 2
Views: 72

### 4.19 6th Edition

The question asks to predict the shapes and estimate the bond angles. For BH2- the e- arrangement is tetrahedral and the molecular shape is bent/angular (AX2E2) so I thought the bond angle for H-B-H would be less than 109.5 degrees. But the solution manual says the H-B-H angle is slightly less than ...
Wed Nov 21, 2018 3:34 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Electron pair arrangement, actual structure
Replies: 3
Views: 71

### Re: Electron pair arrangement, actual structure

Electron arrangement is the arrangement of the electron groups, so it includes the lone pairs as part of the structure.
Molecular shape is the actual structure of the molecule, so it refers to the arrangement of atoms, not including lone pairs.
Wed Nov 21, 2018 12:56 pm
Forum: Electronegativity
Topic: electron density
Replies: 4
Views: 129

### Re: electron density

You count the number of bonded pairs and lone pairs around the central atom to find the total regions of electron density. Double and triple bonds count as one region of electron density.
Wed Nov 21, 2018 12:38 pm
Forum: Hybridization
Topic: s character
Replies: 1
Views: 36

### Re: s character

S character refers to the percentage of s in a hybridization. So for example, sp3 = 25% s-character and 75% p-character because s makes up 1/4 of the hybridization. The s character will decrease as the number of orbitals increase. If there is more s character, the bond will be stronger.
Wed Nov 14, 2018 6:42 pm
Forum: Lewis Structures
Topic: 7th Edition #2E.5(a)
Replies: 2
Views: 78

### Re: 7th Edition #2E.5(a)

The Lewis structure will have Cl in the middle (with 4 lone e-) with one O single bonded (with 6 lone e- around it) to Cl and the other O double bonded (with 4 lone e-) to Cl. Then add brackets around the Lewis structure and put the negative sign. And yes, chlorine can have an expanded octet so it c...
Wed Nov 14, 2018 1:09 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Bent vs Angular
Replies: 3
Views: 96

### Re: Bent vs Angular

Bent and angular are synonyms and are used interchangeably to refer to the same types of structure.
Mon Nov 12, 2018 2:08 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Bond Angles
Replies: 4
Views: 110

### Re: Bond Angles

Linear structures will always be 180 degrees. But trigonal planar can be 120 degrees or if it is trigonal planar bent then it will be less than 120 degrees. Tetrahedral is 109.5 degrees but if bent or trigonal pyramidal then it will be less than 109.5 degrees. So if the molecular geometry is not lin...
Fri Nov 09, 2018 10:29 am
Forum: SI Units, Unit Conversions
Topic: Angstrom
Replies: 5
Views: 429

### Re: Angstrom

Angstrom is not in SI units but it can be converted to SI units. 1 Angstrom = 1*10^-10m
Thu Nov 08, 2018 8:52 pm
Forum: Electronegativity
Topic: Hydrogen electronegativity
Replies: 3
Views: 90

### Hydrogen electronegativity

Why is the electronegativity of fluorine higher than hydrogen? but hydrogen has a high electronegativity right?
Thu Nov 08, 2018 1:12 pm
Forum: Bond Lengths & Energies
Topic: HW 2.25 part a
Replies: 3
Views: 257

### Re: HW 2.25 part a

The CN bond in H3CNH2 is longer. If you were to draw out the Lewis structures, CN in HCN forms a triple bond, while CN in H3CNH2 forms a single bond. As the number of bonds increases between atoms, the length of the bond becomes shorter. So a single bond would be longer than a triple bond. Bond leng...
Fri Nov 02, 2018 1:45 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Tungsten
Replies: 2
Views: 46

### Re: Tungsten

I think it has to do with when filling electrons, the orbitals with the lowest energy/principal quantum number would fill up first so 6s would fill up first, then, 4f, then 5d would fill up last. [Xe]6s^2 4f^14 5d^4 But then you would have to rewrite the electron configuration in order of energy, wh...
Wed Oct 31, 2018 9:57 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Is electron removed from 3d or 4s first?
Replies: 5
Views: 6178

### Re: Is electron removed from 3d or 4s first?

Orbitals are filled in the lower energy orbitals first, then fill higher energy orbitals. The 3d orbitals have a slightly higher energy than the 4s orbitals. So because the 4s orbitals has the lower energy, it gets filled first. When 3d orbitals are filled, 4s is no longer lower in energy. Hence ele...
Wed Oct 31, 2018 9:43 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Shielding Question
Replies: 1
Views: 89

### Re: Shielding Question

True: Electrons in an s-orbital are more effective than those in other orbitals at shielding other electrons from the nuclear charge because an electron in an s-orbital can penetrate to the nucleus of the atom. This is because the s orbitals are located closer to the nucleus of an atom, so the attra...
Thu Oct 25, 2018 8:05 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: quiz 2: 5-s orbital
Replies: 3
Views: 106

### Re: quiz 2: 5-s orbital

You would need to know that the s-orbital only has 1 orbital. Each orbital can only hold 2 electrons.
So if there is only 1 orbital, it can only hold a maximum of 2 electrons.
Thu Oct 25, 2018 7:51 pm
Forum: Quantum Numbers and The H-Atom
Topic: 2.29 Sixth Ed. (c)
Replies: 1
Views: 62

### Re: 2.29 Sixth Ed. (c)

To find how many electrons in an atom based on its quantum numbers, you need to determine how orbitals there are, meaning you have to first find l. You can find l given n. l = 0,1,2,n-1 n=2 so l=0,1 If l=0, then ml=0, so there is 1 orbital. If l=1, then ml= -1,0,+1, so there are 3 orbitals. Each orb...
Thu Oct 25, 2018 7:45 pm
Forum: *Shrodinger Equation
Topic: Help on how to find l, ml, and ms
Replies: 6
Views: 2514

### Re: Help on how to find l, ml, and ms

You will use n (principal quantum number) to find l. l will be a range of values starting from 0 up to n-1. If n=3, l=0,1,2 (because from n-1, you know the last value l can be is 2) If n=2, l=0,1 Once you know l, you will use l to find ml. ml= -l ,..., +l. If l=0, ml=0 if l=1, ml= -1, 0, +1 if l=2, ...
Wed Oct 24, 2018 1:05 pm
Forum: Trends in The Periodic Table
Topic: Helium
Replies: 2
Views: 74

### Re: Helium

Helium has the electron configuration 1s2 so it should belong to the s-block, but helium is placed with the p-block because helium is a noble gas (it has a full electron shell) so it is placed with the other noble gases in group 18. The periodic table is also sorted based on chemical properties, hen...
Fri Oct 19, 2018 4:30 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Shielding in Multi-Electron Atoms (1E.5)
Replies: 1
Views: 45

### Re: Shielding in Multi-Electron Atoms (1E.5)

Yes. Electrons with lower values of l are better at shielding. The lower the number of l, the closer it is to the nucleus. The electrons closer to the nucleus block (or shield) the outer electrons from getting close to the nucleus. The electron density of s- or p- orbital is much higher than the f- ...
Fri Oct 19, 2018 1:52 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Cartesian Axes
Replies: 1
Views: 64

### Re: Cartesian Axes

The cartesian axes are the x-, y-, and z- axes to represent three dimensions. This helps to describe the orientation of orbitals. So for what we would use cartesian axes for in this class, then the answer to your question would yes. It would be used to refer to the axes of the 3D orbitals.
Fri Oct 19, 2018 1:32 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Rydberg Equation
Replies: 7
Views: 216

### Re: Rydberg Equation

You would use the Rydberg Equation when calculating the wavelength of light emitted when an electron moves between energy levels. In terms of what we have learned so far in class, this equation can be applied to find the wavelength of the lines in the hydrogen emission spectrum (refer to the Balmer ...
Thu Oct 11, 2018 11:22 am
Forum: Empirical & Molecular Formulas
Topic: Homework question M.5 part b [ENDORSED]
Replies: 1
Views: 140

### Re: Homework question M.5 part b[ENDORSED]

In part (a) of this question, we determined that BrF3 is the excess reactant and ClO2 is the limiting reactant. So to find the moles of product we use ClO2 (the limiting reactant) and mole ratios (refer to the given chemical equation) to calculate the amount of product that can form: 12 mol ClO2 (6 ...
Thu Oct 11, 2018 2:14 am
Forum: Limiting Reactant Calculations
Topic: M5 [ENDORSED]
Replies: 1
Views: 180

### Re: M5[ENDORSED]

You have to solve for the limiting reactant using mole ratios: 12 mol ClO2 (6 mol ClO2F/6 moleClO2) = 12 mol ClO2F 5 mol BrF3 (6 mol ClO2F/2 mol BrF3) = 15 mol ClO2F ClO2 is the limiting reactant because it produced the lesser amount of product (only 12 mol ClO2F). Hence, BrF3 is the excess reactant...
Wed Oct 10, 2018 11:45 pm
Forum: Molarity, Solutions, Dilutions
Topic: Question G.11 (Sixth Edition)
Replies: 2
Views: 344

### Re: Question G.11 (Sixth Edition)

The question provides you with the molarity and moles of solute, and asks you to find the volume. So you will be using the formula for molarity to solve for the volume: rearrange M=n/V to V=n/M to solve for the volume. M= 0.278 M (or 0.278 mol/L) n= 0.00450 mol (the question gives you 4.50 mmol but ...
Wed Oct 03, 2018 9:23 pm
Forum: Limiting Reactant Calculations
Topic: limiting reactant problem
Replies: 3
Views: 70

### Re: limiting reactant problem

A would be the limiting reactant because in the given equation there is a ratio of 2 mol A : 1 mol B.
But since there is only 1 mol A, then you would only need 0.5 mol B.
So all 1 mol of A is consumed, however there is still 0.5 mol B excess since not all of B was needed to react with A.
Wed Oct 03, 2018 9:16 pm
Forum: Molarity, Solutions, Dilutions
Topic: E.25 part a
Replies: 2
Views: 75

### Re: E.25 part a

"Formula units" are another way to say "atoms" or "molecules."
So 6.022 x 10^23 atoms = 6.022 x 10^23 formula units
Wed Oct 03, 2018 7:42 pm
Forum: Limiting Reactant Calculations
Topic: Post-module #33
Replies: 1
Views: 104

### Post-module #33

Answer the following 4 questions for the balanced equation: 2A + 1B ---> 3C Same equation. 1 mol B reacts with 2 mol A, but there is only 1 mol A present, so it is the limiting reagent. 33. What is the maximum amount of product that can be produced? So I understood the 3 previous questions to this p...