Monday
March 2, 2015

Posts by DrBob222


Total # Posts: 47,353

chemistry
same as # mols NaOH mols NaOH = M x L
March 1, 2015

Chemistry
4. 2Na + Cl2 ==> 2NaCl is one way. 6 & 7. I don't know the cost of making it; it is almost dirt cheap to buy. I can buy a box of salt for less than a dollar. In most cases I don't think it's made. Much of it is mined as NaCl in salt mines in Louisiana (among ...
March 1, 2015

Chemistry
Yes, and I would convert 2.00 g azomethane to mols first.
March 1, 2015

Chem
A = ebc where e is the molar absorptivity = slope of the line. 0.455 = 3163.9*c Calculate c.
March 1, 2015

Chemistry Need Help
Thanks for letting us know.
March 1, 2015

Chemistry
dHrxn = (n*dHf products) - (n*dHf reactants)
March 1, 2015

oops---chemistry
http://en.wikipedia.org/wiki/Standard_state
March 1, 2015

chemistry
I would think this would be covered in your text/notes. Here is a reference but IUPAC rules don't always follow through in chemistry texts/journals. For example, most U.S. based chem text use 1 atm as standard pressure but IUPAC suggest 100 kPa (1 bar).
March 1, 2015

chemisty
......2NO2 (aq) ==== 2NO (aq) + O2(aq) I....0.750............0..........0 C......-2x.............2x.........x E....0.750-2x..........2x.........x So the problem tells you x [which is (O2)] = 1.25M which makes 2x (the concn of NO) just twice that. NO2 then is 0.75-2x
March 1, 2015

chemistry
I don't understand the problem you'r having. Write K exprssion for the reaction, substitute the numbers as shown and punch them into your calculator.
March 1, 2015

chem
It's the product of the right side over the product of the left side and each coefficient becomes an exponent for that substance.
March 1, 2015

chem
I don't see a question. Probably it's to evaluate Z but what in the world is Z. By the way, Does your equation have an extra Fe in it?
March 1, 2015

science
I don't see a question. That's CaCl2
March 1, 2015

Chemistry
I calculated the same number as you; I suspect the problem is that you're reorting too many significnt figures. If you posted the problem EXACTLY as stated, then you're allowed only two places by the 2.1 and I would round the answer to 2.2E3 J. If that 2.1 is really 2....
March 1, 2015

Chemistry
[mass milk x specific heat milk x (Tfinal-Tinitial)] + [mass coffee x specific heat coffee x (Tfinal-Tinitial)] = 0 Substitute and solve for mass milk.
March 1, 2015

chemistry
dHrxn = (n*dHformation products) - (n*dHformation reactants)
March 1, 2015

Chemistry
The pH is determined by the hydrolysis of the salt at the equivalence point. If we call the acid HA (the problem doesn't identify it) nor does it say it is monoprotic or not (but only one pKa suggests monoprotic) so the anion will be A^-. Volume of NaOH to reach the ...
March 1, 2015

Chemistry
You don't have enough information. mols Na3PO4 = M x L but I don't see a L. Same for the Co salt, no M
March 1, 2015

chemistry
mols AgNO3 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols AgNO3 to mols AgCl. Now convert mols AgCl to grams. grams AgCl = mols AgCl x molar mass AgCl.
March 1, 2015

Chemistry combined gas law
I answered this below that it was not correct and what was wrong as well as how to correct it.
March 1, 2015

science
(NH4)2SO4 + CuSO4.5H2O==> no reaction
March 1, 2015

chemistry
mols CH4 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols CH4 to mols H2O. Now convert mols H2O to grams. That is grams H2O = mols H2O x molar mass H2O.
March 1, 2015

Chemistry
From what I can gather from that long string of instructions the idea is to determine the solubility of Co3(PO4)2. I believe the idea is that Co(NO3)2 is colored. So when it reacts with the Na3PO4 it ppts the Co ion as the phosphate (solid) and settles (or is centrifuged) to ...
March 1, 2015

Chemistry
See your previous post. I'm overwhelmed!
March 1, 2015

Chemistry
#1. Exactly how would you like us to help you on this assignment? #2. I don't see a question. #3. Although the instructions are there I don't see any observations. No results. No calculations. #4. Are you trying to dry lab this experiment; write up the results/...
March 1, 2015

Chemistry
KHP + NaOH ==> NaKP + H2O mols KHP = grams KHP/molar mass KHP Since 1 mol KHP reacts with 1 mol NaOH, you know mols NaOH = mols KHP Then M NaOH = mols NaOH/L NaOH. You know mols NaOH and L NaOH, calculate M NaOH
March 1, 2015

Quick chemistry help
Change 566 g to mols. mols = grams/molar mass Change mols to # molecules. You know there are 6.02E23 molecules/ 1 mol so # mols x 6.02E23 = # molecules C12H22O11 Then the formula tells you there are 6 C atoms in 1 molecule so # C atoms must be 12 x the number of molecules.
March 1, 2015

Chemistry combined gas law
No, you have the right formula and your algebra is right but you didn't assign the values correctly. You problem states you have 21.5L at STP so V1 is 21.5, T1 is 273 and P1 is 101.325. You are changing the conditions to 93.2 kPa so that is P2 and to temp of 333 K so that ...
March 1, 2015

Chemistry
Yes, the balanced equation is VERY necessary. Otherwise you don't know that they are reacting in a 1:1 ratio. And you want to be careful when comparing the numbers. For example, suppose we had the same problem and numbers EXCEPT we used H2A and NH4OH. .......H2A + 2NH4OH...
March 1, 2015

Chemistry
mols HA = M x L = 1.20 x 0.050 = 0.06 mols NH4OH = M x L = 0.92 x 0.05 = 0.046 ..........HA + NH4OH ==> NH4A + H2O I.......0.06..0.046.......0.......0 C....... E....... These two materials react in a 1:1 ratio so which one looks as if it is the limiting regent; i.e., which ...
March 1, 2015

chemistry
2NH3 + H2SO4 ==>(NH4)2SO4 mols H2SO4 = M x L = ? mols NH3 required = 2 x mols H2SO4 which you find by using the coefficients in the balanced equation; i.e., mols NH3 = molx H2SO4 x (2 mols NH3/1 mol H2SO4) = ? Then 1 mol NH3 gas at STP occupies 22.4 L; therefore, ? mols NH3...
March 1, 2015

Organic Chemistry
You need to rephrase the question. It makes no sense to me as is.
March 1, 2015

gen chem
I would do this. .......NH4^+ + OH^- ==> NH3 + H2O I......0.3......0.......0.2....... add............0.1................ C,....0.3-0.1..-0.1.....+0.1 E......0.2......0........0.3 Convert the mols in the E line to M (mols/L) and substitute into the Henderson-Hasselbalch ...
March 1, 2015

CHemistry
.........F^- + H^+ ==> HF I.......0.2....0........0 add...........0.05........ C......-0.05..-0.05.....0.05 E......0.15....0.......0.05 Use the Henderson-Hasselbalch and calculate pH.
March 1, 2015

chemistry
Tartatic acid = H2T ........H2T ==> H^+ + HT- I......0.25......0.....0 C........-x......x.....x E......0.25-x....x.....x Substitute the E line into Ka expression and solve for x = (H^+), then convert to pH. This looks like the wrong Ka for the acid.
March 1, 2015

Chemistry
You don't provide enough details to even know what you're doing.
March 1, 2015

Chemistry
q = mass Au x specific heat Au x delta T. 228 = 45.3 x 0.131 x delta T. Solve for delta T, the check the answers and see which matches delta.
February 28, 2015

chem
m = mols/kg solvent Substitute solvent kg and m, solve for mols. Then mols = grams/atomic mass You know atomic mass and mols, solve for grams.
February 28, 2015

chemistry
........FeO(s) + CO(g) = Fe(s) + CO2(g) I......solid....2.00......solid..0 C......solid......-x......solid..x E......solid....2.00-x....solid..x Keq = pCO2/pCO = 0.800 Substitute the E line into Keq and solve for x and 2.00-x
February 28, 2015

Chemistry
Doesn't that depend upon how it is weighed and the volumetric equipment used?
February 28, 2015

chemistry
.........PbI2 ==> Pb^2+ + 2I^- I........solid.....0.......0 C........solid.....x.......2x E........solid.....x.......2x Ksp = (Pb^2+)(I^-)^2 0.00027 = (x)(2x)& = 4x^3 so x = solubility PbI2 = about 0.04M WHICH IS THE MAXIMUM AMOUNT OF PbI2 in solution. Frankly, I don't ...
February 28, 2015

chemistry
Look in your text/notes for the definition of equilibrium, then choose your answer. It is NOT A.
February 28, 2015

chemistry
Yes, absolute temperature is measured in kelvins.
February 28, 2015

Chemistry
a. k=0.693/t1/2. Substitute and solve for half life. b. ln(No/N) = kt No = 100 N = 60 (that's 100-40 = 60) k = from above. t = solve for this. c. See b.
February 28, 2015

Chemistry
ln 2.3E14 = -Ea/RT Solve for Ea
February 28, 2015

Chemistry
Use less energy to form products.
February 28, 2015

Chemistry
I thought I did #1 last night. All of the others are off shoots of that. Remember that any component in an equation (in mols) can be converted to any other component in the reaction by using the coefficients in the balanced equation.
February 28, 2015

Chemistry
What is the Ea change?
February 28, 2015

Chemistry
pCHCl3 = XCHCl3*PoCHCl3 pCCl4 = XCCl4*PoCCl4 Total vapor Pressure = pCHCl3 + pCCl4 X each = 1 mol/2 mols = 0.5 for X In the vapor state, XCHCl3 = pCHCl3/Ptotal XCCl4 = pCCl4/Ptotal
February 28, 2015

Chemistry
Use the Henderson-Hasselbalch equatiion.
February 28, 2015

Chemistry Help
I think your error is caused by not balancing the equation. ALWAYS start with a balanced equation. 2NH3 ==> N2 + 3H2 0.250mols NH3 x (1 mol N2/2 mol NH3) = 0.250 x 1/2 = ? mols N2 0.250 mols NH3 x (3 mols H2/2 mols NH3) = 0.250 x 3/2 = ? mols H2
February 28, 2015

chemistry
HCl + NaOH ==> NaCl + H2O q = heat generated = mass solution x specific heat solution x (Tfinal-Tinitial) mass solution is 150 mL x 1.02 g/mL(NOTE: I think this is a typo and should be 1.02 g/mL and not 1.02 mg/mL) That's q for how many mols? That's 13.6 kcal/mol x...
February 28, 2015

chemistry 2
Is that rate = k[AB]^2? And I assume the units are L/mol*s. If so then it is a second order reaction and (1/A) - (1/Ao) = kt Substitute and solve for A
February 28, 2015

science
Takes the object 6s to get where? 60m/6s = 10 m/s and since it's going up I would put a - sign on it.
February 28, 2015

Chem
A is not true. You would LIKE about 10^5 difference in k1 and k2 but usually we can live with 10^3(1000 times) and in some cases even 10^2 (100 time) but never as close as 10^1 (10 times) I don't believe B is true but the statement isn't really all that clear to me. C ...
February 27, 2015

Chemistry
The problem doesn't ask for pH. I would redo it as a quadratic. 0.316 = (H^+) with the assumption that 2.0-x = 2.0 But it really is 2.0-x and it doesn't appear to me that the x can be ignored with respect to 2.0. That looks like about 15%. I would feel better if the ...
February 27, 2015

Chemistry
...........H3PO3 ==> H^+ + H2PO3^- I..........2.0.......0......0 C...........-x.......x......x E..........2.0-x.....x......x Ignore the contribution to H^+ of k2 then substitute the E line into k1 expression and solve for x = (H^+)=(H2PO3^-)
February 27, 2015

Chemistry
I got x^2 + 6.25E-3 - 2.06E-4 = 0 and when I solved that I came up with 0.0116M (vs 0.0144 before) for x. That gave me 1.93 for pOH and 12.06 for pH.
February 27, 2015

Chemistry
You have assumed 0.033-x = 0.033 and I'm not sure you can do that. If we compare the answer of 0.0144M for OH^- (rounded) is with the assumption but 0.033-0.0144 appears it isn't = 0.033. So I think you need to go through the quadratic equation. A quick guess is that ...
February 27, 2015

Chemistry
I think Kb for the ascorbate ion is kw/k2 instead of kw/k1. Everything else (the procedure that is) looks ok to me. You can prove this is you wish by writing the k1 and k2 expression as well as Keq for the hyrolysis equation for ascorbate ion.
February 27, 2015

Chemistry Help
Ba(OH)2(aq) + 2HBr(aq) → BaBr2(aq) + 2H2O(l) By the way, I see I made a typo late last night. I said your 0.0759 mols HBr was right; I should have said your 0.00759 mols HBr. Here is how you get the 1/2. Use the coefficients in the balanced equation. 0.00759 mols HBr x (...
February 27, 2015

Chemistry Help
You are right with 0.0759. That is mols HBr. Then mols Ba(OH)2 is 1/2 that and not 2x that. Finally, the third is an error also. It should be M Ba(OH)2 = mols Ba(OH)2/L Ba(OH)2. You know mols and M, solve L and convert to mL.
February 27, 2015

Chemistry
Ba(OH)2 ionizes 100%; therefore, ..........Ba(OH)2 ==> Ba^2+ + 2OH^- I.........0.00988.....0.........0 C.......-0.00988....0.00988..2*0.00988 E.........0.........0.00988..0.0198 .........H2O --> H^+ + OH^- I........liquid...0....0.0198 C........liquid...x.....x E...........
February 27, 2015

Chemistry
LiOH = 0.0474M (OH^-) = 0.0474M (H^+)(OH^-) = 1E-13 Substitute the OH^- from LiOH and solve for H^+.
February 27, 2015

Chemistry
Remember that you can convert any material in mols in the equation to any other material in mols by using the coefficients in the balanced equation. 0.65 mole NH3 x (1 mol N2/2 mol NH3) = 0.65 x 1/2 = 0.325 mols N2 reacted and I know that's too many s.f. Important: Note ...
February 27, 2015

Chemistry
Use the periodic table. There are 5 elements that are gases; H2, N2, O2, F2. Cl2 (plus of course the noble gases but we usually ignor those anyway). There are some common compounds that yu will learn as you go along; i.e., CO2, SO2, SO3, CO, NO, NO2, CH4(methane), C2H6(ethane...
February 27, 2015

Chemistry
There are 6.02E23 atoms in a mole of atoms or 6.02E23 molecules in a mole o molecules; actually, 6.02E23 anythings in a mole of anythings. Just as there are 12 in a DOZEN, 144 in a GROSS, 500 sheets in a REAM of paper,etc, there are 6.02E23 molecules in a mole. It's just ...
February 27, 2015

Physical Science
right
February 27, 2015

Chemistry
Let HBz be bezoic acid ..........HBz ==> H^+ + Bz^- I.........2.45....0......0 C..........-x.....x......x E.......2.45-x....x......x Substitute the E line into Ka for HBz (you will need to look that up in your text/notes) and solve for x Then %ionization = [(H^+)/2.45]*100 = ?
February 27, 2015

science
And you know how to do which ones? What do you not understand. The definitions is where you start.
February 27, 2015

Chemistry
Can you explain what you don't understand about this? Have you tried setting up an ICE chart?
February 27, 2015

Chemistry
See your post above.
February 27, 2015

quadratic formula
First you should factor out the 2 to leave X^2 + 4x -5 = 0. You should not be getting an i. Getting an i means you are a negative under the square root. Your problem is that you're not doing the part under the square root sign correctly. sqrt(b^2-4*a*c) sqrt(16-4*1*-5) ...
February 27, 2015

chemistry
dHrxn = (n*dHformation products) - (n*dHformation reactants) dSrxn = (n*dSformation products) - (n*dSformation reactants) Then dGrxn = dHrxn - TdSrxn
February 27, 2015

Chemistry
They do not.
February 27, 2015

Chemistry Need Help
This is a limiting reagent problem (LR) and you know that because amounts are given for BOTH reactants. Your 1.50 mols Cl2 will produce 2*1.50 = 3.00 mols ClF if all of the Cl2 reacts and you have an excess of F2. Your 1.75 mols F2 will produce 2*1.75 = 3.50 mols ClF if you ...
February 27, 2015

Chemistry
pH = pKa + log (base)/(acid) 5.25 = pKa + log (base)/(M x L) Solve for base and since the acid is in mols, the base will be in mols. Then mols = grams/molar mass. You know molar mass and mols, solve for grams. That will be the number of grams of the salt to add to the 532.9 mL...
February 27, 2015

Chemistry Help
No, if the problem states it correctly, the theoretical yield is 12.2g. You collected only 8.82 g so %yield = (8.82/12.2)*100 =
February 27, 2015

Chemistry
Yes. The limiting reagent will always produce the LEAST product (whichever one you choose)
February 27, 2015

Chemistry
2BCl3 + 3H2 ==> 2B + 6HCl 0.5 mol BCl2 will produce 0.5 mol B if you have all of the H2 needed. 1.20 mols H2 gas will produce 1.20 x (2 mol B/3 mols H2) = 1.20 x 2/3 = 0.8 mol B if you hadd all of the BCl3 needed. Both answers can't be right; the correct answer in ...
February 27, 2015

chemistry
mols PbCl2 = grams/molar mass = approx 6E-3 but you need a better number than that. I've estimated as well as all of the calculations that follow. That's mols in 2.00 mL out of 84.8 mL. Convert to mols in 100 mL (the original volume) by 6E-3 x (100/84.8) = approx 7E-3 ...
February 26, 2015

chemistry
Calculate k. k = 0.693/t1/2 k = 0.693/0.34 = ? Then ln(No/N) = kt. Use No = 100 and N = 12.5 (that's 100%-87.5% = 12.5%) k from above. Substitute and solve for t(in hours).
February 26, 2015

chemistry
A quickie fix. That is Cu(OH)2. I'll try. You are right that there is no reaction. When you talk about the activity series you are talking about single replacement reactions. This is not a SR reaction so none of what you suggested applies. That takes care of that. When ...
February 26, 2015

Chemistry help
How would what be affected (or effective)?
February 26, 2015

science
I don't get the response.
February 26, 2015

science
The hot tea cools, the spoon and air heat up but the spoon heats more than the room full of air. The final temperature will be somewhat less than 65 F but not much less (it all depends upon the size of the room). You could calculate the final temperature if you had the masses ...
February 26, 2015

Chem, Oxidation States
Rule 1 is that the oxidation states must add to zero for a compound or to the charge on an ion. Rule 2 is representative elements have their "normal" charge; i.e., group I has +1, group II has +3, group III has +3, etc. Rule 3 is that H is +1 (except in unusual cases...
February 26, 2015

Chemistry, HELP!
X(H2MoO4) = (H^+)^2/D X(HMoO4^-) = k1(H^+)/D X(MoO4^2-) = k1k2/D where D is denominator and that = (H^+)^2 + k1(H^+) + k1k2 when you finish calculating each mole fraction, then X*5.1E-4 = (that specie) There is another way to do this without mole fractions. It's about as ...
February 26, 2015

Chemistry, HELP!
Are you in to mole fractons or alpha zero, alpha 1 and alpha 2?
February 26, 2015

science and calculation
heat lost by metal + heat gained by H2O = 0 [mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute the numbers and solve for specific heat metal which is the only unknown in th equation. BTW, centigrade is out...
February 26, 2015

Chemistry Equilibrium
Please look at your post and your problem. If ALL of the 1.15 g H2O decomposed to 2H2O --> 2H2 + O2 you would have (1.15g H2O x (1 mol H2O/18 g H2O) x (1 mol O2/2 mol H2O) = about 0.064/2 = about 0.032 mols O2 formed. The problem states that 0.15 mol O2 were found at ...
February 26, 2015

Chemistry to DrBob222
Thank you but I really don't deserve any credit. All of the homework helpers on this site are volunteers (no pay in dollars but much pay in satisfaction) and we do this because we like to do it. Most of us are retired teachers. I taught at a university for almost 40 years ...
February 26, 2015

solution making/ chemistry/biotech
Doesn't that depend upon the concentration of the 750 mL final solution?
February 26, 2015

physics help needed---Please help and explain engineering science n3
x
February 26, 2015

hi
wok soup?
February 26, 2015

Science
q = mass water x specific heat H2O x (Tfinal-Tintial) Substitute and solve for Tfinal.
February 26, 2015

Chemistry
Ok, Need some more help... This is what I have so far. 4. Urea (NH2)2CO is prepared by reacting ammonia with carbon dioxide. The byproduct is water. 637.2g of ammonia are reacted with 787.3 g of carbon dioxide. molar mass:NH3=17 CO2=44 CO(NH2)2=60.1 a. (10 points) Write a ...
February 26, 2015

Chemistry Drbob222
a. (10 points) Calculate the theoretical yield of Ti(s). In order to find TY I have to find the limiting agent.. ?? ugh.. Ok, Ti=47.88 Cl4=35.45*4... To find moles in TiCl4 I take 3.54*10^7 * 1moleTiCl4/189.68 = 184,521.89 moles? But I need TY of Ti?? 7.91*10^6 * 1mole Ti/47....
February 26, 2015

clarification please--Chemistry! Please Help!
For a, do you have a Kc or Kp? Or does the question assume ALL of the CaCO3 decomposes?
February 26, 2015

inorganic
This looks like a fishing expedition. Why don't you list the chemicals required for Ag^+ (and confirm Ag^+), then tell me what you know about those materials.
February 26, 2015

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