Sunday
March 29, 2015

Posts by DrBob222


Total # Posts: 47,990

Chemistry Help
If they dissociate they are electrolytes. Na2CO3 yes. O2 no
March 3, 2015

Chemistry help!! Fast
I really don't know but if we go with the "like dissolves like" slogan, I would think it false since the larger molecules have a higher molar mass and they should dissolve better is solvents with high molar mass.
March 3, 2015

Chemistry
yes if you make that (182/4780)*100 = ? What you want is (182*100/4780) = ?
March 3, 2015

Chemistry Basics- Please Help!
I agree with what you have.
March 3, 2015

CHEMISTRY
You have two/three problems in one here. First I would determine the base and acid concentrations in the initial buffer. To do that use pH = pKa + log (base)/(acid) and solve for base/acid ratio. Then you know acid + base = 0.2 Solve those two equation simultaneously to ...
March 3, 2015

Chemistry
density = mass/volume. Plug and Chug.
March 3, 2015

chenistry
You're still trying to make grams go to grams. You CAN'T do that. You MUST change grams to mols, convert mols of what you have to mols of what you want, then change back to grams.
March 3, 2015

Chemistry Help
yes
March 3, 2015

Chemistry Help
You have mols H2SO4 = M x L = 0.178 x 0.2 = ? Since there are 2 H atoms per 1 molecule H2SO4, then there will be twice as many H atoms as there are mols H2SO4.
March 3, 2015

Chemistry Dr. Bob
You don't convert anything to grams. Use mols to convert from what you have to what you want. Then convert to grams. Reactions go by mols, not grams.
March 3, 2015

Chemistry
At the beginning you have pure NH3 solution. ......NH3 + H2O ==> NH4^+ + OH- I.....0.1............0.......0 C......-x............x.......x E.....0.1-x...........x......x Plug the E line into the Kb expression and solve for x = OH, then convert to pH. For part b. You started...
March 3, 2015

Chemistry
You stuck in numbers with the Na2O question without showing the units and you reversed the factor. The units are not there and that's why you can't find the error. First I would not have rounded the 0.84 but kept 0.838. Why throw a perfectly good digit away. You have ...
March 2, 2015

college chemistry
I found a conversion factor to convert 0.737 g/mL to pounds/gallon. 0.737 g/cc = 6.15 pounds/gallon. So 23 gallons x 6.15 pounds/gallon = ?
March 2, 2015

Chemistry
20 uci/kg x 50 kg = ?uci and convert to mci.
March 2, 2015

Chemistry
Yes, it matters what is being neutralized AND chemicals reactions don't occur between grams but between mols so you can't make the statement you made (at least and be right).
March 2, 2015

Science
You don't give us much to go on. NaOH + H2CO3 will make Na2CO3 + H2O
March 2, 2015

Science // Chemistry
Start by asking yourself how many mols you need? Moles KOH you want = 1.00 L x (1.3 mols/L) = ? mols and that is 1.3 moles KOH. Then 1.3 moles KOH x (56.1 grams KOH/mol KOH) = grams KOH you need for 1 L of 1.3M KOH.
March 2, 2015

Chem-mystery
You meant Ca(OH)2. This is new to me but I believe this is an acid/base reaction between HN3 and Ca(OH)2. I think the equation is 2HN3 + Ca(OH)2 ==> Ca(N3)2 + 2H2O mols HN3 = M x L = 0.2 x 0.05 = 0.01 mols Ca(OH)2 = M x L = 0.2 x 0.05 = 0.01 We start with this probably ...
March 2, 2015

Chem-mystery
You don't need M Ca(OH)2??? I don't think you can work the problem without it. .
March 2, 2015

Science
It depends upon how many significant figures you are reporting. You are allowed 4 and I get 24.32 x 100.09 = 2434.189 so you can legitimately report 2434 g with 4 places. There is nothing wrong with reporting it as 2.434E3 grams.
March 2, 2015

Science
You aren't missing anything. You are right on. Thanks for showing your work. When you do that it takes just a glance to know if things are ok.
March 2, 2015

Chemistry
I worked this below.
March 2, 2015

Chemistry
I re-thought that problem (all night long) and there is nothing wrong with the first part; i.e., (H^+) must be 0.0018 to make it work and you must add 2.83 mL of the 0.2 M acid to get that. When I tried to check it to see if all that worked the best I could do was 0.99% and ...
March 2, 2015

oops--Chemistry
I made a typo part way down the page. Instead of retyping the whole things, since it is so long, here is the line in bold. I show it as multiplied and it should be divided. If you want (Ac^-)/(HAc) to be 0.01, then (H^+) must be what?
March 2, 2015

Chemistry
I would do this. ........HAc ==> H^+ + Ac^- I.......0.1......0.....0 C........-x......x.....x E.....0.1-x......x......x Plug the E line into the Ka expression and solve for x = (H^+). I obtained 1.34E-3M for (H^+) .........HAc ==> H^+ + Ac^- What you want to do now is to...
March 2, 2015

Chemistry
When you ask a question like this you should tell us what you used for Ka. Our tables may not give us the same value as your table(s). Therefore, we don't know if you've done it right or not. I used 6.2E-10 for Ka for HCN and using that value Kb for CN^- is 1.6E-5.
March 2, 2015

Chemistry
.........Ba(OH)2 ==> Ba^2+ + 2OH^- I.......3.5E-4........0.......0 C......-3.5E-4.....3.5E-4..2*3.5E-4 E........0.........3.5E-4...7.0E-4 So the (OH^-) from Ba(OH)2 = 7.0E-4. We want to know how much CN^- it will take to produce OH^- = 7.0E-4 M. .........CN^- + HOH ==> ...
March 2, 2015

CHEM
This is a limiting reagent (LR) problem and you know that because amounts are given for all of the reactants instead of just one. NH3 + CO2 + H2O ==> NH4HCO3 I do these the long way. 1. How much product can we get from 50.0 g NH3 if all of the others are in excess? mols NH3...
March 2, 2015

Chemistry
mass = volume x density density is given volume = 4 cm*4 cm*4 cm = ? cm^3.
March 2, 2015

Chemistry
You need to copy that format I gave you earlier. All of these stoichiometry problems are worked the same way. 2C4H10 + 13O2-->10H2O + 8CO2 Step 1. Write and balance the equation. You have that. Step 2. Convert what you have (in this case butane) into mols. mols = grams/...
March 2, 2015

Chemistry
You need to copy that format I gave you earlier. All of these stoichiometry problems are worked the same way. 2C4H10 + 13O2-->10H2O + 8CO2 Step 1. Write and balance the equation. You have that. Step 2. Convert what you have (in this case butane) into mols. mols = grams/...
March 2, 2015

Science
Be careful with the units. If you use the water as 100 g I would change the metal from kg to grams.
March 2, 2015

Science
heat lost by mtal + 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 the equation.
March 2, 2015

Science
heat lost by Cu + heat gained by Al + heat gained by water = 0 [mass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 I would convert all to grams and use specific heat in J/g*C
March 2, 2015

Science
See your other H2O/Cu/Al problem just above.
March 2, 2015

Chemistry
Models give us a visual idea of how to imaging the atoms/molecules/bonding, etc. The real understanding of atoms and how reactions take place is in advanced math, wave mechanics, etc. Many of us simply don't have the math background, especially in high school and lower ...
March 2, 2015

chemistry
This is no different than those problems we've been doing. You can convert any mole in an equation to any mole of something else in the same equation just by using the coefficients in the balanced equation CO2(g)+4H2(g)→CH4(g)+2H2O(g) Step 1. You have the balanced ...
March 2, 2015

Science for ms. Sue
There is no way this statement is true. It get's stronger as you get closer and if you get close enough the high voltage will jump a short distance through the air to you.
March 2, 2015

chemistry
A lot of work here. I don't want to just give you the answers but I'll be happy to help you through them. Exactly what do you not understand?
March 2, 2015

chemistry
g = mols x molar mass
March 2, 2015

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

Pages: <<Prev | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>

Members