Wednesday
June 29, 2016

Posts by drbob222

Total # Posts: 52,597

chemistry 12u
Sure it will. 2.73 = -log(H3O^+) -2.73 = log*H3O^+) Hit the 10^x button and punch in -2.73.
May 10, 2016

chemistry 12u
pH = -log(H3O^+) The nitrous acid bit has nothing to do with the problem. ANY acid or base with a pH of 2.73 will have this H3O^+.
May 10, 2016

chemistry
Another one is LEO the lion goes Grr. LEO is loss electrons oxidation
May 9, 2016

chemistry
backwards. The first one is ADDING electrons. The second one is LOSING electrons.
May 9, 2016

Chemistry
5% what. weight/volume? weight/weight? volume/volume? I assume you mean w/v since you want 300 mL of solution. 5% means 5 g sucrose/100 mL solution. You want 300 so 5g x (300/100) = 15g So you need 15 g sucrose/300 mL solution.
May 9, 2016

Check my answers? Science
I'm serious. You must be kidding.
May 9, 2016

Electroplating in Chem
The Cu strip is going into solution. That one is Cu ==> Cu^2+ + 2e Then the Cu^2+ ions in solution are plated onto the penny as pure copper. That one is Cu^2+ + 2e ==> Cu If you remember the definitions you can't go wrong with which is which. Oxidation is the loss of...
May 9, 2016

Chemistry
mols Al = grams/atomic mass Al Using the coefficients in the balanced equation, convert mols Al to mols AlCl3. Now convert mols AlCl3 to grams. g AlCl3 = mols AlCl3 x molar mass AlCl3.
May 9, 2016

Chemistry
I answered this question earlier. http://www.jiskha.com/display.cgi?id=1462663618
May 9, 2016

Chem
PbCl2 ==> Pb^++ + 2Cl^- AgCl --> Ag^+ + Cl^- Using Ksp for PbCl2, calculate what the (Cl^-) must be to ppt PbCl2. You know (Pb^++) and Ksp, solve for (Cl^-). Do the same for AgCl. You know Ag^+ and Ksp, solve for (Cl^-). The one with the lower Cl^- will appear first so ...
May 9, 2016

Chem
.......Ag2SO4 ==> 2Ag^+ + SO4^= I......solid.......0........0 C......solid.......2x.......x E......solid.......2x.......x Substitute the E line into the Ksp expression and solve for x = solubility.
May 9, 2016

chemistry
I have no idea what tetraoxosulphate(6) is. Perhaps you are referring to sulfuric acid, H2SO4.
May 9, 2016

Chemistry
Google as I did for this one. https://www.google.com/search?q=3-methylpentanoic+acid&ie=utf-8&oe=utf-8
May 8, 2016

chemistry
P*molar mass = density*R*T Substitute P, d, R and T and solve for molar mass.
May 8, 2016

chemistry
How many mols do you need? That's mols = grams/molar mass. You know molar mass and grams, solve for mols. Then M HCl = mols HCl/L HCl. You know mols and M, solve for L and convert to mL.
May 8, 2016

Chemistry
I don't see a question here.
May 8, 2016

Chemistry
g BaSO4 x (molar mass sulfa stuff/3*molar mass BaSO4)*100 = ?
May 8, 2016

Chemistry- Dr.Bob222
The first thing wrong is you are using Ksp. You still haven't answered my question about the Kf. Kf is the formation constant for the complex. Ag^+ + 2NH3 ==> [Ag(NH3)2]^+ The literature shows that as about 1.7E7 and your question gives it as 0.0031. I think that 0.0031...
May 8, 2016

Chemistry- Dr.Bob222
Basically I said the definition for Kf is for Ag^+ + 2NH3 ==> [Ag(NH3)2]^+ and that is something like 1.7E7. I don't think your Kf you quoted of 0.0031 is Kf; I think it is for AgCl + 2NH3 ==> [Ag(NH3)2]^+ + Cl^- and that makes it Keq for that reaction and that is ...
May 7, 2016

chem
Technically I would think M+ ion is the Cu^2+ ion that comes from the bpy complex; i.e., if the ionization is Cu(bpy)2^2+ ==> Cu^2+ + 2bpy So you might expect three separate peaks but then you haven't told us that this is; i.e., infrared, mass spec, magnetic resonance, ...
May 7, 2016

Chemistry- Dr.Bob222
I don't get that answer. Are you sure you have typed in the correct values? I think that 0.0031 is closer to the Knet for the reaction and not Kf for the formation of the complex. I think that number is very much higher; i.e., something like 10^7 or so. If I use 0.0031 for...
May 7, 2016

Chemistry
You did the most important first two. The next one is the easy one. So you want to neutralize 6 mols H2SO4, that will take 12 mols NaOH as you correctly stated. Then M NaOH = mols NaOH/L NaOH. You know mols NaOH and M NaOH, solve for L NaOH. Convert to mL if needed.
May 7, 2016

chemistry
AgNO3 + KCl ==> AgCl + KNO3. NaNO3 + KCl ==> no reaction. So you obtained 3.97 g AgCl. Convert that to g AgNO3 this way. 3.97 x (molar mass AgNO3/molar mass AgCl) = ? g AgNO3. %AgNO3 = (g AgNO3/mass sample)*100 = ?
May 7, 2016

Chemistry
Use PV = nRT with n = grams/molar mass
May 7, 2016

Chemistry
2C2H6 + 7O2 ==> 4CO2 + 6H2O When gases are involved one can take a shortcut and use liters and mols interchangeably. So 4.00 L C2H6 x (7 mols O2/2 mols C2H6) = 4.00 x 7/2 = ? L O2
May 7, 2016

Chemistry
This sounds like a take off on soda lime glass. You can read more about that here. https://en.wikipedia.org/wiki/Soda-lime_glass
May 7, 2016

AlphaPlus
Alpha Plus is the name of the subject? L1 x M1 = L2 x M2 L1 x 4.85 = 7 x 1.25 Solve for L1
May 7, 2016

Chemistry
48.5% x (0.75L/5.60L) = ?%
May 7, 2016

Chemistry
How many mols Ba(NO3)2 do you have? That's mols = grams/molar mass = ? approx 0.02 but that's just an estimate. Then M = mols/L You know M(1.25) and mols (estimated 0.02), solve for L.
May 7, 2016

oops---Chemistry
I don't believe the above answer is right. It should be A2B ==> 2A^+ + B^2- Ksp = (A^+)^2(B^2-) = (0.00147)^2(0.000735) = ?
May 6, 2016

chemistry
Mg + 2HCl ==> MgCl2 + H2 mols Mg = grams/atomic mass = ? Using the coefficients in the balanced equation convert mols Mg to mols HCl. In this case mols HCl = 2 x mols Mg. Now that you know mols HCl, M HCl = mols HCl/L HCl. You know M and mols, solve for L and convert to mL...
May 6, 2016

Chemistry
C7H8 is toluene. 92.14 is molar mass.
May 6, 2016

Chemistry
delta T = kf*m Substitute and solve for m = molality. You know delta T and Kf. m = mols/kg solvent You know kg solvent and m. mols = grams/molar mass. Substitute and solve for molar mass. Yo know grams and mols.
May 6, 2016

chem 161
This is a limiting reagent problem. You know that because amounts are given for BOTH reactants. AgNO3 + NaCl ==> AgCl + NaNO3 mols AgNO3 = M x L = approx 0.013 but need that more accurately. mols NaCl = M x L = ? approx 0.0083 So the amount of AgCl that can be formed from ...
May 6, 2016

chem 161
H3PO4 + 3NaOH ==> Na3PO4 + 3H2O mols H3PO4 = M x L = ? Using the coefficients in the balanced equation, convert mols H3PO4 to mols NaOH. That will be mols NaOH = 3 * mols H3PO4. Then M NaOH = mols NaOH/L NaOH. YOu know mols NaOH and M NaOH, solve for L NaOH and convert to ...
May 6, 2016

Chemistry
This is a two part question. First you set up to see how much of the complex is formed, then you run the problem backwards to see how much of the Cd and others components are formed/or remain. Here is the first part and you assume, with such a huge K value that ALL of the Cd^2...
May 5, 2016

Chemistry
right. See above for my response.
May 5, 2016

Chemistry
If you answered the question I asked I might be able to help.
May 5, 2016

Chemistry
I must be missing something. With no CN^- there how does any [Cd(CN)4]^2- get formed?
May 5, 2016

chemistry
You are both right and wrong. The second equation is wrong. The first one is right. Pb^2+ ==> Pb02 + 2e. The lead ion from lead lactate is +2 and it is going to PbO2 where Pb is +4. That is a loss of electrons which is oxidiation. And the PbO2 will deposit at the anode ...
May 5, 2016

Chemistry
2AgNO3 + CaCl2 ==> 2AgCl + 2NaNO3 mols CaCl2 = M x L = ? Using the coefficients in the balanced equation, convert mols CaCl2 to mols AgNO3. mols AgNO3 - 2x mols CaCl2. Then M AgNO = mols AgNO3/L AgNO3. Ou know mols and M, solve for L and convert to mL.
May 5, 2016

chemistry
I don't see a question here.
May 5, 2016

chemistry
mols NaOH = grams/molar mass ? M = mols NaOH/dm^3 solution
May 5, 2016

Chemistry
I think you are trying to make this too hard. To start the 0.01 PbCl2 is in solution. ......PbCl2 ==> Pb^2+ + 2Cl^- I.....0.01......0.01....0.02 You can show this is in solution. Qsp = (0.01)(0.02)^2 = 4E-6 which is < Ksp so it hasn't pptd and all is in solution. ...
May 5, 2016

Chemistry
Poor question for it has 3 answers and another type is not listed.. Usually c (for acid/base titrations) but sometimes a or b. And the Mohr titration depends upon the Ksp of Ag2CrO4 which is none of the aboe.
May 4, 2016

Chemistry
Stop playing screen names games. This is a limiting reagent problem. Do you know how to do them?
May 4, 2016

Chemistry
Stop playing games with screen names. 1 mol contains 6.022E23 molecules. So if 38 g contain 3E23 molecules what would be the weight of 6.022E23 molecules?
May 4, 2016

Chemistry
That's not my question. I want to know what you don't understand so you can do these yourself.
May 4, 2016

Chemistry
What do you not understand about how to get this answer?
May 4, 2016

Chemestry
I don't see a question here.
May 4, 2016

Chemistry
The correct answer is 11.69. 50 x 0.01 = 0.5 millimols each. ....Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O I....0.5......0.5.......0.......0 C..-0.25.....0.25.......0.5 C...0.25.......0........0.5 So [Ca(OH)2] = 0.25 mmols/100 mL = 0.0025 and (OH^-) = 2*that = 0.005M You can take it ...
May 4, 2016

Chemistry
I don't get the ideal gas part but the equation is 2Mg + O2 ==> 2MgO
May 4, 2016

Chemistry
I assume you meant CuCl2(aq) CuCl2(aq) + Na2CO3(aq) ==> CuCO3(s) + 2NaCl(aq)
May 4, 2016

Chemistry
You must have made a typo; is that 0.157 M? If so then pH = -log(HCl) = ?
May 4, 2016

Chemistry
Ray, James et al. We could help you better if you didn't change screen names. You made another typo or you may not know better. There is a difference between m and M How many moles do you have. That's mols = M x L = ? Then mols = grams/molar mass. You have molar mass ...
May 4, 2016

Chemistry
Ptotal = pHe + pNe + pAr
May 4, 2016

chemistry
(P1/T1) = (P2/T2)
May 4, 2016

chemistry
M = atomic mass M O = atomic mass O S = atomic mass S MO2 = molar mass MO2 M2S3 = molar mass M2S3 ---------------------- M2S3 + 5O2 ==> 2MO2 + 3SO2 4.00.........4.00-0.277 = 3.72 4.00 x (2*MO2/M2S3) = 3.72 4.00 x (2*M+4*amO/2*M + 3*S) 4.00 x (2*M + 64) = 3.72(2*M + 3*96) ...
May 4, 2016

oops-error--chemistry
equation 1 I wrote for you is not right. That should be mols HCl used for Mg(OH)2 + mols HCl used for Al(OH)3 = total mols HCl = M x L = ? Equation 1 should read, (X/2*mm Mg(OH)2) + (Y/3*mm Al(OH)3) = M HCl x L HCl Sorry about that.
May 4, 2016

chemistry
This is a two unknowns/two equation problem. The equations are solved simultaneously. Difficult to explain on a screen but I can show it in parts and leave most of the work to you. I should note that there is a piece of information missing; i.e., you don't have the mass of...
May 4, 2016

chemistry
You have not posted the entire problem.
May 4, 2016

chemistry
This is done the same way as the M2S3/MO2 problem. See that solution above.
May 4, 2016

Chemistry
That should work. Post your work and we can look for the error. Probably you just punched in a wrong number or hit the wrong key on your calculator.
May 4, 2016

chemistry
Where does Xmas come from? Look at your MO2/M2S3 problem above.
May 4, 2016

chemistry
Again, look at your MO2/M2S3 problem.
May 4, 2016

chemistry
This is another set of simultaneously solved equations but it is more complicated. You must separate, somehow, the Fe used to form FeO from the Fe used to form Fe2O3 and/or separate the oxygen used for form FeO and oxygen used to form Fe2O. Then solve stoichiometry problem to ...
May 4, 2016

chemistry
This is almost the same as the MO2/M2S3 problem but I would convert 12.13160 g Ag to the mass of AgBr. Post your work if you get stuck.
May 4, 2016

chemistry
Use PV = nRT
May 4, 2016

Chemistry
I will assume that 0.200m is a typo and you mean 0.200 M. mols NaOH = grams/molar mass = ? Then M NaOH = mols NaOH/L NaOH. YOu know mols and M, solvle for L and convert to mL.
May 4, 2016

chemistry
pH = -log(HCl)
May 4, 2016

chemistry
I assume that 250 mL is 250 mL of solution. mols urea = grams/molar mass = ? Then M = mols/L
May 4, 2016

chemistry
Is there a question here? Is there any absorbance data here? I don't see either.
May 3, 2016

Chemistry
6.0 x 455/2500 =
May 3, 2016

Chemistry
How may what of methane? liters? grams? tons? What is the temperature of the water from which the steam comes?
May 3, 2016

oops--typo--Chemistry
What is the molarity of the HCl? That's 1.2 g/cc x 1000cc x 0.65 x (1/36.5) = ? That line should read 1.2 g/cc x 1000 cc x 0.365 x (1/36.5) = ? It's the .65 that is not right. It should be 0.365 and that comes from the 36.5%
May 3, 2016

Chemistry
What is the molarity of the HCl? That's 1.2 g/cc x 1000cc x 0.65 x (1/36.5) = ? Then mols HCl = M x L = ? mols NaOH = mols HCl Then M NaOH = mols NaOH/L NaOH. You know mols NaOH and M NaOH, solve for L NaOH. Convert to mL if needed.
May 3, 2016

Chemistry
3Fe + 2O2 ==> Fe3O4 mols Fe = grams/atomic mass Fe = ? Using the coefficients in the balanced equation, convert mols Fe to mols Fe3O4. Now convert mols Fe3O4 to grams. g = mols x molar mass = ?
May 3, 2016

Chemistry
right
May 3, 2016

Chemistry
Ca(OH)2 ==> Ca^2+ + 2OH^- dGorxn = (dGo products)-(dGo reactants) Then dGo = -RTlnKsp
May 3, 2016

Chemistry
Look up (or calculate) the heat of combustion for methane. I'll call that Z. Then Z = mass H2O x specific heat H2O x (Tfinal-Tinitial) Keep mass and specific heat units consistent.
May 3, 2016

Chemistry
See your other post.
May 3, 2016

Chemistry
I don't either. With no grams listed, there is no way to know how much water is required to form a saturated solution.
May 2, 2016

Chemistry
You didn't list any quantities of pure water. From the problem 1 L would be a saturated solution but 1 L may not be listed as an answer and it may not be the smallest quantity.
May 2, 2016

Chemistry
A. You're right; Ba^2+ is reduced because it gains electrons. That makes it an oxidizing agent. B. So Co^2+ + 2e = Co and that's the same as A. C. Se. You have the choice of Se --> Se^4+ + 4e with a loss of electrons and that is oxidation or reducing agent. Or. Se...
May 2, 2016

Science
Science?
May 2, 2016

Chemistry
......2H2 + 2NO -> N2 + 2H2O I.....1.2..0.95....0......0 C.....-2x...-2x....x.....2x 5sec..1.02.0.95-2x..x....2x So 1.20-2x = 1.02 Solve for x and evaluate the other values you want to calculate.
May 2, 2016

chemistry
You had 25 g solution and obtained 4g anhydrous salt; therefore, you must have had 21 g H2O. So the saturated solution was 4g/21g solvent. Convert to g/100 mL. 4g salt x (100/21) = ?
May 2, 2016

Chemistry
It requires a little imagination and patience but it can be done. I don't think this will win a Pulitzer prize but here goes. Real chemicals often mesh, Eyes, noses, and other flesh, Digits, toes, and nails galore, On hands and feet that much more, Xenon free to the core.
May 2, 2016

chemistry
You must have meant BeCl2.
May 1, 2016

chem
HC2H3O2 = HAc ..........HAc + H2O--> H3O^+ + Ac^- I.......0.095...........0......0 C.........-x............x......x E.....0.095-x...........x......x Substitute the E line into the Ka expression for HAc and solve for x = (H3O^+).
May 1, 2016

chemistry
I don't get it?
May 1, 2016

Chemistry
You didn't like the answer I gave you earlier? What's wrong with it.
May 1, 2016

chemistry
What do you mean by a 10 DM vessel. Is that 10 dm^3 vessel. If I assume that then the (PCl5) = 0.1. If that isn't right correct accordingly. ......PCl5 ==> PCl3 + Cl2 I......0.1......0......0 C......-x.......x......x E......0.1-x....x......x Substitute the E line into ...
May 1, 2016

chemistry
dS = 100 x 4.2 x ln(T2/T1) Remember to use kelvin for T2 and T1.
May 1, 2016

Chemistry
I don't know the exact procedure you're using but if the original solution is in a volumetric flask you simply take an aliquot and go from there.
May 1, 2016

chemistry
See your other post. M = mols/L
May 1, 2016

chemistry
This is just a coefficient ratio. 0.00001 mol HSO3^- x (1 mol IO^-)/ mol HSO3^-) = 0.00001 x 1/3 = ?
May 1, 2016

science
(H^+) = sqrt(Kw/KaKb) and convert H^+ to pH. ammonium acetate = NH4Ac NH4Ac + HOH ==> NH4OH + HAc Plug in H^+ to Ka expression for HAc and calculate HAc.
May 1, 2016

Chemistry
How many mols H2CO3 are in 16 g? That's mols = grams/molar mass = ? Then there are 6.022E23 molecules in 1 mol.
May 1, 2016

Chemistry
Kayla, I answered this for you last night. You didn't like my answer?
May 1, 2016

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