Search found 60 matches
- Fri Mar 15, 2019 5:51 pm
- Forum: Second Order Reactions
- Topic: 15.39a - "0.37[A]o"
- Replies: 2
- Views: 543
15.39a - "0.37[A]o"
15.39 Determine the time required for each of the following second-order reactions to take place: (a) 2 A -> B + C, for the concentration of A to decrease from 0.10 mol/L to 0.080 mol/L, given that k = 0.015 L-1 mol min for the rate law expressed in terms of the loss of A. In the 6th edition solutio...
- Fri Mar 15, 2019 5:42 pm
- Forum: Balancing Redox Reactions
- Topic: Test 2
- Replies: 2
- Views: 510
Re: Test 2
You would have to refer to the standard reduction potential chart. The more positive the value of E⁰, the stronger the oxidation power.
- Wed Mar 13, 2019 5:30 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Liquid Conductors [ENDORSED]
- Replies: 1
- Views: 276
Liquid Conductors [ENDORSED]
In 14.23a, the galvanic cell is constructed in the solutions manual as: Hg(l) | Hg 2 2+ || NO 3 -(aq), H+(aq) | NO(g) | Pt(s) I understand that liquids may be used as conductors. Can a liquid in the anode/cathode always be used as a conductor? If not, how can it be determined whether it can be used ...
- Sat Mar 09, 2019 10:47 am
- Forum: Administrative Questions and Class Announcements
- Topic: Midterm Review
- Replies: 3
- Views: 483
- Sat Mar 09, 2019 10:46 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Test 2 Gibbs Free Energy
- Replies: 5
- Views: 708
Re: Test 2 Gibbs Free Energy
Why would the cold/heat pack activation be considered spontaneous?
- Sat Mar 09, 2019 10:44 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Slow Step
- Replies: 5
- Views: 680
Re: Slow Step
What would happen if you have 2 slow steps? Is the slower step still the determining rate? Is having more than 2 slow steps possible? You can still have multiple "slow" reactants, but ultimately the slowest one will still determine the overall rate law. (An extremely small bottleneck will...
- Sun Mar 03, 2019 9:19 am
- Forum: Administrative Questions and Class Announcements
- Topic: Final Topics
- Replies: 6
- Views: 684
Re: Final Topics
The only things you should have to remember from 14A are any "fundamentals" that are applied throughout various sections in 14B (i.e. balancing a reaction)
- Sun Mar 03, 2019 9:12 am
- Forum: Van't Hoff Equation
- Topic: Question 11.111 (Sixth Edition)
- Replies: 1
- Views: 591
Re: Question 11.111 (Sixth Edition)
Use the formula deltaGr⁰ = -(RT)lnK; solving for K gives K = e^-(deltaGr⁰/RT) Solve for the first step reaction: K1 = 10K2 = e^-(G1⁰/RT) = e^(2.00*10^5/RT) For the second step reaction: K2 = e^-(G2⁰/RT) Combining the two steps gives the reaction 10 = e^(deltaG1⁰+deltaG2⁰/RT) = e^(2.00*10^5+deltaG2⁰/...
- Wed Feb 27, 2019 3:05 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagrams
- Replies: 4
- Views: 414
Re: Cell Diagrams
Conducting solids (electrodes) should be on the outside of the cell, and aqueous should be on the inside next to the salt bridge.
See viewtopic.php?f=140&t=43085
See viewtopic.php?f=140&t=43085
- Fri Feb 22, 2019 11:09 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: salt bridge
- Replies: 5
- Views: 564
Re: salt bridge
Why is it favorable for the ions to flow through the salt bridge to the anode from the cathode? Is it because the electron flow from the anode to the cathode makes the cathode negatively charged the cathode negatively charged and the anode positively charged, thus making the salt bridge ion flow fav...
- Fri Feb 22, 2019 11:04 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Explanation
- Replies: 3
- Views: 482
- Fri Feb 22, 2019 10:57 am
- Forum: Student Social/Study Group
- Topic: Midterm Results
- Replies: 4
- Views: 540
Re: Midterm Results
Tony Ong 3K wrote:does anyone know if we can argue our midterm scores?
Try to contact your TA about it
- Fri Feb 15, 2019 11:28 am
- Forum: Student Social/Study Group
- Topic: last problem on midterm
- Replies: 4
- Views: 859
Re: last problem on midterm
Athena L 1B wrote:What was the problem for this one?
This was question 8A for the Fall 2018 Chem 14A Midterm
- Fri Feb 15, 2019 11:26 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: HOTDOG problem
- Replies: 3
- Views: 2167
Re: HOTDOG problem
You do flip the enthalpies when the equation is flipped. The final equation should be (2*-345kJ) + (2*21kJ) + (150kJ), where the first equation is multiplied by 2, the second equation is flipped and multiplied by 2, and the final equation is flipped as to give the final reaction 2GingerAle + 5RootBe...
- Tue Feb 12, 2019 3:13 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 23911
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
Madeline Ho 1C wrote:Can someone explain how to do #3?
I got deltaHrxn= 1mol(1560 kJ/mol) - [1mol(-1300 kJ/mol)+2mol(-286 kJ/mol)] = 3432 kJ
Using Hess' Law will give the reaction
deltaHrxn = (1mol)(-1300kJ/mol) + (2mol)(-286kJ) + (1mol)(1560kJ) = -312kJ
- Sun Feb 10, 2019 5:33 pm
- Forum: Ideal Gases
- Topic: Calculating Pressure of a Gas Mixture Using PV=nRT
- Replies: 3
- Views: 573
Calculating Pressure of a Gas Mixture Using PV=nRT
It is given that the gas mixture has 2.250 mol of helium gas and 1.492 mol of krypton gas. The temperature of the system is 348.15K and the volume of the system is 11.0L. To calculate the total pressure of the system, do we first calculate the pressure using mol of helium gas, then calculate the pre...
- Fri Feb 08, 2019 4:36 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: If not Gibbs Free Energy then what is the last topic for midterm?
- Replies: 4
- Views: 449
Re: If not Gibbs Free Energy then what is the last topic for midterm?
According to Lavelle's important midterm information document on his website, the midterm will cover "Chemical Equilibrium, Acid and Base Equilibria, Thermochemistry, Thermodynamics to the end of entropy."
- Fri Feb 01, 2019 2:28 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Open, closed, or isolated systems
- Replies: 10
- Views: 39230
Re: Open, closed, or isolated systems
Some examples:
Open - An unsealed flask of liquid; an open soda can
Closed - A sealed flask of liquid, an unopened soda can
Isolated - The universe; A high-quality sealed thermos*; a bomb calorimeter*
* (these are not 100% isolated, but for our reference the heat transfer is approximated to 0)
Open - An unsealed flask of liquid; an open soda can
Closed - A sealed flask of liquid, an unopened soda can
Isolated - The universe; A high-quality sealed thermos*; a bomb calorimeter*
* (these are not 100% isolated, but for our reference the heat transfer is approximated to 0)
- Fri Feb 01, 2019 2:24 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Midterm
- Replies: 4
- Views: 402
- Fri Feb 01, 2019 2:18 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Isolated Systems
- Replies: 6
- Views: 783
Re: Isolated Systems
The bomb calorimeter is also an example of an "isolated system." There is no truly 100% isolated system other than the universe (arguably). But we refer to a high-quality sealed thermos or a bomb calorimeter as an isolated system because the heat transfer is so slow that it can be approxim...
- Fri Jan 25, 2019 10:00 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: adding two reactions together
- Replies: 2
- Views: 272
Re: adding two reactions together
From the lecture on 1/25/19:
N2(g) + O2(g) --> 2NO(g) (ΔHrxn = 180kJ)
2NO(g) + O2(g) --> 2NO2(g) (ΔHrxn = -112 kJ)
Adding the above reactions gives the net reaction:
N2(g) + 2O2(g) --> 2NO2(g) (Hnet = 68kJ)
N2(g) + O2(g) --> 2NO(g) (ΔHrxn = 180kJ)
2NO(g) + O2(g) --> 2NO2(g) (ΔHrxn = -112 kJ)
Adding the above reactions gives the net reaction:
N2(g) + 2O2(g) --> 2NO2(g) (Hnet = 68kJ)
- Fri Jan 25, 2019 9:54 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Homework
- Replies: 5
- Views: 783
Re: Homework
@504909207
Probably Outline 3 (Thermochemistry and The First Law of Thermodynamics) since we have already started covering that material and the 1st and 2nd outlines have already been tested on
Probably Outline 3 (Thermochemistry and The First Law of Thermodynamics) since we have already started covering that material and the 1st and 2nd outlines have already been tested on
- Fri Jan 25, 2019 9:52 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Stable carbon
- Replies: 3
- Views: 300
Re: Stable carbon
It can also be seen in this carbon phase diagram that carbon is in the form of granite at standard conditions
- Sat Jan 19, 2019 6:49 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Studying for First Discussion Test
- Replies: 10
- Views: 830
Re: Studying for First Discussion Test
Lavelle emphasizes to do the homework as a way to prepare for the tests.
- Sat Jan 19, 2019 6:41 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Post Assessment Part 2
- Replies: 1
- Views: 149
Re: Post Assessment Part 2
If 18.3% of the BrCl gas remains at equilibrium and we have the initial concentration of BrCl, it is possible to calculate the amount of BrCl left at equilibrium. 0.183 * (1.84 * 10^-4) = 3.36 * 10^-5 M of BrCl at equilibrium. Then use the ICE box and plug in this value: 2BrCl(g) ⇌ Br2 (g) + Cl2(g) ...
- Tue Jan 15, 2019 9:28 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Negative pH
- Replies: 12
- Views: 1575
- Fri Jan 11, 2019 9:29 am
- Forum: Ideal Gases
- Topic: What does K represent?
- Replies: 9
- Views: 2540
Re: What does K represent?
K is the equilibrium constant, which is derived from dividing the concentration of product by the concentration of reactant. (K=[P]/[R]). When a reaction is at equilibrium, products and reactants are being formed at equal rates, so the concentration of either remains constant.
- Fri Jan 11, 2019 9:26 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Help with Q 11.7.c (6e)
- Replies: 1
- Views: 220
Re: Help with Q 11.7.c (6e)
In part C, K = (Partial pressure of X)^2 / (Partial Pressure of X2)
Plug in the mole fractions of each molecule (X = 12 molecules/17 molecules) and (X2 = 5 molecules/17 molecules)
Plugging those fractions to K and multiplying by the initial pressure of X2 (0.10 bar) will give the answer 0.17 .
Plug in the mole fractions of each molecule (X = 12 molecules/17 molecules) and (X2 = 5 molecules/17 molecules)
Plugging those fractions to K and multiplying by the initial pressure of X2 (0.10 bar) will give the answer 0.17 .
- Wed Jan 09, 2019 2:20 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Reaction Quotient
- Replies: 5
- Views: 434
Re: Reaction Quotient
K refers to the [P]/[R] ratio at equilibrium. Although Q also refers to the [P]/[R] ratio, it can be measured at any point in time in a reaction--including equilibrium. If Q<K, then the forward reaction (products forming) is favored. If Q>K, then the reverse reaction (reactants forming) is favored. ...
- Sat Dec 08, 2018 2:06 pm
- Forum: Administrative Questions and Class Announcements
- Topic: FINAL PRACTICE - Lyndon's Churro Review Session [ENDORSED]
- Replies: 118
- Views: 21416
Re: FINAL PRACTICE - Lyndon's Churro Review Session [ENDORSED]
Reese - Dis 1G wrote:Can someone explain 27? I thought you only added -ate when the compound was negative.
If K2 has a +2 charge, then [Ni(CN)4] must have a -2 charge since the overall compound is neutral. It's called "nickelate" because the compound in the brackets has the -2 charge.
- Sat Dec 08, 2018 1:51 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Energy Levels
- Replies: 2
- Views: 704
Re: Energy Levels
The electron gains energy when it jumps from a lower energy level to a higher energy level.
- Fri Dec 07, 2018 9:00 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: Review questions
- Replies: 3
- Views: 1377
Re: Review questions
You are also already told initially that H2Se has a higher boiling point than H2S, so you don't have to worry about determining which one has the higher boiling point. All you have to do is explain why H2Se is the one with the higher boiling point, and in this case the only possible way for H2Se to ...
- Fri Dec 07, 2018 2:13 am
- Forum: Amphoteric Compounds
- Topic: Amphiprotic vs. amphoteric
- Replies: 3
- Views: 498
Re: Amphiprotic vs. amphoteric
I understand that all amphiprotic compounds can be amphoteric compounds. But what is an example of an amphoteric compound that is not amphiprotic?
- Fri Nov 30, 2018 6:09 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Acid vs base [ENDORSED]
- Replies: 15
- Views: 1566
Re: Acid vs base [ENDORSED]
Lewis Acids accept electrons while Lewis Bases donate electrons.
This is not to be confused with Bronsted:
Bronsted acids donate protons while Bronsted Bases accept protons.
This is not to be confused with Bronsted:
Bronsted acids donate protons while Bronsted Bases accept protons.
- Fri Nov 30, 2018 5:59 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Physical properties of acids and bases
- Replies: 1
- Views: 170
Re: Physical properties of acids and bases
Hydronium ions trigger taste buds that send a signal of "sourness" to the brain.
Hydroxide ions break down fatty acids and oils on the skin which reduces friction on your skin, which is why they feel soapy.
Hydroxide ions break down fatty acids and oils on the skin which reduces friction on your skin, which is why they feel soapy.
- Fri Nov 30, 2018 5:55 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Pentagonal bipyramidal
- Replies: 4
- Views: 1332
Re: Pentagonal bipyramidal
ZrF73- and HfF73 also have pentagonal bipyramidal shape.
- Sat Nov 24, 2018 3:41 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 4.25 6th edition: Polarity of SF4
- Replies: 4
- Views: 742
Re: 4.25 6th edition: Polarity of SF4
SF4 has the AXE notation AX4E. Therefore it has the see-saw molecular shape which is asymmetric and the dipole moments do not cancel out.
- Sat Nov 24, 2018 3:31 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Incomplete octet
- Replies: 2
- Views: 284
Re: Incomplete octet
Aluminum can also sometimes have an incomplete octet for molecules like AlCl3. However, AlCl3 only exists at high temperatures--in normal temperatures, two molecules of AlCl3 combine to form Al2Cl6 where both Al atoms obey the octet rule.
- Sat Nov 24, 2018 3:26 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: CH 4 HW 4.41 6TH EDITION
- Replies: 3
- Views: 271
Re: CH 4 HW 4.41 6TH EDITION
In acrylonitrile, yes, the C-N bond is a triple bond. However not every C-N bond has to be a triple bond (eg. Methylamine)
- Fri Nov 16, 2018 1:01 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: CH 4 4.91 HW 6TH EDITION
- Replies: 1
- Views: 100
- Fri Nov 16, 2018 1:00 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: CH 4 HW 4.41 6TH EDITION
- Replies: 3
- Views: 271
- Fri Nov 16, 2018 12:36 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Linear for AX2E3
- Replies: 2
- Views: 191
Re: Linear for AX2E3
The 3 lone electrons form a trigonal shape in the equatorial plane as it is the most favorable in terms of electron repulsion, so the two other bonded atoms will take on a linear shape.
- Sun Nov 11, 2018 11:56 am
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3620069
- Sun Nov 11, 2018 11:47 am
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: Dipole vs Induced Dipole
- Replies: 3
- Views: 15914
Re: Dipole vs Induced Dipole
A dipole is a molecule with opposite charges on both ends (positive on one side, negative on the other side). For example, H2O is a dipole as the hydrogens have a partial positive charge and the oxygen has a partial negative charge. An induced dipole is the result of two molecules interacting and ca...
- Wed Nov 07, 2018 11:13 pm
- Forum: Ionic & Covalent Bonds
- Topic: Covalent or ionic?
- Replies: 4
- Views: 4133
Re: Covalent or ionic?
Be and Br have similar electronegativities. Since the electronegativity values of Be and Br are so close, they both pull the electron with nearly the same strength, meaning the electron is closer to the middle and is shared between both Be and Br
- Thu Nov 01, 2018 6:23 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron Configuration of Ruthenium
- Replies: 1
- Views: 845
Electron Configuration of Ruthenium
The electron configuration of Ru is [Kr] 4d^7 5s^1.
Why is the electron configuration not [Kr] 4d^6 5s^2? Wouldn't it make more sense to have a filled 5s and a partially filled 4d rather than a half-filled 5s and partially filled 4d?
Why is the electron configuration not [Kr] 4d^6 5s^2? Wouldn't it make more sense to have a filled 5s and a partially filled 4d rather than a half-filled 5s and partially filled 4d?
- Thu Nov 01, 2018 5:56 pm
- Forum: Ionic & Covalent Bonds
- Topic: Electron Configuration of a Cation (hw problem 3.21 part d)
- Replies: 4
- Views: 382
Re: Electron Configuration of a Cation (hw problem 3.21 part d)
Ag has an electron configuration of [Kr] 4d^10 5s^1 because the element is much more stable when the electrons are configured in the element to have a full d shell and a half-full s shell. When Ag is ionized (Ag+), the electron is removed from the outermost shell 5s. Therefore, the electron configur...
- Thu Nov 01, 2018 4:26 pm
- Forum: Lewis Structures
- Topic: Determining Lewis Structure
- Replies: 3
- Views: 261
Re: Determining Lewis Structure
The correct Lewis structure is the one with as many atoms having a formal charge of 0. Sometimes for compounds like SO3 2- there are variations of the Lewis structure that all have the same overall formal charge which is where resonance comes into play (multiple possible lewis structures)
- Thu Oct 25, 2018 11:40 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: The values of L and ML
- Replies: 3
- Views: 314
Re: The values of L and ML
l refers to the subshells of n. The allowed values of l depend on the value of n and range from 0 to n-1. For example, if n=3, then l can = 0, 1, or 2.
- Thu Oct 25, 2018 11:31 pm
- Forum: Trends in The Periodic Table
- Topic: CH 2 6TH EDITION 2.85
- Replies: 2
- Views: 229
Re: CH 2 6TH EDITION 2.85
In the heavier transition-metal elements, the energy levels of the subshells get so close that their energy levels actually depend on how many electrons are occupying them. For example, 4s is filled before 3d because 4s has a lower energy level (rather than 3d filling before 4s like we might normall...
- Thu Oct 25, 2018 11:13 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: H spectrum
- Replies: 3
- Views: 448
Re: H spectrum
The Rydberg equation does not take into account electron-electron repulsion, which is why it's okay to use for Hydrogen and H-spectrum (only one electron to worry about). It will not work for atoms with multiple electrons because of electron-electron repulsion.
- Fri Oct 19, 2018 9:45 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electrostatic Attraction
- Replies: 5
- Views: 4941
Re: Electrostatic Attraction
Electrostatic attraction refers to the attraction between a positively charged nucleus and a negatively charged electron.
- Mon Oct 15, 2018 11:11 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Negative v. Positive when calculating energy
- Replies: 8
- Views: 2523
Re: Negative v. Positive when calculating energy
When delta E is negative, it means that energy has been released. Positive delta E refers to the absorption of energy.
- Mon Oct 15, 2018 11:03 pm
- Forum: Student Social/Study Group
- Topic: Note Taking
- Replies: 145
- Views: 16871
Re: Note Taking
Handwriting notes is very helpful because I feel that I comprehend the material more, but it's all a matter of personal preference.
- Fri Oct 12, 2018 12:10 am
- Forum: Einstein Equation
- Topic: Symbol for frequency
- Replies: 12
- Views: 2289
Re: Symbol for frequency
Both are commonly used as symbols for frequency. Perhaps you could use whichever one is given to you in the problem but ultimately it should not matter.
- Wed Oct 10, 2018 11:42 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Question G.11 (Sixth Edition)
- Replies: 2
- Views: 730
Re: Question G.11 (Sixth Edition)
Use the formula M = n/V. In the context of this problem, solve for V.
V = n/M = (4.50 * 10^-3 mol)/(0.278 mol*L^-1) = 1.62 * 10^-2 L, or 16.2 mL
V = n/M = (4.50 * 10^-3 mol)/(0.278 mol*L^-1) = 1.62 * 10^-2 L, or 16.2 mL
- Wed Oct 10, 2018 5:07 pm
- Forum: Empirical & Molecular Formulas
- Topic: Problem L39
- Replies: 1
- Views: 380
Re: Problem L39
The crucible and product together weigh 28.35g. Subtract the mass of the crucible (26.45g) from the mass of the crucible and the product (28.35g) to get the mass of the product (1.900g). Since we know we started with 1.500g of Sn, you can subtract that mass from the mass of the product (1.900g) to f...
- Fri Oct 05, 2018 11:52 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Fraction of total mass due to oxygen?
- Replies: 1
- Views: 566
Re: Fraction of total mass due to oxygen?
To clear up any confusion, it's referring to the oxygen itself, not oxygen gas molecules. To find the fraction of the total mass of the sample due to oxygen, divide the molar mass of the oxygen by the molar mass of the entire compound. (4 * 15.999 g mol^-1) / (206.53 g mol^-1) (The 4 comes from the ...
- Fri Oct 05, 2018 11:37 am
- Forum: Molarity, Solutions, Dilutions
- Topic: Water Added to Solution Placed in a New Flask
- Replies: 3
- Views: 615
Re: Water Added to Solution Placed in a New Flask
Since only 20 mL of the first solution is diluted, you should first find the molarity of the first solution and using that molarity and 20 mL volume as your M1V1 later on. Start by finding the molarity of the original solution. Convert the 5.00g of KMnO 4 into moles by dividing it by the molar mass....
- Fri Oct 05, 2018 11:21 am
- Forum: General Science Questions
- Topic: Periodic Table
- Replies: 16
- Views: 1664
Re: Periodic Table
I will typically use the exact atomic weight listed on the periodic table and round my final answer if needed. Future problems should supposedly have given atomic weights