you use this equation in order to solve for calorimeter problems because you want the energy of the system, not surroundings
q=CΔT
q=heat energy
C=heat constant
ΔT=change in temperature
Search found 104 matches
- Tue Mar 10, 2020 10:43 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Steps to solve calorimeter problem
- Replies: 3
- Views: 908
- Tue Mar 10, 2020 10:34 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 7D.1
- Replies: 4
- Views: 311
Re: 7D.1
we use ln(k prime/k) = (Ea/R)(1/T-1/T prime) = (Ea/R)((T prime-T)/T(T prime))
ln (k prime/k)=ln(0.87 s^-1/0.76 s^-1)
=(Ea/8.31x10^-3 kJ/K.mol)((1030K-1000K)/(1030Kx1000K))
Ea=((8.31x10^-3kJ/K.mol)(1000K)(1030K)/(1030K-1000K))ln(0.87 s^-1/0.76 s^-1)
=39 kJ/mol
ln (k prime/k)=ln(0.87 s^-1/0.76 s^-1)
=(Ea/8.31x10^-3 kJ/K.mol)((1030K-1000K)/(1030Kx1000K))
Ea=((8.31x10^-3kJ/K.mol)(1000K)(1030K)/(1030K-1000K))ln(0.87 s^-1/0.76 s^-1)
=39 kJ/mol
- Tue Mar 10, 2020 10:29 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: 7.11
- Replies: 2
- Views: 229
Re: 7.11
a. The objective is to reproduce the observed rate law. If step 2 is the slow step, if step 1 is a rapid equilibrium, and if step 3 is fast also, then our proposed rate law will be rate=k2[N2O2][H2]. Consider the equilibrium of step 1: k1[NO]^2=k prime[N2O2] [N2O2]=k1/k prime [NO]^2 Substituting in ...
- Tue Mar 10, 2020 10:19 pm
- Forum: First Order Reactions
- Topic: 7.27 half life
- Replies: 1
- Views: 149
Re: 7.27 half life
use equation for half-life of a first order rxn t1/2=ln 2/k k=ln 2/t1/2 k=0.154 h^-1 ln([A]0/[A]t)=kt [A]/[A]0=e^-kt after two hours this ratio is, [A]/[A]0=e^-(0.154 h^-1)(2 h)=0.735 the mass of phenobarbital that remains after 2 hours is 0.735 x 150 mg = 110 mg. to restore the initial amount 40 mg...
- Tue Mar 10, 2020 10:15 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 7.11
- Replies: 1
- Views: 97
Re: 7.11
the objective is to reproduce the observed rate law. if step 2 is the slow step, if step 1 is a rapid equilibrium, and if step 3 is fast also, then our proposed rate law will be rate= k2[N2O2][H2]. consider the equilibrium of step 1: k1[NO]^2=k prime [N2O2] [N2O2]=k1/k prime[NO]^2 substituting in ou...
- Tue Mar 03, 2020 10:44 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: cell diagram question
- Replies: 3
- Views: 216
Re: cell diagram question
sorry, I meant instead not inside
- Tue Mar 03, 2020 10:44 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: cell diagram question
- Replies: 3
- Views: 216
cell diagram question
Pb(s)|Ni^3+,Ni^2+,H^+(aq)||I^2+(aq)|I(s)|Pb(s)
In this case, should the solids on the right side also have a comma inside of a line like the aqueous on the left?
In this case, should the solids on the right side also have a comma inside of a line like the aqueous on the left?
- Tue Mar 03, 2020 10:40 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: calculating K2?
- Replies: 3
- Views: 601
calculating K2?
The ionic dissociation of water is given by the following rxn: The ΔH° for the rxn is 58 kJ/mol. The Kw for the rxn at 25 celsius is 10^-14. Is a pH of 7 acidic or basic at 10 celsius? 2H2O(l)⇄H3O^+(aq)+OH^-(aq) ln K2/K1=-ΔH°/R [1/T2-1/T1] 25 celsius=298.15K 10 celsius=283.15K 58kJ/mol=58,000 J/mol ...
- Tue Mar 03, 2020 10:30 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: calculate largest E cell
- Replies: 1
- Views: 265
calculate largest E cell
Please write the two species that would give the largest E°cell if assembled in an electrochemical cell. (a) Pt^2+(aq)+2e-→ Pb(s): E°=1.20V Cu^2+(aq)+2e-→ Cu(s) E°=0.34V Pt^2+(aq)+2e-→ Pt(s) E°=1.20V (b) Ti^2+(aq)+2e-→ Ti(s) E°=-1.63V Sn^2+(aq)+2e-→ Sn(s) E°=-0.14V Mg^2+(aq)+2e-→ Mg(s) E°=-2.36V I d...
- Tue Mar 03, 2020 10:18 pm
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: change in gibbs free energy
- Replies: 4
- Views: 322
change in gibbs free energy
Choose the one that best describes the change in Gibbs free energy for the process described; briefly justify your answer using a short sentence. The photosynthesis of glucose inside the chloroplast of an old growth Coastal Redwood tree. ΔG<0 ΔG=0 ΔG>0 I don't understand why the answer isn't ΔG>0?
- Sun Mar 01, 2020 11:50 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: calculating pH [ENDORSED]
- Replies: 2
- Views: 230
calculating pH [ENDORSED]
If cell potential is 0.67 V at 25 celsius, what is pH? [Fe^2+]=0.25M, [MnO4^-]=0.30 M, [Mn^2+]=0.01 M, [Fe^3+]=0.05M 5Fe^(2+)+MnO4^(-)+H^(+) → Mn^(2+)+4H2O+5Fe^(3+) E°=0.74 V The answer should be H^+=0.04347 and pH=1.36, however I get different numbers. Can someone tell me what I'm doing wrong? E=E°...
- Sun Mar 01, 2020 11:00 pm
- Forum: Balancing Redox Reactions
- Topic: 6K.5 part b
- Replies: 4
- Views: 688
Re: 6K.5 part b
rxn of bromine w/ itself (disproportionation) in aqueous solution: Br2(l)→BrO3^-(aq)+Br^-(aq) Br2(l)+2e-→2Br^-(aq) (balanced half rxn) Br2(l)+6H2O(l)→2BrO3^-(aq) (O balanced) Br2(l)+6H2O(l)+12OH^-(aq)→2BrO3^-(aq)+12H2O(l) (H balanced) Br2(l)+12OH^-(aq)→2BrO3^-(aq)+6H2O(l)+10e- (e-s balanced, the H2O...
- Tue Feb 25, 2020 9:32 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: what is value of E°?
- Replies: 2
- Views: 155
what is value of E°?
Consider the following unbalanced half reaction.
Ag(s)+I^-(aq) →AgI(s)+e-
E°=-0.15 V
AgCl(s)+e- →Ag(s)+Cl^-(aq)
E=+0.22V
What is value of E°?
E°=E°cath-E°anode
I'm confused as to whether it should be 0.22-(-0.15)=0.37V or 0.22-0.15=0.07V.
Ag(s)+I^-(aq) →AgI(s)+e-
E°=-0.15 V
AgCl(s)+e- →Ag(s)+Cl^-(aq)
E=+0.22V
What is value of E°?
E°=E°cath-E°anode
I'm confused as to whether it should be 0.22-(-0.15)=0.37V or 0.22-0.15=0.07V.
- Tue Feb 25, 2020 9:26 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: predict std. rxn for this galvanic cell
- Replies: 1
- Views: 124
predict std. rxn for this galvanic cell
Predict the standard potential for the following galvanic cells: Cu(s)|Cu^2+(aq)||Au^2+(aq)|Au(s) Cu^2+(aq)+2e- → Cu(s) E°=0.34V Au^+(aq)+e- → Au(s) E°=1.69V 2Au^+(aq)+2e-→ 2Au(s) Cu(s) → Cu^2+(aq)+2e- E°=-0.34 V (e-s are cancelled out) 2Au^+(aq)+Cu(s) →2Au(s)+Cu^2+(aq) E°=E°cath-E°anode The solutio...
- Tue Feb 25, 2020 9:15 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: determining unknown quantity in cell
- Replies: 2
- Views: 181
determining unknown quantity in cell
Determine the unknown quantity in the cell. E=0.050 V Pb(s)|Pb^2+(aq,?)|Ni^2+(aq, 0.20 mol/L)|Ni(s) Pb^2+(aq)+2e-→Pb(s) E°=-0.13 V Ni^2+(aq)+2e-→Ni(s) E°=-0.23 V Pb(s)→Pb^2+(aq)+2e- Ni^2+(aq)+2e-→Ni(s) e-s are cancelled out Pb(s)+Ni^2+(aq)→Pb^2+(aq)+Ni(s) E°cell=E°cath - E°anode =-0.23V-(+0.13V) =-0...
- Sun Feb 23, 2020 3:24 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 6L.7 C
- Replies: 2
- Views: 194
Re: 6L.7 C
Cd(OH)2(s)+2e- →Cd(s)+2OH^-(aq) E°(anode)=-0.81V Ni(OH)3(s)+e-→ Ni(OH)2(s)+OH^-(aq) E°(cathode)=+0.49V Reversing the anode rxn and multiplying the cathode rxn by 2 yields Cd(s)+2OH^-(aq)→Cd(OH)2(s)+2e- 2Ni(OH)3+2e-→2Ni(OH)2(s)+2OH^-(aq) then, upon addition, 2Ni(OH)2(s)+Cd(s)→Cd(OH)2(s)+2Ni(OH)2(s) o...
- Sun Feb 23, 2020 3:16 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Textbook question 4.51
- Replies: 1
- Views: 151
Re: Textbook question 4.51
b. the value at 100 celsius should be exactly 0, because this is the normal boiling point of water c. the discrepancy arises because the enthalpy and entropy values calculated from the tables are not rigorously constant with temperature. better values would be obtained using the actual enthalpy and ...
- Sun Feb 23, 2020 3:03 pm
- Forum: Balancing Redox Reactions
- Topic: 6K.5 a
- Replies: 1
- Views: 96
Re: 6K.5 a
O3 (g) →O2(g) O3(g) →O2(g)+H2O(l) (balances O) 2H2O(l)+O3(g) →O2(g)+H2O(l)+2OH^-(aq) (balances H) H2O(l)+O3(g) →O2(g)+2OH^-(aq) (cancels H2O) H2O(l)+O3(g)+2e- →O2(g)+2OH^-(aq) (balances charge) Br^-(aq) →BeO3^-(aq) 3H2O(l)+Br^-(aq) →BrO3^- (balances O) 6OH^-(aq)+3H2O(l)+Br^-(aq) →BrO3^-(aq)+^H2O(l) ...
- Sun Feb 23, 2020 2:51 pm
- Forum: Balancing Redox Reactions
- Topic: 6k. d
- Replies: 1
- Views: 115
Re: 6k. d
3[P4(s)+8OH-(aq) →4H2PO2^-(aq)+4e-]
P4(s)+12H2O(l)+12e- →4PH3(g)+12OH^-(aq)
4P4(s)+12H2O(l)+24OH^-(aq)+12e- →12H2PO2^-(aq)+4PH3(g)+12OH^-(aq)+12e-
4P4(s)+12H2O(l)+12OH^-(aq) →12H2PO2^-(aq)+4PH3(g)
or
P4(s)+3H2O(l)+3OH^-(aq) →3H2PO2^-(aq)+PH3(g)
P4(s) is both the oxidizing and the reducing agent.
P4(s)+12H2O(l)+12e- →4PH3(g)+12OH^-(aq)
4P4(s)+12H2O(l)+24OH^-(aq)+12e- →12H2PO2^-(aq)+4PH3(g)+12OH^-(aq)+12e-
4P4(s)+12H2O(l)+12OH^-(aq) →12H2PO2^-(aq)+4PH3(g)
or
P4(s)+3H2O(l)+3OH^-(aq) →3H2PO2^-(aq)+PH3(g)
P4(s) is both the oxidizing and the reducing agent.
- Sun Feb 23, 2020 2:43 pm
- Forum: Balancing Redox Reactions
- Topic: 6K.3
- Replies: 1
- Views: 88
Re: 6K.3
Cl2 (g) +2e- → 2Cl^- (aq)
2H2O (l)+2Cl2(g) →2HOCl (aq) +2H^+(aq)+2e-
2H2O(l)+2Cl2(g)+2e- →2HOCl(aq)+2H^+(aq)+2Cl^-(aq)+2e-
or
H2O(l)+Cl2(g) →HOCl(aq)+H^+(aq)+Cl^-(aq)
Cl2 is both the oxidizing and the reducing agent.
2H2O (l)+2Cl2(g) →2HOCl (aq) +2H^+(aq)+2e-
2H2O(l)+2Cl2(g)+2e- →2HOCl(aq)+2H^+(aq)+2Cl^-(aq)+2e-
or
H2O(l)+Cl2(g) →HOCl(aq)+H^+(aq)+Cl^-(aq)
Cl2 is both the oxidizing and the reducing agent.
- Wed Feb 19, 2020 5:32 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: midterm Q4
- Replies: 4
- Views: 624
midterm Q4
The drug amphetamine, C6H5CH2CH(CH3)NH2 (pKb=3.11), is usually marketed as the hydrogen bromide salt, C6H5CH2CH(CH3)NH3^+Br^-, because it is more stable in this solid form. When the drug is ingested as a salt (solid form), it enters the stomach, which contains digestive fluids at pH=1.7. Will this d...
- Wed Feb 19, 2020 5:24 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: midterm 3C
- Replies: 4
- Views: 328
midterm 3C
A buffer solution made with NH3 and NH4Cl has a pH of 10.1. Which procedure(s) could be used to lower the pH? 1.adding HCl 2. adding HN3 3. adding NH4Cl a. 1 only, b. 2 only, 3. 1 and 3 only, 4. 2 and 3 only I know HCl can be used but I'm a little confused as to why NH4Cl can be used? Does it have a...
- Wed Feb 19, 2020 5:16 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: midterm Q1A
- Replies: 7
- Views: 449
midterm Q1A
2NO(g)+O2(g)⇄2NO2(g)
Does the equilibrium constant change if this reaction is heated, if so, how?
I'm confused as to why K would decrease in this case.
Does the equilibrium constant change if this reaction is heated, if so, how?
I'm confused as to why K would decrease in this case.
- Wed Feb 19, 2020 4:58 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Textbook Problem 4J.17
- Replies: 2
- Views: 369
Re: Textbook Problem 4J.17
a. deltaH=2deltaH(BF3,g)+3deltaH(H2O,l)-(deltaH(B2O3)+6deltaH(HF,g)) =2(-1137kJ/mol)+3(-285.83kJ/mol)-((-1272.8kJ/mol)+6(-271.1kJ/mol)) =-232.1kJ/mol deltaS=2Sm(BF3,g)+3Sm(H2O,l)-(Sm(B2O3,s)+6Sm(HF,g) =2(254.12J/K.mol)+3(69.91J/K.mol)-(53.97J/K.mol+6(173.78J/K.mol)) =-378.68J/K.mol deltaG=-232.1J/K....
- Mon Feb 17, 2020 10:54 pm
- Forum: Van't Hoff Equation
- Topic: 5.55b [ENDORSED]
- Replies: 1
- Views: 441
Re: 5.55b [ENDORSED]
5.20 x 10^3 g C x 1molC/12.011 g C = 433 mol C
125 g H2O x 1 mol H2O/18.016 g H2o = 6.94 mol H2O
H2O is limiting. Conc. of H2O = 6.94 mol/10 L = 0.694 mol/L
Kc=[CO][H2]/[H2O] = (x)(x)/(0.694-x)=0.403
x^2=0.280-0.403x
x^2+0.403x-0.280=0
x=+0.364 or -0.766
125 g H2O x 1 mol H2O/18.016 g H2o = 6.94 mol H2O
H2O is limiting. Conc. of H2O = 6.94 mol/10 L = 0.694 mol/L
Kc=[CO][H2]/[H2O] = (x)(x)/(0.694-x)=0.403
x^2=0.280-0.403x
x^2+0.403x-0.280=0
x=+0.364 or -0.766
- Mon Feb 10, 2020 3:40 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: q=mCdeltaT vs q=nCdeltaT
- Replies: 2
- Views: 97
Re: q=mCdeltaT vs q=nCdeltaT
use q=mCdeltaT when the mass is in grams. use q=nCdeltaT when mass is in moles.
- Mon Feb 10, 2020 3:39 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: entropy change due to T
- Replies: 1
- Views: 75
entropy change due to T
Container A is filled with 1.0 mol of the atoms of an ideal monatomic gas. Container B has 1.0 mol of atoms bound together as diatomic molecules that are not vibrationally active. Container C has 1.0 mol of atoms bound together as diatomic molecules that are vibrationally active. The containers all ...
- Mon Feb 10, 2020 3:28 pm
- Forum: Calculating Work of Expansion
- Topic: w=-nRT(ln V2/V1) Derivation
- Replies: 3
- Views: 171
Re: w=-nRT(ln V2/V1) Derivation
I don't think so, I believe the equations will be given.
- Mon Feb 10, 2020 3:26 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Example 4.c.1
- Replies: 2
- Views: 65
Re: Example 4.c.1
I'm pretty sure you use Cv = 3/2R and Cp=5/2R if it's an ideal gas (monatomic) and you use Cp=7/2R and Cv=5/2R if it's diatomic.
- Mon Feb 10, 2020 12:54 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: heat capacity question
- Replies: 1
- Views: 109
heat capacity question
The enthalpy of combustion of benzoic acid, C6H5COOH, which is often used to calibrate calorimeters, is -3227 kJ/mol. When 1.453 g of benzoic acid was burned in a calorimeter, the temperature increased by 2.265 celsius. What is the heat capacity of the calorimeter? The answer says to use q=CΔT to so...
- Mon Feb 10, 2020 12:38 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: q=CΔT
- Replies: 5
- Views: 301
q=CΔT
Is the specific heat capacity used primarily for calorimeter calculations?
- Fri Jan 31, 2020 11:46 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 4D.21 c.
- Replies: 1
- Views: 98
Re: 4D.21 c.
delta H= delta H[K2S + 2(H2O)] - delta H[H2S + 2(KOH)]
=[-417.5kJ/mol+2(-285.83kJ/mol)] - [-39.7kJ/mol + 2(-482.37 kJ/mol)]
=[-417.5kJ/mol+2(-285.83kJ/mol)] - [-39.7kJ/mol + 2(-482.37 kJ/mol)]
- Fri Jan 31, 2020 11:42 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 4E.5
- Replies: 3
- Views: 78
Re: 4E.5
For the benzene ring, you are supposed to use the single/double resonance bond enthalpy (518 kJ) for all six of the bonds.
- Fri Jan 31, 2020 11:40 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Homework 4C.9
- Replies: 2
- Views: 84
Re: Homework 4C.9
The heat change will be made up of two terms: one term to raise the temperature of the copper and the other to raise the temperature of the water:
q=(400.0g)(4.18J/celsius.g)(100.0celsius - 22.0celsius) + (500.0g)(0.38J/celsius.g)(100.0celsius-22.0celsius)
q=(400.0g)(4.18J/celsius.g)(100.0celsius - 22.0celsius) + (500.0g)(0.38J/celsius.g)(100.0celsius-22.0celsius)
- Fri Jan 31, 2020 11:38 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Test 1
- Replies: 1
- Views: 113
Re: Test 1
I think the tests for slightly different depending on what discussion you were in. This is #5 on my test. The pKb of fluoride is 10.8. A 0.12 M solution of hydrogen fluoride was made for silicon wafer processing. What is the pOH of this solution at equilibrium? HF(aq)+H2O⇋F^-(aq)+H3O^+ pKa=14 - 10.8...
- Fri Jan 31, 2020 11:28 pm
- Forum: Phase Changes & Related Calculations
- Topic: 4C.13
- Replies: 2
- Views: 155
Re: 4C.13
The heat gained by the water in the ice cube will be equal to the heat lost by the initial sample of hot water. The enthalpy change for the water in the ice cube will be composed of two terms: the heat to melt the ice at 0 celsius to the final temperature. heat (ice cube) = (50g/18.02 g.mol^-1)(6.01...
- Fri Jan 31, 2020 5:14 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 4A.11
- Replies: 3
- Views: 91
4A.11
A calorimeter was calibrated with an electric heater, which supplied 22.5 kJ of energy as heat to the calorimeter and increased the temperature of the calorimeter and its water bath from 22.45 celsius to 23.97 celsius. What is the heat capacity of the calorimeter? The answer says to use: C = 22.5 kJ...
- Fri Jan 31, 2020 12:57 am
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Reversible Reactions
- Replies: 5
- Views: 211
Re: Reversible Reactions
I think a reversible reaction is a reaction where the products are able to react together to form the reactant and vice versa.
- Fri Jan 31, 2020 12:50 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: benzene mean bond
- Replies: 1
- Views: 89
benzene mean bond
Does the 518 kJ/mol bond enthalpy for the carbon double/single apply for only benzene or does it also apply for other possible structures?
- Fri Jan 31, 2020 12:29 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 4D.9
- Replies: 2
- Views: 105
4D.9
The enthalpy of formation of trinitrotoluene (TNT) is -67 kJ/mol, and the density of TNT is 1.65 g/cm^-3. In principle, it could be used as a rocket fuel, with the gases resulting from its decomposition streaming out of the rocket to give the required thrust. In practice, of course, it would be extr...
- Thu Jan 30, 2020 9:51 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Ka and percent ionization problem
- Replies: 2
- Views: 180
Ka and percent ionization problem
The pH of a 0.010 M solution of hydrofluoric acid is 2.63. What are Ka and percent ionization? [H^+]=10^-2.63= 2.344 x 10^-3 HF + H2O ⇋ F^- +H3O^+ I 0.010 / 0 0 C 0.010 - 2.344 x 10^-3 / E 7.656 x 10^-3 / 2.344x10^-3 Ka = (2.344 x 10^-3)^2/(7.656x10^-3) =7.18x10^-4 % ionization: (2.344x10^-3)^2/(7.6...
- Thu Jan 16, 2020 8:37 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: equilibrium concentration
- Replies: 1
- Views: 58
equilibrium concentration
A reaction mixture is prepared by mixing 0.100 mol SO2, 0.200 mol NO2, 0.100 mol NO, and 0.150 mol SO3 in a 5.00 L reaction vessel. The reaction SO2 (g) + NO2 ⇌ NO(g) + SO3 (g) if allowed to reach equilibrium at 460 celsius, when Kc = 85.0. What is the equilibrium concentration of each substance? Th...
- Wed Jan 15, 2020 5:53 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: solubility
- Replies: 3
- Views: 117
solubility
How do you use equilibrium constants to predict solubility?
- Wed Jan 15, 2020 5:52 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: biological examples
- Replies: 2
- Views: 166
biological examples
What do we need to know about the biological examples: osmotic pressure and ATP hydrolysis?
- Wed Jan 15, 2020 5:50 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: effect on K
- Replies: 3
- Views: 154
effect on K
Calculate the effect on K of multiplying the chemical equation by a factor.
I'm confused as to what this is asking about/referring to?
I'm confused as to what this is asking about/referring to?
- Wed Jan 15, 2020 5:47 pm
- Forum: Ideal Gases
- Topic: approximate ideal gas law
- Replies: 3
- Views: 145
approximate ideal gas law
Identify reactions where the ideal gas law can be used as an approximation.
I'm a bit confused as to what this is referring to?
I'm a bit confused as to what this is referring to?
- Fri Jan 10, 2020 9:43 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: predicting effects
- Replies: 9
- Views: 394
predicting effects
The two air pollutants SO2 and NO2 can react in the atmosphere as follows: SO2 (g) + NO2 (g) ⇌ SO3 (g) + NO (g) Predict the effect of the following changes to the amount of NO when the reaction above has come to equilibrium in a stainless steel bulb equipped with entrants for chemicals. i. the amoun...
- Thu Jan 09, 2020 3:58 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: equilibrium and forming/decomposing
- Replies: 2
- Views: 60
equilibrium and forming/decomposing
The equilibrium constant, Kp, for the reaction SO2 (g) + O2 (g) ⇌ SO3 (g) at 700 K is 3 x 10^4. A mixture of SO2, O2, and SO3, each at 65 bars was introduced into a container at 700 K. Is the reaction at equilibrium? If not, does SO3 tend to form or decompose? I'm a little confused as to how to dete...
- Thu Jan 09, 2020 3:47 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: determining shift in equilibrium
- Replies: 4
- Views: 216
determining shift in equilibrium
Determine the shift in equilibrium position, if any, which will occur when the temperature is increased. The hydrolysis of ATP: ATP(aq) + H2O (l) ⇌ ADP + PO4^-2(aq) delta H°= -30 kJ/mol The answer says that it shifts to the left. Do I just take into account the delta to determine the shift? Also, in...
- Thu Jan 09, 2020 3:39 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: calculating the equilibrium concentrations of products and reactants
- Replies: 1
- Views: 153
calculating the equilibrium concentrations of products and reactants
A vial of SO2 (0.522 mol/L) and O2 (0.633 mol/L) react and reach equilibrium. Calculate the equilibrium concentrations of the products and reactants given that Kc= 5.66 x 10^-10 for this reaction: 2SO2 (g) + O2 (g) ⇌ 2SO3 (g) concentration SO2 O2 SO3 initial 0.522 0.633 0 change -2X -X +2X equilibri...
- Thu Jan 09, 2020 3:28 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: calculating reaction quotient
- Replies: 3
- Views: 63
calculating reaction quotient
Calculate the reaction quotient, Qc, from the following equilibrium data collected in a 3.00 L sealed reaction vessel for the reaction: AsH3 = 5.55 x 10^-4 mol, As= 3.31 x 10^-3 mol, H2= 1.23 x 10^-3 mol. I keep getting the wrong answer when I try to solve this problem. Can someone show me how to ac...
- Wed Nov 27, 2019 11:25 pm
- Forum: Lewis Acids & Bases
- Topic: Conjugates
- Replies: 3
- Views: 225
Re: Conjugates
the more electroneg. it is, the more acidic it tends to be
- Wed Nov 27, 2019 11:23 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Polydentate
- Replies: 3
- Views: 150
Re: Polydentate
polydentate lignads can occupy more than one binding site simultaneously
- Wed Nov 27, 2019 11:10 pm
- Forum: Amphoteric Compounds
- Topic: distinguishing oxides
- Replies: 2
- Views: 150
distinguishing oxides
State whether the following oxides are acidic, basic, amphoteric: (a) BaO; (b) SO3; (c) As2O; (d) Bi2O3.
I'm not sure how to distinguish between the three?
I'm not sure how to distinguish between the three?
- Tue Nov 26, 2019 10:46 pm
- Forum: Lewis Acids & Bases
- Topic: writing chemical equation
- Replies: 1
- Views: 120
writing chemical equation
Either the cation or anion is a weak acid or weak base in water. Write the chemical equation for the proton transfer equation of the cation or anion with water: c. C5H5NHCl The answer says that C5H5NH + H2O ➔ C5H5N +H3O. I'm guessing the forgotten Cl would remain part of the C5H5NH? So, C5H5NHCl + H...
- Mon Nov 25, 2019 10:09 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: coordination number
- Replies: 3
- Views: 134
coordination number
Determine the coordination number of the metal ion in each of the following complexes:
a.[PtCl2(en)2]^2+ and b.[Cr(edta)]^+
The answer for a. is 6 (en is bidentate) and b. is 6 (EDTA is hexadentate). I don't understand how it results in six?
a.[PtCl2(en)2]^2+ and b.[Cr(edta)]^+
The answer for a. is 6 (en is bidentate) and b. is 6 (EDTA is hexadentate). I don't understand how it results in six?
- Mon Nov 25, 2019 5:58 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: polydentate
- Replies: 2
- Views: 233
polydentate
Would something that is polydentate mean that it would have a lot of lone e- pairs?
So between HN(CH2CH2NH2)2, CO3^-2, oxalate, the polydentates would be CO3^-2 and oxalate?
So between HN(CH2CH2NH2)2, CO3^-2, oxalate, the polydentates would be CO3^-2 and oxalate?
ferrate
Writing the formula of this compound: sodium bisoxalato (diaqua) ferrate (III)
What does ferrate refer to?
What does ferrate refer to?
- Mon Nov 25, 2019 5:41 pm
- Forum: Naming
- Topic: -ide naming
- Replies: 1
- Views: 72
-ide naming
chloride and bromide mean that they're Cl2 and Br2 right?
- Mon Nov 25, 2019 5:36 pm
- Forum: Naming
- Topic: OH2 naming
- Replies: 4
- Views: 504
OH2 naming
For [Co(CN)5(OH2)]^-2, would the OH2 be named aqua or would it remain hydroxide?
So, would the name be penta cyano hydroxide cobalt (II) or penta cyano aqua cobalt (II)?
So, would the name be penta cyano hydroxide cobalt (II) or penta cyano aqua cobalt (II)?
- Mon Nov 25, 2019 5:24 pm
- Forum: Naming
- Topic: roman numerals
- Replies: 7
- Views: 678
roman numerals
I'm confused to what the roman numeral refers to.
In [Ni(CN)4]^-2, the name is tetracyanonickelate (II), so I'm guessing it refers to the -2?
In [Ni(CN)4]^-2, the name is tetracyanonickelate (II), so I'm guessing it refers to the -2?
- Sun Nov 17, 2019 9:12 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: bond angles
- Replies: 1
- Views: 125
bond angles
In a hydromium ion, I think it would be T-shaped w/ bond angles of <90?
However, the answer is <120, did I identify the shape wrong?
If I did, what is the actual shape and what does it look like?
However, the answer is <120, did I identify the shape wrong?
If I did, what is the actual shape and what does it look like?
- Sat Nov 16, 2019 11:03 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: identifying pi & sigma bonds
- Replies: 3
- Views: 215
identifying pi & sigma bonds
How do I tell if a bond is a sigma or pi bond?
- Sat Nov 16, 2019 10:56 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VSEPR notation
- Replies: 8
- Views: 510
VSEPR notation
For NO3-, would the VSEPR notation be AX3 or AX4 and why?
- Sat Nov 16, 2019 10:54 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: VSEPR notation
- Replies: 6
- Views: 400
VSEPR notation
In CH2Cl2, would the VSEPR notation would be AX4?
In PCl3, would the VSEPR notation would be AX3E?
If not, please explain.
In PCl3, would the VSEPR notation would be AX3E?
If not, please explain.
- Sat Nov 16, 2019 10:11 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: sigma and pi bonds
- Replies: 2
- Views: 140
sigma and pi bonds
What are sigma and pi bonds and how do I label them?
- Mon Nov 11, 2019 3:03 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: london forces
- Replies: 6
- Views: 270
london forces
Do all molecules possess London forces?
- Mon Nov 11, 2019 2:52 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: shape of molecule/intermolecular forces
- Replies: 3
- Views: 102
Re: shape of molecule/intermolecular forces
what does the molecular structure of dimethylpropane look like? I have a hard time visualizing it.
- Mon Nov 11, 2019 2:46 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: hydrogen bonding
- Replies: 1
- Views: 68
hydrogen bonding
Suggest, giving reasons, which substances in each of the following pairs is likely to have the higher normal melting point (Lewis structure may help your arguments): (c) C2H5OC2H5 (diethyl ether) or C4H9OH (butanol) The answers says that butanol has a higher normal melting point due to hydrogen bond...
- Mon Nov 11, 2019 1:43 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: shape of molecule/intermolecular forces
- Replies: 3
- Views: 102
shape of molecule/intermolecular forces
Account for the following observations in terms of the type and strength of the type and strength of intermolecular forces. (c) The boiling point of pentane, CH3(CH2)CH3, is 36.1 degrees celsius, whereas that of 2,2-dimethylpropane (aka neopentane) C(CH3)4 is 9.5 degrees celsius. The answer says tha...
- Mon Nov 11, 2019 12:57 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: intermolecular forces/polarity
- Replies: 2
- Views: 130
intermolecular forces/polarity
Explain the difference in the boiling points of AsF3 (63 celsius) and AsF5 (-53 celsius).
The answer says that it is clear that AsF3 is a polar molecule while AsF5 is not from the Lewis structures. How do you tell whether a structure is polar or not?
The answer says that it is clear that AsF3 is a polar molecule while AsF5 is not from the Lewis structures. How do you tell whether a structure is polar or not?
- Mon Nov 11, 2019 12:54 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: types of intermolecular forces
- Replies: 5
- Views: 145
types of intermolecular forces
Identify the types of attractive intermolecular interactions that might arise between molecules of each of the following substances: (a) NH2OH; (b) CBr4; (c) H2SeO4; (d) SO2.
I'm not sure how to know whether or not a molecule has dipole-dipole intermolecular forces.
I'm not sure how to know whether or not a molecule has dipole-dipole intermolecular forces.
- Tue Nov 05, 2019 4:41 pm
- Forum: Lewis Structures
- Topic: 7 lewis structure
- Replies: 1
- Views: 67
7 lewis structure
How important is a 7 lewis structure for CO2 where there's one triple bond and one single bond to the actual structure of CO2?
I don't understand what the question os asking for?
I don't understand what the question os asking for?
- Tue Nov 05, 2019 4:28 pm
- Forum: SI Units, Unit Conversions
- Topic: converting J to kJ/mol
- Replies: 2
- Views: 9165
converting J to kJ/mol
converting 4.89x10^-17 J to kJ/mol
first convert J to kJ by multiplying by 0.001.
I'm not sure how to do the next part? Do I divide or multiply the number by avogadro?
first convert J to kJ by multiplying by 0.001.
I'm not sure how to do the next part? Do I divide or multiply the number by avogadro?
- Tue Nov 05, 2019 4:19 pm
- Forum: Photoelectric Effect
- Topic: binding energy
- Replies: 3
- Views: 369
binding energy
An x-ray photon of wavelength 0.989 nm strikes a surface. The emitted electron has a kinetic energy of 969 eV. (1 eV = 1.602 10^-19 J)
Is the binding energy in the question referring to the electron's threshold energy or something else?
Is the binding energy in the question referring to the electron's threshold energy or something else?
- Sun Nov 03, 2019 4:40 pm
- Forum: Dipole Moments
- Topic: drawing dipole moments
- Replies: 5
- Views: 621
drawing dipole moments
how do you find a dipole moment and how would you draw it?
- Sun Nov 03, 2019 4:10 pm
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: polarizing power
- Replies: 3
- Views: 169
polarizing power
How do you figure out which ion has the most ion polarizing power? For example, with Li^+, Na^+, K^+.
- Tue Oct 29, 2019 8:07 pm
- Forum: Ionic & Covalent Bonds
- Topic: finding chemical formula based on charges of ions
- Replies: 1
- Views: 135
finding chemical formula based on charges of ions
On the basis of the expected charges on the monatomic ions, give the chemical formula of each of the following compounds.
a. magnesium sulfide
b. indium (III) sulfide
c. aluminum hydride
d. hydrogen telluride
e. bismuth (III) fluoride
I'm confused on where to start.
a. magnesium sulfide
b. indium (III) sulfide
c. aluminum hydride
d. hydrogen telluride
e. bismuth (III) fluoride
I'm confused on where to start.
- Fri Oct 25, 2019 9:00 pm
- Forum: Ionic & Covalent Bonds
- Topic: ion charge
- Replies: 3
- Views: 130
Re: ion charge
For d. Ga, I don't understand why the answer is Ga^3+?
the ground state configuration I got was [Ar]3d^10 4s^2 4p^1, so I assumed that it would be best to just get rid of the 4p^1, so that it would be [Ar]3d^10 4s^2 and the charge would be Ga^+.
the ground state configuration I got was [Ar]3d^10 4s^2 4p^1, so I assumed that it would be best to just get rid of the 4p^1, so that it would be [Ar]3d^10 4s^2 and the charge would be Ga^+.
- Fri Oct 25, 2019 8:05 pm
- Forum: Ionic & Covalent Bonds
- Topic: ion charge
- Replies: 3
- Views: 130
ion charge
Write the most likely charge for the ions formed by each of the following elements:
a. S
b. Te
c. Rb
d. Ga
I don't know where to start.
a. S
b. Te
c. Rb
d. Ga
I don't know where to start.
- Thu Oct 24, 2019 10:04 pm
- Forum: Ionic & Covalent Bonds
- Topic: ground-state configuration of ions
- Replies: 4
- Views: 141
ground-state configuration of ions
Give the ground-state electron configuration expected for each of the following ions: a. Cu^+ b. Bi^3+ c. Ga^3+ d. Tl^3+ For a, I got [Ar]3d^9 4s^2, which becomes [Ar]3d^10 4s^1. I don't understand why the answer is just [Ar]3d^10? As for b, the answer is [Xe}4f^14 5d^10 6s^2. I'm not sure how to ge...
- Thu Oct 24, 2019 4:27 pm
- Forum: Ionic & Covalent Bonds
- Topic: valence electrons
- Replies: 6
- Views: 423
valence electrons
Give the number of valence electrons (including d electrons) for each of the following elements:
a. Sb b. Si c. Mn d. B
I think I understand how to find the valence electrons for b, c, and d. Si: 4, Mn: 7, B: 3. I don't understand how the answer for a. (5 electrons) came to be?
a. Sb b. Si c. Mn d. B
I think I understand how to find the valence electrons for b, c, and d. Si: 4, Mn: 7, B: 3. I don't understand how the answer for a. (5 electrons) came to be?
- Thu Oct 17, 2019 10:26 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: heisenberg, calculating uncertainty in speed
- Replies: 1
- Views: 98
heisenberg, calculating uncertainty in speed
You are caught in a radar trap and hope to show that the speed measured by the radar gun is in error due to the Heisenberg uncertainty principle. If you assume that the uncertainty in your position +/- 5m when your speed was measured, and that the car has a mass of 2150 kg, what is your calculated u...
- Thu Oct 17, 2019 10:05 pm
- Forum: Heisenberg Indeterminacy (Uncertainty) Equation
- Topic: heisenberg, calculating kinetic energy
- Replies: 3
- Views: 171
heisenberg, calculating kinetic energy
Use the uncertainty in velocity (3.4x10^10 m/s) to calculate the electron's uncertainty in kinetic energy. Then calculate the uncertainty in kinetic energy per mole of electrons (that is, per mole of hydrogen atoms). I'm not quite sure where to start. I tried plugging the velocity and mass of electr...
- Thu Oct 17, 2019 9:57 pm
- Forum: DeBroglie Equation
- Topic: deBroglie, finding wavelike properties
- Replies: 2
- Views: 122
deBroglie, finding wavelike properties
Calculate the wavelength of a 275 kg single-seat electric car traveling at a speed of 125 km/hr. Do electric cars have wavelike properties? The first thing I do is convert 125 km/hr to m/s. 125 km/hr x 1000m/1km x 1 hr/60s = 2083.33 m/s lambda= (6.63x10^-34 Js)/((275 kg)(2083.33m/s) At the end, I ge...
- Thu Oct 17, 2019 9:42 pm
- Forum: Properties of Light
- Topic: atomic spectra
- Replies: 2
- Views: 137
atomic spectra
In 1.0 s, a 60 W bulb emits 11 J of energy in the form of infrared radiation (heat) of wavelength 1850 nm. What is the energy per photon of light emitted? How many photons of infrared radiation does the lamp generate in 1.0 s?
I have no idea where to begin/which equations to use.
I have no idea where to begin/which equations to use.
- Thu Oct 17, 2019 9:37 pm
- Forum: Properties of Light
- Topic: atomic spectra module
- Replies: 1
- Views: 146
atomic spectra module
The meter was defined in 1963 as 1,650,763.73 wavelengths of radiation emitted by krypton-86 (it has since been redefined). What is the wavelength of this krypton-86 radiation? What is the wavelength of this krypton-86 radiation? To what region of the electromagnetic spectrum does this wavelength co...
- Sun Oct 13, 2019 10:42 pm
- Forum: Properties of Electrons
- Topic: E(photon) equation help
- Replies: 2
- Views: 85
Re: E(photon) equation help
So after I convert 1509.6 kJ/mol to 150600 J/mol, I convert J/mol to 2.5008 x 10^-19 J.
However when I plug the numbers into the equation,
E(photon)-2.5008x10^-19=1.99x10^-19, I end up with 4.4908x10^-19 J, which is different from the answer 2.501x10^-19 J.
Am I doing something incorrect?
However when I plug the numbers into the equation,
E(photon)-2.5008x10^-19=1.99x10^-19, I end up with 4.4908x10^-19 J, which is different from the answer 2.501x10^-19 J.
Am I doing something incorrect?
- Sun Oct 13, 2019 9:00 pm
- Forum: Properties of Light
- Topic: photoelectric effect
- Replies: 3
- Views: 135
photoelectric effect
Which one of following is not describing the photoelectric effect?
A. E (photon) – E (remove e-) = E (excess)
B. E (photon) – E (remove e-) = EK (e-)
C. hv - work function = 1/2mv2
D. λv = c
E. None of the above
I don't understand which of these are not describing the effect and why.
A. E (photon) – E (remove e-) = E (excess)
B. E (photon) – E (remove e-) = EK (e-)
C. hv - work function = 1/2mv2
D. λv = c
E. None of the above
I don't understand which of these are not describing the effect and why.
- Sun Oct 13, 2019 8:48 pm
- Forum: Properties of Electrons
- Topic: kinetic energy help
- Replies: 1
- Views: 55
kinetic energy help
If molybdenum is irradiated with 194 nm light, what is the maximum possible kinetic energy of the emitted electrons?
min. freq.: 1.09x10^15 Hz
min. energy: 7.22x10^-19 J
I don't have a clue where to start from this.
min. freq.: 1.09x10^15 Hz
min. energy: 7.22x10^-19 J
I don't have a clue where to start from this.
- Sun Oct 13, 2019 8:34 pm
- Forum: Properties of Electrons
- Topic: E(photon) equation help
- Replies: 2
- Views: 85
E(photon) equation help
How much energy is required to remove an electron from one sodium atom? velocity of ejected electron: 6.61x10^5 m/s work function of sodium: 150.6 kJ/mol kinetic energy of ejected electron: 1.99x10^-19 J I assume that the equation to use is E(photon)-work function=KE(ejected e-) So, E(photon)-150.6 ...
- Sun Oct 13, 2019 7:21 pm
- Forum: Properties of Electrons
- Topic: kinetic energy equation help
- Replies: 2
- Views: 56
kinetic energy equation help
Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1. Answer the following three questions. What is the kinetic energy of the ejected electron? I know to use the kinetic energy equation KE=1/2mv^2 but I'm not s...
- Sat Oct 05, 2019 7:08 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Solution problem
- Replies: 3
- Views: 148
Re: Solution problem
molarity=moles of solute/volume of solution
the moles of solute would be the numerator, volume would be denominator
the moles of solute would be the numerator, volume would be denominator
- Sat Oct 05, 2019 7:05 pm
- Forum: Molarity, Solutions, Dilutions
- Topic: Molarity and Dilution G23
- Replies: 3
- Views: 302
Re: Molarity and Dilution G23
just assume that NaCl and KCl are the only two you're using
- Sat Oct 05, 2019 6:56 pm
- Forum: SI Units, Unit Conversions
- Topic: How units of mass effect Scientific notation
- Replies: 3
- Views: 158
Re: How units of mass effect Scientific notation
if you are converting to grams, then it should be 10^3
- Sat Oct 05, 2019 2:45 pm
- Forum: Significant Figures
- Topic: Sig Figs When Adding and Subtracting
- Replies: 3
- Views: 173
Re: Sig Figs When Adding and Subtracting
For example, let's say these three numbers: 34.07 g, 70 g, and 60.0 g create the number 7070.3 g 34.07 g has 4 sig figs 70 g has 1 sig fig 60.0 g has 3 sig figs 7070.3 g would be rounded down to 7000 g because you always want to choose the number that has the least amount of sig figs. In this case, ...
- Sat Oct 05, 2019 2:40 pm
- Forum: Properties of Light
- Topic: Electrical Field
- Replies: 1
- Views: 79
Re: Electrical Field
the electrical field corresponds to the amplitude; as the frequency decreases the waves broaden and the extent of the change (slope of the wave) decreases
- Sat Oct 05, 2019 12:47 pm
- Forum: Properties of Light
- Topic: Classical Mechanics vs. Quantum Mechanics
- Replies: 4
- Views: 89
Re: Classical Mechanics vs. Quantum Mechanics
In classical mechanics the events tend to be continuous, which is to say they move in smooth, orderly and predictable patterns. In quantum mechanics, the events are unpredictable, which is to say "jumps" occur that involve seemingly random transitions between states.
- Sat Sep 28, 2019 10:48 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Molar Mass
- Replies: 2
- Views: 203
Re: Molar Mass
Subtract the molar mass of the (OH)2 from the metal hydroxide.
74.1 g/mol - 34.02 g/mol = 40.08 g/mol
You get the molar mass of Ca. This is the metal.
So, the molar mass of calcium sulfide (CaS) = 72.15 g/mol
74.1 g/mol - 34.02 g/mol = 40.08 g/mol
You get the molar mass of Ca. This is the metal.
So, the molar mass of calcium sulfide (CaS) = 72.15 g/mol
- Sat Sep 28, 2019 8:03 am
- Forum: Administrative Questions and Class Announcements
- Topic: Formatting homework
- Replies: 12
- Views: 613
Re: Formatting homework
I'm just writing down my answers onto a piece of binder paper. As long as you indicate which question is which, I think that's fine.