Search found 60 matches
- Mon Mar 11, 2019 1:58 pm
- Forum: General Rate Laws
- Topic: 6th edition 15.17
- Replies: 1
- Views: 349
Re: 6th edition 15.17
When you want to figure out the reaction order of one reactant (for example: for A), choose the two experiments in which the initial concentrations of the other reactants (B in this case) stay the same. For the two experiments that you choose to compare to find the order with respect to A, only the ...
- Mon Mar 11, 2019 12:33 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Frequency factor
- Replies: 3
- Views: 499
Re: Frequency factor
Adding to the previous comment, the frequency factor is how many times the molecules will collide with the correct orientation, meaning that specific places on one molecule must collide with another molecule in a certain location. If the frequency factor is high, there will be more collisions result...
- Mon Mar 11, 2019 12:27 pm
- Forum: Zero Order Reactions
- Topic: Concept
- Replies: 4
- Views: 840
Re: Concept
A zero-order reaction occurs when the rate of the reaction in unaffected by the concentration of the reactants. Rather, it could depend on another variable like enzymes or catalysts. Zero-order reactions are typically found when a material that is required for the reaction to proceed, such as a surf...
- Mon Mar 04, 2019 8:21 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Concentration cells
- Replies: 6
- Views: 1631
Re: Concentration cells
A concentration cell is made up of two half-cells with the same electrodes but with different concentrations. This difference in ion concentration drives the flow of electrons from the more concentrated solution to the lower concentrated solution, which creates a voltage as the cell reaches equilibr...
- Mon Mar 04, 2019 8:07 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Units for rate laws
- Replies: 2
- Views: 455
Re: Units for rate laws
The reaction rate units is usually in the form mol L^-1 s^-1 (the time unit varies - s, min, hr, etc). However, the rate constant, k, is determined by the using the rate law for the order of reaction and the unit of rates and concentration. For example, the first order reaction has rate=k[A], or k=r...
- Mon Mar 04, 2019 7:50 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Order of Reaction
- Replies: 5
- Views: 812
Re: Order of Reaction
The order of a reaction determines how the concentrations and the rate of reaction are related. By examining the overall order and the individual orders of the components of a reaction, we can see which concentrations influence the rate of the reaction. We can see if the concentration of one reactan...
- Mon Feb 25, 2019 12:24 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Using commas in cell diagram
- Replies: 3
- Views: 769
Re: Using commas in cell diagram
You use a single line to separate chemicals in different phases. However, if you have multiple chemicals in the same phase (for example, 2 aqueous solutions) you can separate them with a comma. A double line, which represents a salt bridge, can be used to separate the cathode and anode. Hope this he...
- Mon Feb 25, 2019 12:13 pm
- Forum: Balancing Redox Reactions
- Topic: Redox in Basic compounds
- Replies: 2
- Views: 420
Re: Redox in Basic compounds
From the textbook, there is a general set of procedures you should follow to balance redox reactions. 1. First, separate the half reactions 2. Balance the non Hydrogen and Oxygens (usually the metal) 3. Balance the Oxygens by adding H20 on the opposite side 4. Balance the Hydrogens by adding protons...
- Mon Feb 25, 2019 12:05 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Pt in a cell diagram
- Replies: 1
- Views: 299
Re: Pt in a cell diagram
Pt is added when there is no conducting metal available to carry the electrion and lacks a solid conductor. For example, if one half-reaction consists only of aqueous species, then that sides needs Pt(s). In the textbook problem, the half-reaction of I2(s) / I- (aq) requires Pt(s) because I2(s) is a...
- Tue Feb 19, 2019 9:28 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: difference between gibbs and entropy
- Replies: 3
- Views: 449
Re: difference between gibbs and entropy
Gibbs free energy is a measure of the change in free energy of a reaction. In other words, it shows how much free energy is available to do nonexpansion work. Delta s is a measure of the change in entropy (relates to the possible number of states a system could be in). Gibbs free energy, entropy, an...
- Tue Feb 19, 2019 8:28 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Homework 9.65 6th Edition
- Replies: 2
- Views: 440
Re: Homework 9.65 6th Edition
You can use delta S (entropy) to find out whether it is stable or unstable as T increases. Using the equation delta G= delta H - TdeltaS, if delta S is negative, then -TdeltaS will be positive as T increases, resulting in a positive delta G, which causes the compound to be less stable . If delta S i...
- Tue Feb 19, 2019 8:18 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Homework 6th Edition 9.63
- Replies: 3
- Views: 422
Re: Homework 6th Edition 9.63
The question asks for the stability of the compounds. If the formation energy is positive, the compound form has more energy and is thus more unstable. If the formation energy is negative, the compound is thermodynamically stable.
Hope this helped!
Hope this helped!
- Tue Feb 12, 2019 12:00 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: enthalpy units
- Replies: 1
- Views: 330
Re: enthalpy units
If it's written as kJ/mol, it's usually used when you are describing the enthalpy of a reaction for 1 mole of a reactant/product. You also use kJ/mol when dealing with various types of enthalpies associated with individual molecules (heat of vaporization, heat of fusion, q = mcT, etc.) Usually your ...
- Tue Feb 12, 2019 11:34 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: free expansion
- Replies: 2
- Views: 353
Re: free expansion
Free expansion, (expansion in a vacuum/expansion against zero external pressure), is when no work is expended (equal to 0) when a gas increases in volume because the pressure-volume work is reliant on the external pressure.
Hope this helped!
Hope this helped!
- Tue Feb 12, 2019 11:26 am
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: 9.25 6th edition
- Replies: 4
- Views: 902
Re: 9.25 6th edition
SO2F2 has six possible configurations, which means that the degeneracy of this molecule is W=6. The change in entropy in this instance is equal to Kb(ln(W^Na)), where Kb=Boltzmann's constant, W= degeneracy, and Na=the number of particles within the molecule. To find Na, you take the one mole of SO2F...
- Mon Feb 04, 2019 11:36 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: Gas taking up more states
- Replies: 2
- Views: 402
Re: Gas taking up more states
Gases take up more states than liquids or solids due to the spacing of particles. Gases can occupy more possible positions because it occupies a larger volume compared to liquids and solids, resulting in a higher entropy. Because of the larger volume, gas molecules can freely move in a space and are...
- Mon Feb 04, 2019 11:09 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Bomb Calorimeter
- Replies: 4
- Views: 616
Re: Bomb Calorimeter
Here is a link to a visual representation of a bomb calorimeter and how it works. Basically, the ignition unit start the combustion, and since there is no change in volume, all of the energy is released as heat. This released heat raises the temperature of the surrounding water and the temperature o...
- Mon Feb 04, 2019 10:56 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.19
- Replies: 3
- Views: 870
Re: 8.19
For this problem, you would have to add the heat for copper and heat for water to find the total heat. We have to calculate the temperature rise for copper and water separately because they have different specific heats. Your equation for part a should look like this: Total heat= m x c x delta T (fo...
- Tue Jan 29, 2019 9:48 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: open, closed, isolated
- Replies: 4
- Views: 570
Re: open, closed, isolated
A thermodynamic system should be one of the three types: open, closed, or isolated Here are the definitions of each type that can help you decipher which system fits with each type of system. An isolated system cannot transfer energy or matter with its surroundings. A closed system can transfer ener...
- Tue Jan 29, 2019 8:35 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond Enthalpy
- Replies: 2
- Views: 337
Re: Bond Enthalpy
We use the bond enthalpy to calculate delta H.
This is the equation to find delta H and gives the change in enthalpy for the reaction.
ΔH= (energy needed to break the reactant bonds) - (energy released by forming the product bonds)
This is the equation to find delta H and gives the change in enthalpy for the reaction.
ΔH= (energy needed to break the reactant bonds) - (energy released by forming the product bonds)
- Tue Jan 29, 2019 6:06 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Calorimeter
- Replies: 4
- Views: 551
Re: Calorimeter
A calorimeter is an instrument that is used to measure the amount of energy transferred as heat during a chemical reaction. We can use it to calculate the specific heat of a substance.
- Wed Jan 23, 2019 4:18 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sig Figs
- Replies: 3
- Views: 502
Re: Sig Figs
When calculating pH and pOH, count the number of sig figs given in the concentration and the pH/pOH should have that number of sig figs after the decimal. For example, if the pH=-log(1.4x 10^-6) then pH = 5.85 (two sig figs in 1.4 x 10^-6 and two after the decimal in 5.85).
Hope this helped!
Hope this helped!
- Wed Jan 23, 2019 4:02 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: pKa, Ka, and relative acid strength
- Replies: 3
- Views: 779
Re: pKa, Ka, and relative acid strength
You are correct!
I believe it is the same for bases; decreasing pKb and increasing Kb will correspond to increasing base strength.
An example of this concept can be found in 12.39 in the 6th edition.
I believe it is the same for bases; decreasing pKb and increasing Kb will correspond to increasing base strength.
An example of this concept can be found in 12.39 in the 6th edition.
- Wed Jan 23, 2019 3:53 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 6th Edition 12.27
- Replies: 2
- Views: 691
Re: 6th Edition 12.27
For part a, you plug in 0.025 directly into the equation [pH= negative log (H3O+) ] For part b, you use the dilution equation M1V1 = M2V2 (you are solving for M2 in this case) in order to find the actual concentration of HCl, which will then give you the H30+ concentration to plug into the equation....
- Wed Jan 16, 2019 8:52 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Q vs K [ENDORSED]
- Replies: 4
- Views: 610
Re: Q vs K [ENDORSED]
K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage of a reaction. The direction of a reaction can be determined by comparing Q and K. The difference is that equilibrium constant (K) is the ratio between the ...
- Wed Jan 16, 2019 8:29 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Acid and Base Equilibria
- Replies: 7
- Views: 836
Re: Acid and Base Equilibria
Like you said, weak acids do not dissociate 100%, so they will produce less H3O+, resulting in a higher pH value. The concentration of H3O+ will be lower than that of a stronger acid, since stronger acids almost always dissociates completely.
- Wed Jan 16, 2019 8:12 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: partial pressure v concentration
- Replies: 4
- Views: 369
Re: partial pressure v concentration
My TA mentioned that we can just write K, unless stated specifically to use partial pressure or concentration. But, I would just write Kp or Kc just to make sure!
- Wed Jan 09, 2019 8:36 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Q vs. K
- Replies: 2
- Views: 413
Re: Q vs. K
When Q=K, the reaction is in equilibrium and there is no shift. This link goes into more detail about the relationships between K and Q. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemi...
- Wed Jan 09, 2019 7:37 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th edition, 11.23
- Replies: 4
- Views: 427
Re: 6th edition, 11.23
I believe the answer should have two sig figs because the Kc given in the problem is 0.031, which is 2 sig figs. Even though the other given values have three sig figs, you always want to use the one with the least amount of sig figs, so the final answer for Br2 would have two sig figs. Hope this he...
- Tue Jan 08, 2019 9:16 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.7
- Replies: 5
- Views: 560
Re: 11.7
The question asks for when it reaches equilibrium, so it would be the third flask because there is no dissociation in flasks 3 and 4 (they are both 5; no change after the third flask). The diatomics in each flask from left to right is 11, 8, 5, 5. Flask 3 is the time at which the reaction reached eq...
- Mon Dec 03, 2018 1:14 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Conjugate
- Replies: 2
- Views: 364
Re: Conjugate
The conjugate seesaw describes the relative strength of an acid or base to its conjugate base or acid. Here are points to remember... stronger acid = conjugate base will be weaker stronger base= conjugate acid will be weaker weaker acid= conjugate base will be stronger weaker base= conjugate acid wi...
- Mon Dec 03, 2018 1:08 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Ligand names
- Replies: 7
- Views: 999
Re: Ligand names
Yes, you should memorize the chart that he provided on the website under "Naming Coordination Compunds." I believe he will test us by giving us a compound and we would have to name it using ligand names.
- Mon Dec 03, 2018 1:03 pm
- Forum: Biological Examples
- Topic: 17.31
- Replies: 1
- Views: 176
17.31
Hi all! I'm having trouble figuring out how to write the formula from the given using the oxidation number.
For example...
for tetraamminediaquacobalt (III) bromide, why is the three placed with Bromine?
Thank you!
For example...
for tetraamminediaquacobalt (III) bromide, why is the three placed with Bromine?
Thank you!
- Mon Nov 26, 2018 8:30 pm
- Forum: Hybridization
- Topic: hybrid orbitals
- Replies: 1
- Views: 310
Re: hybrid orbitals
After drawing the lewis structures, it asks you to find the hybrid orbitals, which you can find by finding out the number of atoms bonded to the atom + the number of lone pairs the atom has. BF3 has three bonds and no lone pairs. There is a total of 3 bonds, which is trigonal. The hybrid orbital is ...
- Mon Nov 26, 2018 8:23 pm
- Forum: Hybridization
- Topic: 4.73
- Replies: 2
- Views: 403
Re: 4.73
After drawing the lewis structures, all of them have paired electrons, meaning they are diamagnetic. Radicals are an atom, molecule, or ion that has an unpaired valence electron. Therefore, none of the structures in the problem are radicals.
- Mon Nov 26, 2018 4:51 pm
- Forum: Hybridization
- Topic: 4.81
- Replies: 1
- Views: 310
4.81
I was wondering if we will be asked to draw something like the structure in question 4.81 on Test #3 (6th edition). I have a hard time trying to draw the structure (B3N3H6) and I was wondering if anyone had any suggestions in trying to attempt to draw the structure.
Thank you!
Thank you!
- Mon Nov 19, 2018 3:59 pm
- Forum: Sigma & Pi Bonds
- Topic: Rotating Pi bonds
- Replies: 3
- Views: 469
Re: Rotating Pi bonds
I also agree with the answers above! Pi bonds cannot rotate the same way as sigma bonds do since the rotation would break the pi bond interaction. Compared to pi bonds which have their electron density above and below the bond axis (bonding sideways), sigma bonds have their electron density along th...
- Mon Nov 19, 2018 3:45 pm
- Forum: Lewis Structures
- Topic: Lewis Structure of BrO3-
- Replies: 3
- Views: 10226
Re: Lewis Structure of BrO3-
Like the answer above, this structure breaks the octet rule to minimize the formal charges. The total number of valence electrons for BrO3- should be 26. The final structure should have formal charges of 0, except for one of the oxygens (which should be -1). To get the most stable structure, There a...
- Mon Nov 19, 2018 3:38 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Formal Charge
- Replies: 8
- Views: 3763
Re: Formal Charge
Instead of using the formula, a shortcut in finding the formal charge is (number of valence electrons - number of the dots/number of lines).
Here is a good youtube video I found explaining how to use the shortcut method!
Hope this helps!
https://www.youtube.com/watch?v=38fSDZypRB0
Here is a good youtube video I found explaining how to use the shortcut method!
Hope this helps!
https://www.youtube.com/watch?v=38fSDZypRB0
- Wed Nov 14, 2018 12:12 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 5
- Views: 844
Re: Bond Angles
https://www.angelo.edu/faculty/kboudrea/index/Handout_Lewis_VSEPR.pdf I had the same question at first, but I used resources online (like the link above) to help me know the bond angles. I think as you do more problems, you start to familiarize yourself with the different angles according to their ...
- Tue Nov 13, 2018 11:57 pm
- Forum: Sigma & Pi Bonds
- Topic: Sigma v. Pi
- Replies: 6
- Views: 1029
Re: Sigma v. Pi
Another key note to keep in mind is that sigma bonds are stronger than pi bonds. This is because the atomic orbitals forming sigma bonds overlap to a greater extent compared to the orbitals forming pi bonds. Here is a link of an image comparing sigma bonds and pi bonds. https://qph.fs.quoracdn.net/m...
- Tue Nov 13, 2018 6:18 pm
- Forum: Lewis Structures
- Topic: VSEPR Model
- Replies: 2
- Views: 387
Re: VSEPR Model
A polar molecule occurs when two atoms do not share electrons equally, one in which one end of the molecule has a slight positive charge and the other end has a slight negative charge. This will occur whenever the molecule is not completely symmetric. Examples of this would be H20, NH3, SO2, H2S, et...
- Tue Nov 06, 2018 1:52 pm
- Forum: Trends in The Periodic Table
- Topic: ionization energy & electron affinity
- Replies: 3
- Views: 446
Re: ionization energy & electron affinity
Ionization energy is the energy required to remove an electron from an atom, whereas electron affinity is the energy involved when an electron adds to an atom.
The difference between the two is exemplified below:
Ca → Ca⁺ + e⁻ [Ionization Energy]
Ca + e⁻ →Ca⁻ [Electron Affinity]
The difference between the two is exemplified below:
Ca → Ca⁺ + e⁻ [Ionization Energy]
Ca + e⁻ →Ca⁻ [Electron Affinity]
- Tue Nov 06, 2018 1:41 pm
- Forum: Lewis Structures
- Topic: Identifying the element
- Replies: 8
- Views: 1078
Re: Identifying the element
Yes, mentioning stable structure and formal charge (how it equals to 0) would be the correct answer for the reasoning. If the formal charge doesn't equal 0, then it probably wouldn't be the correct element.
- Tue Nov 06, 2018 1:33 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Which electron disappears for ion configurations
- Replies: 2
- Views: 455
Re: Which electron disappears for ion configurations
Here is an example: Ga^(3+) The electron configuration for just Ga would be [Ar]4s^2 3d^10 4p^1. Since there is a 3+ attached to Ga (meaning you have to take away 3 electrons), you would get rid of 4p^1 first since it is the highest. Then you would take out 4s^2. Therefore your final answer for Ga^(...
- Tue Oct 30, 2018 4:31 pm
- Forum: Sigma & Pi Bonds
- Topic: Strength of bonds
- Replies: 1
- Views: 586
Re: Strength of bonds
In sigma bonds, linear overlapping occurs whereas in pi bonds, parallel overlapping occurs. In pi bonds, the electron density is parallely concentrated, which results in less stronger bonds. In sigma bonds, electron density is concentrated along the x-axis. The greater the extent of overlapping, the...
- Tue Oct 30, 2018 4:18 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: 2.25 6th Edition
- Replies: 2
- Views: 448
Re: 2.25 6th Edition
Personally, the best way to approach this problem is to remember these key points: s orbital; l=0 p orbital; l=1 d orbital; l=2 For example for 2.25C) The problem gives you 3d. From this, you know that n=3 and that l= 2. The number of electrons would be 10 because ms=-2,-1,0,1,2; there are 5 orienta...
- Tue Oct 30, 2018 4:12 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: degeneracy
- Replies: 2
- Views: 382
Re: degeneracy
Degeneracy means the electron orbitals have the same energy level. To determine the degeneracy of orbitals, Consider the three orbitals s, p, d: s orbital l=0; the ml would be 0, therefore no degeneracy p orbital l=1; the ml would be -1, 0 +1, therefore degeneracy=3 d orbital l=2, ml would be -2, -1...
- Wed Oct 24, 2018 1:18 pm
- Forum: DeBroglie Equation
- Topic: DeBroglie Equation Post Assessment
- Replies: 4
- Views: 711
Re: DeBroglie Equation Post Assessment
Your formula is correct; After you rearrange the DeBroglie Equation to set it equal to the velocity, you get v = h/(wavelength) (mass). When you plug in the numbers into the equation, you should get 1.00 m.s^-1 as the answer. It might be a simple calculation error! h= 6.62608 x 10^-34 J.s wavelength...
- Wed Oct 24, 2018 1:09 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: Heisenberg Indeterminacy Post Assessment #20
- Replies: 2
- Views: 437
Re: Heisenberg Indeterminacy Post Assessment #20
You use the answer of the velocity you found in problem #19. Then you want to use the kinetic energy formula. Make sure you convert the answer you get to energy per mol of electrons (multiply the answer with 6.022x10^23e/mol). When are you asked to find the uncertainty in kinetic energy, just rememb...
- Wed Oct 24, 2018 12:56 pm
- Forum: Quantum Numbers and The H-Atom
- Topic: m with subscript l
- Replies: 4
- Views: 782
Re: m with subscript l
Just like what the others said, m with the subscript l is the magnetic quantum number which describes an orbital's orientation. It can be found from the angular momentum (l). If l was 1 for example, then the possible ml values would be -1, 0, and 1.
- Tue Oct 16, 2018 12:44 pm
- Forum: Properties of Light
- Topic: 1A.3 7th Edition Question
- Replies: 4
- Views: 545
Re: 1A.3 7th Edition Question
I agree with all the comments above. The extent of the change in the electrical field and amplitude correspond with each other. When frequency decreases, the slope of the wave decreases because of the change in amplitude. Therefore, the extent of change also decreases as the frequency of the electro...
- Tue Oct 16, 2018 12:29 pm
- Forum: Einstein Equation
- Topic: 7th edition 1B.5
- Replies: 5
- Views: 727
Re: 7th edition 1B.5
Kiloelectron volt (keV) is one thousand electron volts. To convert keV to joules, you just need to remember the formula that 1 J= 1.6022x 10^-19 keV. Joules and eV is connected in the sense that the potential difference of 1 volt causes an electron to gain an amount of energy.
- Tue Oct 16, 2018 12:14 pm
- Forum: Photoelectric Effect
- Topic: 1B.7
- Replies: 2
- Views: 422
Re: 1B.7
To find b, you have to first convert 5.00 mg Na to atoms. 5.00 mg Na x 1g/1000mg Na = 5.00 x 10^-3 g Na 5.00 x 10^-3 g Na x 1 mol/22.99 g Na x 6.022 x 10^23 atoms/1 mol ... = your answer After you find the atoms, you have to multiply the atoms with the answer you got in part a (3.37 x 10^-19 J. atom...
- Thu Oct 11, 2018 6:42 pm
- Forum: Properties of Electrons
- Topic: Atomic Spectrum
- Replies: 2
- Views: 319
Re: Atomic Spectrum
Just like the comment above, the atomic spectrum shows a series of spectral lines, which are used to identify atoms and molecules. When the atoms are excited, energy or light is emitted, which corresponds to different colored lines. The colored lines produced by the atomic spectra is what helps to i...
- Thu Oct 11, 2018 6:30 pm
- Forum: Empirical & Molecular Formulas
- Topic: Is it possible for the Empirical and Molecular formula to be the same? [ENDORSED]
- Replies: 8
- Views: 1843
Re: Is it possible for the Empirical and Molecular formula to be the same? [ENDORSED]
Yes, it is definitely possible! If the molecular formula cannot be simplified any further, then the empirical formula would be the same as the molecular formula. The empirical formula is the simplest ratio of atoms in a molecule, and sometimes the molecular formula cannot be simplified, explaining t...
- Thu Oct 11, 2018 1:32 pm
- Forum: Properties of Light
- Topic: Memorizing wavelengths and frequencies
- Replies: 5
- Views: 3141
Re: Memorizing wavelengths and frequencies
I feel that it is essential to remember the order of the electromagnetic spectrum to be safe. One trick I use to memorize the order is coming up with an acronym. A common one is "Real Monkeys Insist Very Useful X-mas Gifts" (Radio, Microwaves, Infrared, Visible, Ultraviolet, X-rays, Gammas...
- Wed Oct 03, 2018 1:10 pm
- Forum: Balancing Chemical Reactions
- Topic: Net Number of Moles
- Replies: 3
- Views: 332
Re: Net Number of Moles
After balancing the equation, you get 4C4H10 + 26O2 -> 16CO2 + 20H20. To calculate the net number of moles of gas produced in this question, you just have to remember this simple formula: The sum of the coefficients of the product - sum of the coefficients of reactants So for this question, it would...
- Wed Oct 03, 2018 1:01 pm
- Forum: Significant Figures
- Topic: Sig Figs when finding Mol
- Replies: 4
- Views: 10028
Sig Figs when finding Mol
Hi Everyone! I was wondering if sig figs matter when finding the mol. For Example (Module 1: #5) Mol of C: 54. 82 g/12.01 g. mol^-1= 4.56 mol OR 4.564 mol? Regardless of which one we use, I know that we would get the same answer in the end after dividing by the smallest number, but I was wondering i...
- Wed Oct 03, 2018 12:37 pm
- Forum: Significant Figures
- Topic: Significant figures for molar mass
- Replies: 7
- Views: 5550
Re: Significant figures for molar mass
In my chemistry discussion this morning, my TA informed our class that we will be given a periodic table to use during the test. He said that it would most likely go to two significant figures. So, Carbon would be 12.01. Depends on the periodic table given. Just like the other replies above, I would...