## Search found 64 matches

Sat Jan 16, 2021 9:37 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Sapling HW #2
Replies: 6
Views: 42

### Re: Sapling HW #2

Yes, fill in the information for the ICE table. You have the initial concentration of SO3 and the equilibrium concentration of O2. Using the equilibrium concentration of O2, you can figure out the change and fill out the rest of the table and use the equilibrium values to calculate Kc.
Sat Jan 16, 2021 9:32 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Textbook 6D.13 part b
Replies: 2
Views: 27

### Re: Textbook 6D.13 part b

Since CH3NH3+ is the conjugate acid of a weak base (CH3NH2), it will affect the pH.
CH3NH3+ + H2O <-> CH3NH2 + H3O+
[CH3NH3+] = [CH3NH3Cl] = 0.20 because salts dissociate in water
the Kb of CH3NH2 is given in table 6C.2, but Ka is needed (Ka = Kw/Kb)
Sat Jan 16, 2021 9:20 pm
Forum: Calculating the pH of Salt Solutions
Topic: Textbook Problem 6D.13
Replies: 2
Views: 9

### Re: Textbook Problem 6D.13

the pH of d is larger than b. The value you calculated for d is the pOH, so you just need to convert it to pH.
Sat Jan 16, 2021 9:17 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook Question 6D.15 (b)
Replies: 1
Views: 12

### Re: Textbook Question 6D.15 (b)

Al3+ can interact with water to form Al(H2O)6 3+. Table 6D.1 includes that example and the Ka value. Figure 6D.2 next to the table also explains why some metal ions are acidic.
Sat Jan 16, 2021 9:12 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Textbook Question 6E.1
Replies: 2
Views: 7

### Re: Textbook Question 6E.1

I think we have to calculate the [H3O+] from the second deprotonation because Ka2 is still relatively large and will affect the pH.
Sun Jan 10, 2021 10:39 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Chemical Equilibrium Part 2 Question 29
Replies: 2
Views: 11

### Re: Chemical Equilibrium Part 2 Question 29

2BrCl(g) ⇌ Br2 (g) + Cl2(g)
1.84 x 10-4
-2x +x +x
3.37x10-5 x x
E: [BrCl] = 0.183(1.84 x 10-4) = 3.37x10-5
[Br2] = [Cl2] = x and 2x = 3.37x10-5
Sun Jan 10, 2021 10:33 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Chemical Equilibrium Part 2 Question 27
Replies: 3
Views: 13

### Re: Chemical Equilibrium Part 2 Question 27

C(s) + H2O (g) ⇌ CO (g) + H2 (g) 0.05 M 0 0 -x +x +x 0.05-x x 0.040 I: [H2O] = 2.5mol/50L = 0.05 M; pure solids (C) don't affect K and you can disregard the amount of C C: the change is x and all the coefficients are 1 E: the equilibrium concentration of H2 is given, so you can figure out x and equi...
Thu Jan 07, 2021 5:25 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium Partial Pressure (Problem 5I.3)
Replies: 1
Views: 12

### Re: Equilibrium Partial Pressure (Problem 5I.3)

Your setup is correct, but the Kc value you have identified from the table corresponds to 298K while the problem states that the temperature is 500K. That is probably why your answer doesn't match the solution manual
Thu Jan 07, 2021 3:48 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 5.35
Replies: 2
Views: 26

### Re: 5.35

From the graph, the partial pressures at equilibrium are A=17kPa, B=5kPa, and C=10kPa. The change from B initial to B equilibrium is 5. Assume x = 5, then the change from C initial to C equilibrium is 2x and the change from A initial to A final is also 2x. After checking, x = 5 makes sense. The stoi...
Wed Jan 06, 2021 8:56 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Difference between "sits" and "shifts"
Replies: 5
Views: 52

### Re: Difference between "sits" and "shifts"

I think it would be more appropriate to use "sits" when the system is at equilibrium and you need to identify whether products (R) or reactants (L) are favored. This implies no change to the system and equilibrium doesn't change/move. "Shift" would be more appropriate when a chan...
Wed Jan 06, 2021 8:49 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Stability of reactants vs products
Replies: 5
Views: 23

### Re: Stability of reactants vs products

for reactants <-> products if K is large, then there are more products and the forward rxn is favored. That means reactants are less likely to stay as reactants and become whatever the products are, so the products would be more stable than the reactants. if K is small, then there are more reactants...
Tue Jan 05, 2021 9:08 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Audio-Visual Module Question #15
Replies: 3
Views: 19

### Re: Audio-Visual Module Question #15

a) water is added since H2O (reactant) is added, more products would be formed to use up the extra H2O. Another way of approaching this is by comparing the equilibrium constant and the reaction quotient. If there is an increase in H2O, then Q would be smaller than K. To adjust and reach equilibrium,...
Sat Dec 12, 2020 8:48 pm
Forum: Bronsted Acids & Bases
Topic: Why is H2SO3 an acid when the oxygens have lone pairs?
Replies: 2
Views: 29

### Re: Why is H2SO3 an acid when the oxygens have lone pairs?

SO3^2- has 2 oxygens with a formal charge of -1 and those are the sites where H+ from H2O will form HSO3- or H2SO3. That makes SO3^2- a base.
Sat Dec 12, 2020 8:38 pm
Forum: Naming
Topic: Textbook Question 9C.1
Replies: 5
Views: 45

### Re: Textbook Question 9C.1

OH2: aqua, (CN)5: pentacyano, Co: cobaltate because the complex has a negative charge
aquapentacyanocobaltate(III) ion
Sat Dec 12, 2020 8:34 pm
Forum: Conjugate Acids & Bases
Topic: Sapling 14
Replies: 6
Views: 72

### Re: Sapling 14

Yes, it will be neutral. This is how I like to think of it. When pH>pKa, that means the surrounding is more basic and there are less H+ and more OH- present. Since there are more OH- present, the equilibrium will shift left and produce more B, which is neutral.
B + H2O <--> BH+ + OH-
Sat Dec 12, 2020 8:25 pm
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Prioritizing Polarizing Power of Polarizabilty
Replies: 3
Views: 22

### Re: Prioritizing Polarizing Power of Polarizabilty

It may be easier to approach this problem by comparing electronegativities where the smaller the difference the greater the covalent character. If the difference of electronegativities is high, that means that electrons are more likely to be attracted to the more electronegative atom and display cha...
Sat Dec 12, 2020 8:14 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: HBrO2 vs HClO2
Replies: 7
Views: 68

### Re: HBrO2 vs HClO2

In HBrO2 and HClO2, the H is bonded to the O, so the strength of the acid depends on the stability of the anions (BrO2- and ClO2-). ClO2- is more stable because Cl is more electronegative than Br which means it can stabilize the negative charge on the O by withdrawing e-
Sun Dec 06, 2020 7:44 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Textbook 2E 5
Replies: 2
Views: 25

### Re: Textbook 2E 5

a) what is the shape of a ClO2 + ion
b) what is the expected OClO bond angle
Sun Dec 06, 2020 7:25 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Figuring Out monodentate, bidentate, etc
Replies: 5
Views: 38

### Re: Figuring Out monodentate, bidentate, etc

A ligand is a species bound to the metal ion which means it can be an atom or a molecule. If a ligand is a molecule that has multiple sites with lone pairs and single bonds that allow it to rotate in a way where multiple sites can bind to the metal ion simultaneously, then it is polydentate.
Sun Dec 06, 2020 7:18 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Chelates
Replies: 3
Views: 26

### Re: Chelates

A chelate is formed when the ligand is polydentate.
Sun Dec 06, 2020 7:17 pm
Forum: Formal Charge and Oxidation Numbers
Topic: Sapling #6
Replies: 3
Views: 21

### Re: Sapling #6

The coordination sphere has an overall charge of +1 since the oxidation state of Cl (outside) is -1 and the entire unit is neutral. The oxidation state of NH3 is 0 and the oxidation number of Cl (inside) is -1 and there is only one Cl within the coordination sphere so the oxidation state of Co must ...
Fri Dec 04, 2020 5:25 pm
Forum: Properties & Structures of Inorganic & Organic Acids
Topic: Calculations involving strong acids
Replies: 3
Views: 34

### Re: Calculations involving strong acids

The 100% ionization of HCl means that all 0.1 M of HCl will become H+ and Cl- HCl(aq) --> H+ (aq) + Cl-(aq) or HCl(aq) + H2O(l) --> H3O+(aq) + Cl-(aq) 0.1M of HCl yields 0.1M of H+ and 0.1M of Cl- because of 1:1 between HCl and H+/Cl- In this scenario, [H+]=0.1M so calculating the pH would be simple...
Sun Nov 29, 2020 2:30 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Sapling #3
Replies: 5
Views: 75

### Re: Sapling #3

In an octahedral arrangement, it doesn't matter which position (equatorial or axial) is made the lone pair. If the equatorial position is made the lone pair, there will be four 90 angles. Two between the two adjacent equatorial atoms; two between the two axial atoms. If the axial position is made th...
Wed Nov 25, 2020 1:19 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Chelate Lecture Example
Replies: 1
Views: 17

### Re: Chelate Lecture Example

Yes, the ligand that forms a chelate is considered one ligand. It just has multiple bonding sites. Yes, NH2-CH2-CH2-NH2 is bidentate because there are two sites bonded to Co (i.e. the two N from NH2)
Wed Nov 25, 2020 1:14 pm
Forum: Hybridization
Topic: Chelate example from the lecture
Replies: 1
Views: 13

### Re: Chelate example from the lecture

The ligand is NH2-CH2-CH2-NH2 and the Ns from NH2 are bonded to Co. That particular ligand is bidentate that has two regions bonded to Co. A ring is formed between Co, NH2, CH2, CH2, and NH2. The Cl is a different ligand that's monodentate because it can only donate 1 e- pair at a time.
Wed Nov 25, 2020 1:05 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: EDTA example in 11/25 lecture
Replies: 5
Views: 63

### Re: EDTA example in 11/25 lecture

From the structure shown in lecture, EDTA4- has six atoms with single bonds and lone pairs, which means when it rotates into a certain position, EDTA4- can bond with the metal ion at those six sites
Wed Nov 25, 2020 1:00 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: How to tell whether a ligand is polydentate
Replies: 2
Views: 36

### Re: How to tell whether a ligand is polydentate

From the lecture, I think the ligand needs to have single bonds so it can rotate and have more than one site with lone pairs to form a chelate
Sun Nov 22, 2020 8:40 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: sapling #4
Replies: 3
Views: 16

### Re: sapling #4

From the textbook (2E), square pyramidal has 5 bonding pairs and 1 lone pair.
Sun Nov 22, 2020 8:37 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: polarity from lewis structure
Replies: 7
Views: 28

### Re: polarity from lewis structure

The shape of CO2 is linear, so it is nonpolar because the dipoles cancel.
Fri Nov 20, 2020 8:32 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Textbook 2E.1b
Replies: 2
Views: 15

### Textbook 2E.1b

Below are ball-and-stick models of two molecules. In each case, indicate whether there must be, may be, or cannot be one or more lone pairs of electrons on the central atom. b) O--O--O and the angle is 180 The answer from the book is "may have lone pairs" but I thought there couldn't be lo...
Tue Nov 17, 2020 10:14 pm
Forum: Lewis Structures
Topic: midterm question re: sapling #3
Replies: 3
Views: 36

### Re: midterm question re: sapling #3

My TA said we don't need to memorize the formulas. If the names are given, the formulas would also be provided.
Tue Nov 17, 2020 10:08 pm
Forum: Dipole Moments
Topic: Textbook Question 3F.13
Replies: 1
Views: 23

### Re: Textbook Question 3F.13

For II, the dipole moments are aligned in a way where the slightly positive end of the molecule is closest to the slightly negative end of the molecule +-> +-> +-> For III, the dipole moments are pointing up and down while the molecules are adjacent to each other so the oppositely charged ends of th...
Sun Nov 15, 2020 6:13 pm
Forum: Lewis Structures
Topic: Number of Bonds for Elements
Replies: 3
Views: 23

### Re: Number of Bonds for Elements

For atoms in period 3 and beyond, electrons can be in the d state because n=3 and l=0, 1, 2. This allows S or Cl to have more than 4 bonds and break the octet rule. For nitrogen, it follows the octet rule and can have up to 4 bonds, but it often forms 3 bonds and has 1 lone pair because its formal c...
Sun Nov 15, 2020 5:54 pm
Forum: Ionic & Covalent Bonds
Topic: Sapling Q: Hydrogen Bonds
Replies: 7
Views: 55

### Re: Sapling Q: Hydrogen Bonds

Regarding the second requirement that you mentioned, the "highly electronegative atom" that is bonded to a hydrogen atom can only be N, O, or F. That means a hydrogen bond can only form between a hydrogen atom that is bonded to N, O, or F and an electronegative atom (N, O, or F) with a lon...
Thu Nov 12, 2020 6:16 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: textbook problem 2A #9,11
Replies: 2
Views: 20

### Re: textbook problem 2A #9,11

I'll use 9A as an example I believe you added 2 electrons to 3d, which led you to believe that there were 9 electrons in the 3d state. For the transition metals, 4s needs to be filled before 3d is filled (except for Cr and Cu). So, the 2 electrons should have been removed from the 4s state and not 3...
Thu Nov 12, 2020 6:02 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: e- p
Replies: 1
Views: 14

### Re: e- p

the "p" is rho (Greek) and it means density
Thu Nov 12, 2020 1:01 pm
Forum: Trends in The Periodic Table
Topic: Ionization energy of Copper?
Replies: 2
Views: 24

### Re: Ionization energy of Copper?

The e- configuration of Cu is [Ar]3d^104s^1, so the 4s electron would be removed if it was ionized. The electron in the 4s subshell is further from the nucleus, so the effective nuclear charge is lower due to the "shielding" of electrons in lower energy levels. It would take less energy to...
Sun Nov 08, 2020 12:32 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Textbook Exercise 2A.3
Replies: 2
Views: 14

### Re: Textbook Exercise 2A.3

You wrote it in the correct order. 3d comes before 4s because n=3 is a lower energy state than n=4
You can also check your answers with the answer key on Sapling
Sun Nov 08, 2020 12:11 pm
Forum: Properties of Light
Topic: Equation clarifications
Replies: 4
Views: 43

### Re: Equation clarifications

If you are solving the energy of radiation (no mass), E=hc/λ
If you are solving the energy of e- (has mass) and is given the de Broglie wavelength, Ek = 0.5mv^2 and v = h/mλ where v is velocity
Sun Nov 08, 2020 11:49 am
Forum: Lewis Structures
Topic: Lone Pairs Question
Replies: 22
Views: 119

### Re: Lone Pairs Question

Lone pairs are pairs of valence electrons that are not shared or not bonded. In Lewis structures, lone pairs are represented as two dots (i.e. 2 non-bonding e-)
Fri Nov 06, 2020 2:25 pm
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Polarizability and Polarizing Power trends
Replies: 2
Views: 20

### Re: Polarizability and Polarizing Power trends

It is true that the number of electrons increase across the period, but the number of protons also increases, which results in a higher nuclear charge that would pull the electrons in. So, the atomic radius decreases across the period.
Thu Nov 05, 2020 6:54 pm
Forum: Lewis Structures
Topic: Formal Charge
Replies: 5
Views: 28

### Re: Formal Charge

The structure is more stable if the formal charge is 0. The sum of the formal charge must always add up to the net charge, which helps you check your work.
Sun Nov 01, 2020 6:42 pm
Forum: Student Social/Study Group
Topic: Sappling #25
Replies: 4
Views: 46

### Re: Sappling #25

E(photon): size of bacterium = λ of photon = 2.3 μm To convert μm to m, (2.3 μm)(10^-6) E(photon) = hc/λ E(e-): it's the kinetic energy de Broglie λ = 2.3 μm; Ek = 0.5mv^2 convert μm to m we need the velocity of e- to solve for the kinetic energy of the electron λ=h/mv --> v = h/mλ solve of Ek with ...
Sun Nov 01, 2020 6:33 pm
Forum: Student Social/Study Group
Topic: Sampling HW #7
Replies: 2
Views: 17

### Re: Sampling HW #7

First, you can determine the energy of the radiation with λ=751nm using E = hc/λ. Remember the energy calculated is joules per photon. Then decide how much energy is needed to melt 417g of ice by multiplying the 417 by the enthalpy of fusion (found in the table). To determine how many photons are ne...
Sun Nov 01, 2020 10:41 am
Forum: Properties of Light
Topic: Sapling Hw #5
Replies: 5
Views: 65

### Re: Sapling Hw #5

With the wavelength, you can calculate the energy in joules/photon. When you multiplied the mass of ice with the enthalpy of fusion, you get the total energy in joules to melt the ice. In order to find out how many photons are needed to melt the ice, divide (333.6 x 423) by the energy of the radiati...
Sat Oct 31, 2020 7:33 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Sapling Homework
Replies: 8
Views: 56

### Re: Sapling Homework

Also, the subshells (s, p, d) need to be italicized
Wed Oct 28, 2020 7:45 pm
Forum: Molarity, Solutions, Dilutions
Topic: Dilution module
Replies: 6
Views: 65

### Re: Dilution module

Not necessarily. When you solve for the volume, you would get liters, but the options are in milliliters, so you need to convert your answer in L to mL.
Thu Oct 22, 2020 8:23 pm
Forum: Einstein Equation
Topic: Sapling Homework - electron affinity
Replies: 5
Views: 82

### Re: Sapling Homework - electron affinity

To convert 1.003eV/atom to kJ/mol, I used dimensional analysis
multiply (1.003eV/atom) by (6.022x10^23 atoms/mol) --> get eV/mol
then multiply by (1.602x10^-19 J/1 eV) --> get J/mol
then multiply by (1 kJ/1000 J) --> get kJ/mol
Thu Oct 22, 2020 8:13 pm
Forum: Photoelectric Effect
Topic: Sapling HW Weeks 2, 3, 4 Question #9
Replies: 10
Views: 104

### Re: Sapling HW Weeks 2, 3, 4 Question #9

The maximum wavelength of the radiation that will eject electrons occurs when the energy of the incident light equals the work function. When you are given the work function, you can calculate the wavelength of the radiation.

work function = E(photon) = hc/(wavelength)
Thu Oct 22, 2020 8:07 pm
Forum: DeBroglie Equation
Topic: Sapling Homework - electron affinity
Replies: 9
Views: 107

### Re: Sapling Homework - electron affinity

Yes, the length of the bacterium will be the wavelengths of the photon and electron. To find the energy of the photon, use E = h(frequency) and c = λ(frequency) and solve for E The energy of the electron is its kinetic energy, which involves mass and velocity. To find the velocity of e-, use de Brog...
Thu Oct 22, 2020 8:00 pm
Forum: SI Units, Unit Conversions
Topic: SI Conversions
Replies: 11
Views: 82

### Re: SI Conversions

I don't know if we will be given easy units that don't have to be converted, but it is best to know how to convert units since we already have to do that in Sapling and textbook problems
Thu Oct 22, 2020 7:57 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Indeterminacy Problem
Replies: 3
Views: 28

### Re: Indeterminacy Problem

use ∆p = m∆v, where m=8.00 kg and ∆v=10 m/s
then use ∆p∆x ≥ h/4π to find ∆x
btw, the answer in the back of the textbook is incorrect because they used ∆v=5 m/s, so that may have caused you some trouble
Sat Oct 17, 2020 7:28 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Photon Frequency and Energy Between Ground States
Replies: 3
Views: 13

### Re: Photon Frequency and Energy Between Ground States

The electron will not absorb photons that doesn't match the energy differences between energy levels. The specificity results in unique spectral lines that can be used to identify the atom (or molecule).
Sat Oct 17, 2020 7:24 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Midterm #1
Replies: 3
Views: 33

### Re: Midterm #1

If you are referring to the Heisenberg uncertainty principle, it is part of 1B
Sat Oct 17, 2020 6:59 pm
Forum: Properties of Light
Topic: Textbook Problems (Topic 1B)
Replies: 1
Views: 37

### Re: Textbook Problems (Topic 1B)

This is how I solved the problem:
Convert 140.511keV to joules
Used E = hv, v=c/(wavelength) -> E = hc/(wavelength) to solve for wavelength
convert m to pm using 1m = 1x10^12 pm
Fri Oct 16, 2020 1:57 pm
Forum: Properties of Electrons
Topic: Using empirical equation for h-atom
Replies: 4
Views: 29

### Re: Using empirical equation for h-atom

The -1/16 comes from -1/n^2 of En = -hR/n^2, where n=4
Fri Oct 16, 2020 1:52 pm
Forum: Properties of Light
Topic: rydberg equation
Replies: 7
Views: 61

### Re: rydberg equation

Hi
It's not necessary to know the Rydberg equation, but you do need to know how to solve difference in energy using the professors way
Sun Oct 11, 2020 7:18 pm
Forum: Limiting Reactant Calculations
Topic: Help finding limiting reactant
Replies: 4
Views: 59

### Re: Help finding limiting reactant

One way of finding the limiting reactant is to determine how much product each reactant yields. In this scenario, you also need to find the theoretical yield of H2O, so you can calculate how much H2O is produced from 0.0625 moles of O2, and how much H2O is produced from 0.006795 moles of aspartame. ...
Thu Oct 08, 2020 5:50 pm
Forum: DeBroglie Equation
Topic: Post-module assessment question #35
Replies: 3
Views: 53

### Re: Post-module assessment question #35

The wavelength of the electric car is so small that it is unnoticeable and insignificant so you can say large objects like an electric car don't have wavelike properties.
Thu Oct 08, 2020 5:41 pm
Forum: SI Units, Unit Conversions
Topic: Review HW E23
Replies: 2
Views: 47

### Re: Review HW E23

Your process is not wrong. I tried your method and I got the correct answer, so you may have messed up the calculations somewhere which gave you a wrong answer.

mass % of Cu2+ = 28.45%
3.00(0.2845) = 0.8535g Cu2+
0.8535g Cu2+ /(63.546g/mol Cu) = 0.0134 mol Cu2+
Thu Oct 08, 2020 10:37 am
Forum: Empirical & Molecular Formulas
Topic: Finding the Empirical and Molecular Formulas
Replies: 3
Views: 47

### Re: Finding the Empirical and Molecular Formulas

From the mass of CO2, you can determine how many moles and grams of C were present in the caproic acid. From the mass of H20, you can determine how many moles and grams of H were present in the caproic acid. The remaining mass of the caproic acid sample would be the grams of O present in the caproic...
Thu Oct 08, 2020 10:23 am
Forum: Limiting Reactant Calculations
Topic: Sapling HW Q10
Replies: 2
Views: 46

### Re: Sapling HW Q10

Hello, You do have to write out the formulas so you calculate for the molar masses, but it isn't necessary to figure out the limiting reactant because it states in the prompt that 2-butanone is the limiting reactant and the other reactants are in excess. When you convert milliliters of 2-butanone to...
Sun Oct 04, 2020 9:54 pm
Forum: Limiting Reactant Calculations
Topic: Week 1 Sapling HW Chem 14A Problem 10
Replies: 10
Views: 198

### Re: Week 1 Sapling HW Chem 14A Problem 10

You are correct. 2-butanone is the limiting reactant. To find the theoretical yield, you would use the given mass of 2-butanone and it's density (0.81g/ml) to determine the moles of 2-butanone. Using the mole to mole ratio of 2-butanone and 3-methyl-3-hexanol, you can determine the mole and the mass...
Sun Oct 04, 2020 9:43 pm
Forum: Empirical & Molecular Formulas
Topic: Week 1 Sapling HW Chem 14A Problem 9
Replies: 8
Views: 178

### Re: Week 1 Sapling HW Chem 14A Problem 9

To start, you can determine the moles and masses of carbon and hydrogen from the given mass of carbon dioxide and water. After figuring out the masses of carbon and hydrogen, you can determine the mass of oxygen by subtracting 1.000g of caproic acid by the combined mass of carbon and hydrogen. With ...

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