Thursday
August 25, 2016

Posts by DrBob222

Total # Posts: 52,855

chemestry
I've done all but the water. You can fill that in. 2Cr2S3 + 30PbO2 + 24H2SO4 --> 30PbSO4 + 2H2Cr2O7 + ?H20 Check it out. Make sure I made no typos.
June 22, 2016

chemestry
I REALLY wish you would spell chemistry right. You've posted the last several days with the wrong spelling.
June 22, 2016

chemistry
right
June 21, 2016

Chemistry
The answer is here. https://en.wikipedia.org/wiki/Free_expansion
June 21, 2016

Chemistry
It is symmetrical in three dimensions so it is non-polar.
June 21, 2016

chemistry
See your other post.
June 21, 2016

Chemistry
What does P = 40% mean? You don't need the density. mols needed = M x L = ? Then grams = mols x molar mass. You know mols and molar mass, solve for grams HF.
June 21, 2016

Chemistry
First, a correction. The Henderson-Hasselbalch equation is pH = pKa + log (base)/(acid). That negative sign is either a typo in the book or by you in the post. What you are asking isn't exactly clear. I think you want to know why a. is the solution a buffer b. why it is ...
June 21, 2016

Chemistry
I don't think the molar mass of the 58.3 mg sample is 122.44 nor do I think you can calculate what it is. The NaClO4 contains Cl35, Cl36, Cl37in this sample. You know the percent Cl36 but you don't know the percentages of the other two (except of course in naturally ...
June 21, 2016

Chemistry
It's all here. The last links defines what an isotonic solution is for NaCl. https://en.wikipedia.org/wiki/Tonicity http://differentmedicalcareers.com/hypertonic-hypotonic-isotonic-solutions/ https://www.google.com/search?q=what+concentration+NaCl+is+isotoinic&ie=utf-8&oe=...
June 21, 2016

chemistry
For the same reason that no reaction every completes. For reversible reactions there is always at least one molecule that is "ready" to reverse direction.
June 21, 2016

chemistry
If you are in beginning chemistry, the easy answer is that it is 0.2M in (H^+) so pH = -log(0.2). Then pH + pOH = pKw = 14 and you get pOH from that. However, that isn't quite right since the second H^+ is not a strong acid and more advanced classes will do it slightly ...
June 20, 2016

chemistry
Ba(OH)2(aq) + 2HCl(aq) BaCl2(aq) + 2H2O (g) molBa(OH)2 = M x L = ? Now using the coefficients in the balanced equation, convert mols Ba(OH)2 to mols HCl. Finally, M HCl = mols HCl/L HCl.
June 20, 2016

Chemistry
K2SO4 = potassium sulfate NH4)F = ammonium fluoride
June 20, 2016

Chemistry
KaKb = Kw. You know Ka and Kw, solve for Kb .
June 20, 2016

chemestry
Please explain what the recipient is. Are all gases 2.4 moles or is that CO and H2 = 2.4 mols.
June 19, 2016

chemestry
.......SO2Cl2(g) ⇄ SO2(g) + Cl2(g) I......1.2..........0........0.8' C.......-x..........x.........x E.....1.2-x.........x......0.8+x Kp = 2.4 = (x)(0.8+x)/(1.2-x) Solve for x and evaluate each of the components.
June 19, 2016

chemestry
P.S. Note the correct spelling of chemistry.
June 19, 2016

chemestry
Essentially one ignores the contribution of H^+ from the HCN but you can calculate how much that is. .......HCN ==> H^+ + CN^- I......1........0.....0 C......-x.......x.....x E......1-x......x.....x For HCl, remember it is a strong electrolyte and ionizes completely...
June 19, 2016

chemestry
Note the correct spelling of chemistry. NH3 + HCl ==> NH4Cl mols NH3 = M x L = 0.1 x 0.05 = 0.005. mols HCl 0.2 x 0.025 = 0.005. You see this is an exact neutralizing solution; you have no HCl or NH3 left and have formed 0.005 mols NH3 in 75 mL solution so (NH4Cl) = 0.005/0...
June 19, 2016

@ kandace---chemistry
I should point out here that your problem makes no sense to me. True that the CO2 bubbled into Ca(OH)2 produces CaCO3 BUT you say the mass of the hydrocarbon is 10 g. So did you start with 10 g hydrocarbon or did the CaCO3 weigh 10 g. I don't call CaCO3 a hydrocarbon but ...
June 19, 2016

chemestry
Use the Henderson-Hasselbalch equation. Note the correct selling of chemistry. ALSO, and this is important, do you man m or M. m stands for molality. M stands for molarity. Also, I don't believe Ka for acetic acid is 1.76 x 10^-6. You either looked it up wrong or y you ...
June 19, 2016

Chemistry
I don't know many who have the solubility tables of thousands of chemicals memorized. I know I don't. Surely you have a table that lists the solubility of KClO3 at 80 C and 10 C. Or perhaps it's a graph that gives those numbers. If you will post those perhaps we ...
June 18, 2016

Organic chemistry
vanillan from vanilla
June 16, 2016

chemistry
What you have looks ok to me. Cr will go to Cr^3+.
June 16, 2016

CHEMISTRY
2F^- --> F2 + 2e -2.87 2I^- ==> I2 + 2e -0.535 But somehow I don't think this is the answer your are looking for. As to WHY the potentials are that way-- I suppose that's the nature of the beast. F^- being such a small ion isn't likely to give up electrons. ...
June 16, 2016

Chemistry
Why didn't you show us how you came up with 1.69. If we don't know how you got there how can we tell where you're going wrong? I cam up with approx 1
June 16, 2016

Chemistry
%bu ,ass = (grams solute/total grams)*100 = ? grams solute = 200 g total grams = 200+500 g = ?
June 15, 2016

Chemistry
1.25 g C/12 = mols C Convert to mols soap. grams soap = mols soap x molar mass soap.
June 14, 2016

chemistry lab
I suppose you meant "weigh". I can't answer because I don't know what you're doing.
June 14, 2016

Chemistry
mass % = g solute/100 g sample. M = mols/L solution. So you have 4.6 g C2H5OH/100 g solution. molar mass C2H5OH = 46 mols C2H5OH in 4.6 g is 4.6/46 = 0.1 mol. What is the volume? That's mass = volume x density or volume = mass/density = 100 g/0.93 = 108 mL or 0.108 L. Then...
June 14, 2016

Chem
There is none in what you have written. If you have water of crystallization it would show something like this: Fe(NO3)2.xH2O where x is 1, 2, 3, or some other whole number.
June 14, 2016

Chemistry
The right line of thought but you missed a turn somewhere. That's rate = k[x]^2[y]^1, then rate = k(2)^2(3)^1 = 12*k
June 14, 2016

CHEMISTRY
You don't have enough information to answer this. I assume you are talking about the electron in a H atom. It will different for other elements. How much energy do you get from a photon falling from n = 6 to n = 1? That's delta E = 2.180E-18J(1/1^2 - 1/6^2) = ?J/photon...
June 13, 2016

chemistry
CuO + H2SO4 ==> CuSO4 + H2O The reaction proceeds because the driving force is the formation of a weakly ionized material; i.e,in this case it is water. The reaction is a double replacement.
June 13, 2016

chemistry
The driving force is that PbI2 is a solid that is insoluble in the aqueous solution present with Pb(NO3)2 and KI
June 13, 2016

Chemistry
wavelength = h/mv wavelength = 5E-26. Convert to m. Convert 60 mph to m/s Substitute and solve for mass in grams and convert to pounds.Post your work if you get stuck.
June 13, 2016

Chemistry
Not quite. mols in each = M x L = ? Then after you know (H^+) in mols/L (i.e., molarity) then use the pH formula Bob Pursley gave you in the last line to get pH.
June 13, 2016

Chemistry
The idea here is that you add an excess of HCl to the sea shells and that completely reacts with ALL of the calcium carbonate. The excess HCl is titrated with NaOH. CaCO3 + 2HCl -> CaCl2 + 2H2O mols HCl added initially = M x L = 0.025 x 5 = 0.125. How much HCl was in excess...
June 13, 2016

Chemistry
Instead of posting to get my opinion why don't you OBSERVE this in the lab yourself. Then you will know.
June 12, 2016

chemistry
-COOH is a carboxylic acid.
June 12, 2016

chemistry
Convert g CO2 to grams C. Convert g H2O to grams H. Then grams O = mass sample - grams C - g H. Convert grams C to mols C. Convert grams H to mols H. Convert grams O to mols O. Now find the ratio of C,H,O to each other with the smallest number being one. The easy way to do ...
June 11, 2016

Science
2Mg + O2 ==> 2MgO mols Mg = grams/atomic mass = ? Using the coefficients in the balanced equation, convert mols Mg to mols MgO. In this case that is mols Mg x 1 = mols MgO Then convert mols MgO to grams. g MgO = mols MgO x molar mass MgO = ?
June 11, 2016

Chemistry
You're welcome.
June 11, 2016

Chemistry
C3H8 + 5O2 == 3CO2 + 4H2O Look up the delta H for this reaction or calculate it from dHrxn = (n*dHo products) - (dHo reactants) = ? Then that is the dH for the reaction for 44 g C3H8. Call that Y. Then Y x (355/44) = ? energy available.
June 11, 2016

chemistry
I don't believe there is enough information to answer that. pH2 = XH2*Ptotal. XH2 = 0.5 but you don't have a total pressure and I don't see a way to calculate it.
June 10, 2016

chemistry
I think the question is worded wrong because 500 mL of Pb(NO3)2 and 500 mL Na2CrO4 will ppt PbCrO4 big time. I think the problem is asking for volume of 0.1M Pb(NO3)2 needed to start pptn of PbCrO4 in 500 mL of 0.2M Na2CrO4.l If you start with 0.2M Na2CrO4, then (CrO4^2-) = 0....
June 9, 2016

chemistry
2Fe2O3 + 3C → 4Fe + 3CO2? C:Fe = 3/2. Look at the coefficients.
June 9, 2016

Chemistry
A is correct at 15.77 rounded to 15.8%
June 9, 2016

chemi
Your post isn't quite clear. What DID YOU OBSERVE? I don't know what you saw. ZnCO3 ==> ZnO + CO2 What do you mean by APPARENT loss in mass. It either lost mass or didn't lose mass. There is nothing apparent about it.
June 9, 2016

Chemistry
Your error is in thinking the Ac is the same as HAc. They are equal in a solution of acetic acid but the problem tells you that NaAc has been added. Now the solution has H^+ one number and the Ac is whatever you've added. So make that Ka = (0.0025)(Ac^-)/(0.9975) and solve...
June 9, 2016

chemistry
This is not a buffer problem; therefore, the HH equation is not applicable. ......HCOOH ==> H^+ + HCOO^- I......Y........0......0 C......-x.......x......x E.....Y-x.......x......x You know pKa, convert that to Ka. Ka = (H^+)(HCOO^-)/(HCOOH) The problem gives you pH so you ...
June 8, 2016

Chemistry
4Fe + 3O2→ 2Fe2O3
June 8, 2016

Chemistry
right
June 8, 2016

Chemistry
/both. Taught for 50 years. Now retired.
June 8, 2016

Chemistry
yes
June 8, 2016

Chemistry
You have the best when you have Bob Pursley.
June 8, 2016

Chemistry
In your first attempt I saw two errors. There may be more but I quit looking. One error is that x = (OH^-) but you used it as (H^+). The second is I don't know where the 3E-12 came from. In the second trial you made a math error right off the bat and I quit looking. 5E-4 ...
June 8, 2016

chemistry
Ahhh---but we do that sometimes. It is a GREAT learning experience. We aren't likely to do it twice.:-)
June 8, 2016

chemistry
The easy way. pKa + pKb = pKw = 14 You have pKa of 9.26 which is right but you took the -log of 4.26 which isn't right. pKa = 9.26 = -log Ka Ka = ?
June 8, 2016

Analytical Chemistry
Your question (are these two questions) is too general. What do you want? pH, OH or something else. What are you titrating it with; i.e., you need to know the final volume. Example. 0.1mol NaOH titrated with 0.1M HCl. 0.1 x 0.99 = 0.099 titrated. That leaves 0.1-0.099 = 0.001 ...
June 8, 2016

chemistry
I think of molecular compounds as being made up of molecules with covalent bonds. CO2 for example. NaCl is an ionic compound and consists of discrete Na^+ and Cl^- so it is not molecular while CO2 exists as discrete CO2 molecules. SO2 and SO3 are other molecular compounds.
June 8, 2016

chemistry
So far you are good. So Q = 1E-40 < K = 1.4E-37 which lets you know c and d can't be right. The answer must be a or b. Now write the Keq expression. Keq (which is really Ksp) = Cu3(PO4)2 = (Cu^2+)^3(PO4^3-)^2 = 1.4E-37 Q must be larger. How can it get larger? No way ...
June 7, 2016

chemistry
Reverse eqn 1(change sign of dH), add to eqn 2 and 2 x eqn 3.
June 7, 2016

chemistry
By the way, the Ka value has no units and don't let anyone tell you they do. When they are used, most of the time they are called just units, but often they are called "tentative units". "temporary units", "trial units", or
June 7, 2016

chemistry
You can get the definition of an acid from you text/notes/google. For the HA problem. For pH = 3.50 convert to (H^+). I obtained approx 3E-4 but you need to confirm since that's an estimate. .......HA ==> H^+ + A^- E......x....3E-4...3E-4 Substitute the E line into the ...
June 7, 2016

chemistry
Your text or notes contain all of these definitions. YOu don't need us to do your homework for that. As the for Lewis structure, we can't draw structures on this forum.
June 7, 2016

Chemistry
If the 35 is a superscript and the 15 is a subscript, then C is correct.
June 7, 2016

Chemistry
I think you can heat K2CO3 up to 110 C without decomposition. Spread crystals out on a watch glass and heat at 110 for 2 Hrs. Cool in a desiccator.
June 6, 2016

chemistry
Al ==> Al^+ + 3e Eo ox = ? Ag^+ + e ==> Ag Eo red = 0.8 ----------------------- Al + 3Ag^+ ==> 3 Ag + Al^3+ E=2.46 ? + 0.8 = 2.46 ? in this set up is the oxidation potential. Change the sign to make it the reduction potential.
June 6, 2016

Chemistry
When AgNO3 is added drop wise to the solution, the one with Qsp greater than Ksp will ppt first. You're correct, that is AgI. More drops of AgNO3 results in AgI continuing to ppt until Ksp for AgCl is met. What is (Ag^+) then? Ksp = (Ag^+)(Cl^-). You know Ksp and you know ...
June 5, 2016

Chemistry
mols needed = M x L = ? Then grams = mols x molar mass = ?
June 5, 2016

Analytical Chemistry
Here is a very detailed document. Read through that and see if it helps. I've not seen one this detailed in some time. All of the reactions are there although it is a little light on the stoichiometry part. http://www.lasalle.edu/~prushan/Experiment8-redox%20titration.pdf
June 5, 2016

chem
........CO(g) + Cl2(g)⇌ COCl2(g) E.....1.9E-6...9.3E-7.....x Substitute the E line into the Keq expression and solve for x = (COCL2) in mols/L. Then mols/L x 5.89 L = mols COCl2. Grams COCl2 = mols COCl2 x molar mass COCl2.
June 5, 2016

quimica
millimols NaOH = mL*M = 20*0.175 = 3.5 If pH = 12.55, then pOH = 1.45 and (OH^-) = 0.03548 M so mmols at the end will be 0.03548*(20+xmL) where x mL is the volume of the 0.2M HCl added. .....OH^- + H^+ ==> H2O I...3.5.....0 add.......0.2x.......... C...-x...-0.2x.........s ...
June 4, 2016

chemistry
aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide a. balance the chem eqn for this rxn Al2S3+6H2O = 2Al(OH)3+3H2S right. b.how many g of water are needed in the rxn to produce al(oh)3? what is the formula i use here?? You can't do without ...
June 4, 2016

Chemistry Homework
No. Your first line is in error and I quit reading there. 2.45 dL = 245 cc.
June 3, 2016

Chemistry
3 mols CO2 x (5 mols O2/3 mols CO2) = 3 x (5/3) = ?
June 3, 2016

Chemistry/- Dr.Bob222
Zn + Ni^++ ==> Zn^++ + Ni Ecell = Eocell - (0.05916/n)*Log Q where Q (I didn't have space to write it is) (Zn^2+)(Ni)/(Ni^2+)(Zn) You know of course that (Ni) and (Zn) = 1 since they are solids. Initially (Ni^2+) = 1.5*(Zn^2+) so at the end, since both are divalent, ...
June 2, 2016

chemistry
Did you make a typo and you meant to type in that the teacher obtained 0.30 N? Technically, you can't say. Both could be right but I disagree with Scott as to why that is so. You must know the equation involved. For example, if you titrate one(1) H then mL x N x ...
June 2, 2016

Chemistry
Actually, you don't need any of this other information since Kb for HPO4- = Kw/Ka for H2PO4^-
June 2, 2016

Chemistry
You didn't read my earlier response? I guess it's just easier to repost and hope someone will work the problem for you.
June 2, 2016

chemistry
The question isn't clear. Decomposed to what? KNO2 + O2? K _ N2 + O2, K2O + NO2? I suspect to KNO2 + O2 but I don't know that and, in my opinion, that isn't completely decomposed.
June 2, 2016

Chemistry
I think you should review your notes and answers. I question 64, 66, 67, 69 and I quit reading.
June 2, 2016

Chemistry
Bobby, what do you not understand about this? In detail, please. Make an ICE chart using mols and substitute into the Henderson-Hasselbalch equation.
June 2, 2016

chemistry
Have you used the Arrhenius equation? Show your work; perhaps we can find the error.
June 2, 2016

Chemistry
I believe the original definition for an Arrhenius base was that a base material contained the OH ion. LiOH would be the answer for this question. I believe the definition was expanded somewhat, later, to include anything that produced an OH ion in aqueous solution. As such, ...
June 1, 2016

Chemistry - Dr.Bob
You don't get Na and S and O. You get H2 gas at one electrode and O2 at the other.
June 1, 2016

Chemistry
I think your answer is right but actually it depends upon the energy of the gamma ray.
June 1, 2016

Chemistry
(V1/T1) = (V2/T2) Remember T must be in kelvin
May 31, 2016

Chemistry
Wouldn't it have been better to show your work instead of leading us around to the back of the barn? No you don't need the formula mas of MgCl2.6H2O. If you can change pounds into liters you're a better chemist than I am. Pounds is a weight. L is a volume. Change ...
May 31, 2016

Chemistry
Something is wrong with the Zn/Cu question above. You have answered it correctly but the computer won't show my answer when I try to post it.
May 31, 2016

Chemistry
Did you leave off the electrons? What you have check is the reduction reaction but that isn't right because the electrons have been omitted. Oxidation is the loss of eletrons. Fe^2+ ==> Fe^3+ + e is the oxidation half reaction. Also note that neither A nor D are ...
May 31, 2016

Chemistry
right
May 31, 2016

chemistry
q = m * c * (change in temperature) q = 20.5 * 2.05 * (0+24.3) = 1021.2 J right and I would make that 1.02 kJ. Then you need to melt the ice. q = (20.5/18)*6.02 = ? q = 20.5 * 4.184 * (100-0) = 8577.2 J right and I would make that 8.68 kJ. Then you need to vaporize the water...
May 31, 2016

Chemistry
I wonder why you don't call that silver nitrate? 2AgNO3 + Cu --> Cu(NO3)2 + 2Ag mols Cu = grams/atomic mass = ? Then you get 2 mols Ag for every mol of Cu.
May 31, 2016

Chemistry
.......H2O2 ==> H^+ + HO2^- I.....0.86......0......0 C......-x.......x......x E...0.86-x......x......x Substitute the E line into the Ka expression and solve for x = (H3O^+).
May 31, 2016

Chemistry
(Na^+) = 0.10 M (OH^-) = 0.10 M (H^+) = Kw/(OH^-) = ?
May 31, 2016

Chemistry!!! PLZ PLZ HELP!!
You can't work this problem without knowing the volume. If we assume a volume of 1L, then (H2) initial is 0.1 M and (I2) initial is 0.2 M. .......H2 + I2 ==> 2HI I.....0.1M..0.2M....0 C.....-x.....-x.....2x E...0.1-x...0.2-x...2x Substitute the E line into the Keq ...
May 30, 2016

@ Mayra---Chemistry
The primary reason I said M was not possible is because that definition is mols/L SOLUTION. Your problem gives 400 mL water which is not the same as 400 mL solution. 25 g NaOH + 400 mL water is not 400 mL solution but more than that. By what volume who knows. So you may ...
May 30, 2016

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