Thursday
August 25, 2016

Posts by DrBob222

Total # Posts: 52,855

chemistry
I wonder why phosphoric acid is not as good as hydrophosphoric acid. Also, note that Zn is Zn(II) phosphate etc. 3Zn + 2H3PO4 ==> Zn3(PO4)2 + 3H2 mols Zn = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols Zn to mols Zn3(PO4)2. Then g Zn3(...
April 12, 2016

Chemistry
6 L CO2. Knowing that 1 mol of any gas occupies 22.4 L at STP, then 6/22.4 = ? mols CO2. Then using the coefficients in the balanced equation, that many mols CO2 x 2 = mols HCl used Then M HCl = mols HCl/L HCl = ?
April 12, 2016

Chemistry
Is that an aqueous solution? 1337.6 kJ/0.0125 = ? kJ = heat released by 1 mol MgO I didn't check the mol calculation for MgO.
April 12, 2016

Chemistry
Do you mean 0.052 molar? pH = -log(HCl) = -log(0.052M) = ?
April 12, 2016

Chemistry
http://www.jiskha.com/display.cgi?id=1460431080
April 12, 2016

Chemistry
This is the H2A problem and I worked that for you re the link for the post just above.
April 12, 2016

Chemistry
You have no concns listed so you can't get Keq that way. Use dGo = -nEoF You will need to look up the value of Eo for the reaction. Obviously the one with Eo a positive number will give you a negative dGo and that is the one that is spontaneous. If you want Keq, you can ...
April 12, 2016

Chemistry
Is D density? Zn + 2HCl ==> ZnCl2 + H2 Calculate mols Zn and convert to mols H2. Then use PV = nRT P will be 1.234 atm - vp H2O (in atm)[or convert 1.234 atm to mm Hg and use 21.0 mm for vp H2O) Everything else is as usual and V will be ihn liters.
April 12, 2016

Chemistry
(NaOH) = 1 M NaOH x (100 mL/1,500) = ? The N is the same as the M (in this case).
April 12, 2016

a note for olereb48---chemistry
Using the quadratic (H^+) from ka1 is 0.023 and not 0.027 which makes H3PO4 then 0.10-0.023 = 0.077 M Yes, (H^+) from ka2 is = ka2 and that is insignificant. Yes and No for part ka3. (H^+) is that from part 1; i.e., 0.023 and not = k3. Likewise, (HPO4^2-) is known from ka2 ...
April 12, 2016

chemistry
H3PO4 ==> H^+ + H2PO4^- H2PO4^- ==> H^+ + HPO4^2- HPO4^2- ==> H^+ + PO4^3- Write the expressions for ka1, ka2, and ka3. You solve k1 as if it were a monoprotic acid; ie. ......H3PO4 ==> H^+ + H2PO4^- I......0.1......0......0 C......-x.......x......x E.....0.1-x...
April 12, 2016

Chemistry-Dr.Bob222
H2A + 2NaOH ==> Na2A + 2H2O mols NaOH = M x L = ? mols H2A = 1/2 x mols NaOH (look at the coefficients in the balanced equation). Then mols H2A = grams H2A/molar mass H2A. You know mols and grams, solve for molar mass.
April 11, 2016

Chemistry-Dr.Bob222
Refer to the H2A problem above BUT you only titrate the first H on H2SO4. Post your work if you get stuck.
April 11, 2016

Chemistry
PV = nRT
April 11, 2016

Chemistry
yes. Good work.
April 11, 2016

Chemistry- Dr.Bob222
I would not throw away the places after 5 that are eligible to be retained. I came up with 5.04 to three significant figures.
April 11, 2016

Chemistry
(p1v1/t1) = (p2v2/t2)
April 11, 2016

Chemistry
Robin,you've done a terrific job but you need a couple of changes. Is F^- a weak base; if so it would not attract H^+ to make HF. Then HCl is a strong acid that IONIZES COMPLETELY which makes Cl^- a weak conjugate base of HCl. Since HF is a weak base the added H^+ makes HF...
April 11, 2016

Chemistry
1250.0 kJ/mol x # mol = 352.40 kJ Solve for # mols. Then mol = grams/molar mass. You know mols and molar mass, solve for grams.
April 11, 2016

Chemistry
SO3 + H2O --> H2SO4 mols SO3 = 4.88E24/6.02E23 = approx 8 but you need more than an estimate. Convert mols SO3 to mols H2SO4 using the coefficients in the balanced equation. Since 1 mol SO3 produces 1 mol H2SO4, then mols SO3 = mols H2SO4. Convert mols H2SO4 to grams. g ...
April 10, 2016

chemistry redox reactions
I don't understand the post. For example is R(e-) and R^- the same?
April 10, 2016

Chemistry
Before anyone can answer this for you we need to know the procedure you used in detail.
April 10, 2016

Chemistry
How many mols CH3COOH (HAc) do you have? That's M x L = 1.2 x 1.5 = 1.8 .........HAc + OH^- ==> Ac^- + H2O I........1.8...0........0.......0 add............x.................. C.........-x..-x........+x...... E.......1.8-x...0........x Now substitute the E line into the...
April 10, 2016

Chemistry
Since it's only 92% UO2 then you have only 10 kg x 0.92 = ?kg UO2. Convert that many kg UO2 to ? kg U. That's ?kg UO2 x (atomic mass U/molar mass UO2) = ? kg of U.
April 9, 2016

Chemistry
2SO2 + O2 ==> 2SO3 mols SO3 needed = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols SO3 to mols O2. Now convert mols O2 to grams. g O2 = mols O2 x molar mass O2.
April 9, 2016

Analytical chemistry
Fe2O3 ==> 2Fe^3+ and reduce to Fe^2+. 6Fe^2+ + Cr2O7^2- ==> 6Fe^3+ + 2Cr^3+ I balanced only the redox portion of the equation which is sufficient for what we want to do. mols Cr2O7^2- = M x L = ? mols Fe = 6 x that (look at the coefficients in the equation above). g Fe...
April 9, 2016

Chemistry
You were right on the (H^+) = 10^-1.57 = 0.027 mols/L.But wrong on the rest of it. You want 0.027 mols/L so you will need 1/4 that for mols in 250 mL or 0.027 x (250 mL/1000 mL) = 0.0069 Then mols = grams/molar mass. You know molar mass and mols, solve for grams. I estimated ...
April 9, 2016

Science Please help ASAP
No, cold ocean water want do it. If that were true we would have hurricanes in the winter.
April 8, 2016

Chemistry
p1v1 = p2v2
April 8, 2016

Chemistry
Keq = Kf*Ksp = [Ag(NH3)2]*(Cl^-)/(NH3)^2 From my viewpoint, the 2 for NH3 does play a role.
April 8, 2016

Chemistry
(V1/T1) = (V2/T2) Remember to convert T1 and T2 to kelvin.
April 8, 2016

Chemistry
Would you believe 200.0 mL? Isn't 400 k and 400 K the same? Technically it should be written as 400 K.
April 8, 2016

Chemistry
A fairly long solution. Start with two equations. The first is to find the ratio of base to acid. pH = pK3 + log (base)/(acid) Substitute and solve for b/a. Equation 2 is b + a = 0.01. Solve those two equations simultaneously for a (acid) and b (base). Since those are M concns...
April 7, 2016

Chemistry please help!
I thought Bob Pursley showed you how to do this. Add equation to the reverse of equation 2. That gives you kJ/mol graphite to diamond. If you want 1 kg, then kJ/mol from the above x (1000 g/12.01) = ? heat absorbed to make 1 kg. For part b. ?kg from above = mass H2O x specific...
April 7, 2016

Chemistry
The first number is the atomic number(it's a subscript); the second is the mass number (it's a superscript). 84Po210 = 2He4 + 82X206 Note the subscripts add up on each side; i.e., 94 = 82 + 2 and the superscripts add up. That is 210 = 26 + 4. So X must be Pb.
April 7, 2016

Chemistry
Jessie, Mark, Chris or whomever. Just follow the solution for Mark.
April 7, 2016

chemistry
See your posts above.
April 7, 2016

Chemisty
I wonder how "common lab equipment" is defined. I wonder how accurately this is to be done. Does that mean we can't use any other chemical? I would prepare a 0.1M baking soda solution (NaHCO3) and fill a buret. Add phenolphthalein indicator, then add NaHCO3 from ...
April 7, 2016

Chemistry
Refer to your other posts. Almost all of these can be worked with two thoughts. #1. k = 0.693/t1/2 #2. ln(No/N) = kt where No = initial and N = final, k is from the #1 calculation and t1/2 is the half life.
April 7, 2016

Chemistry
Cool 212 water to what. zero? 10? 99. 211? Pick a number and you'll be right.
April 7, 2016

Chemistry
(g solute/g solution)*100 = % mass/mass
April 7, 2016

Chemistry
% w/w of 20.1% means 20.1 g NaBr in 100 g solution. So g NaBr = 20.1 g solution = 100 g = g NaBr + g H2O or g H2O = 100-20.1 = 79.9 n NaBr = grams/molar mass = ? n H2O = grams/molar mass = ? Total mols = n NaBr + n H2O XNaBr = nNaBr/total mols.
April 7, 2016

chemistry
124 kcal x (11.2 g HF/molar mass H2) = ?
April 7, 2016

Chemistry
2Na + 2H2O -- 2NaOH + H2 mols Na = grams Na/atomic mass Na. You know atomic mass Na and mols Na, solve for grams Na. The 50 cc NaOH has nothing to do with the problem assuming that the water used was in excess.
April 7, 2016

Chemistry
http://www.jiskha.com/display.cgi?id=1459868645
April 6, 2016

Chemistry
Diagram of the solution? We can't do that on this forum.
April 6, 2016

Chemistry
Will you please explain, in detail, exactly what you don't understand about each step?
April 6, 2016

Chemistry
K2O + HOH ==> 2KOH mols KOH needed = grams/molar mass = ? mols K2O = (1/2)*mols KOH from the coefficients in the balanced equation. grams K2O needed = mols K2O x molar mass K2O = ?
April 5, 2016

Chemistry
See your post on Si.
April 5, 2016

Chemistry
657 kJ x (130/atomic mass Si) = kJ released.
April 5, 2016

chemistry 107
You must mean precipitate and not participate. BaSO4 is the ppt that forms.
April 5, 2016

Chemistry
The molecular formula is not 44*0.5. If 0.5 mol weighs 44 g then a whole mole must weigh 88 g. An acid has a -COOH group which is 45 so the rest of the molecule must be 88-45 = 43. A -CH3 group is 15 so 43-15 = 28 and CH2 groups are 14 each so the molecular formula must be ...
April 5, 2016

Chemistry
And the 1.3? What about it?
April 5, 2016

Chemistry
Your addendum makes no sense. That's k = 1.3 for what. Is is Mg(OH)2? If so 1.3 isn't right. You've made a typo surely.
April 5, 2016

Chem
That's the specific heat for H2O. q = m*c*delta T.
April 5, 2016

chemistry
mL x density x (%/100) = 13.37 Solve for mL of the solution.
April 5, 2016

Chemistry
This forum can't handle drawings.
April 5, 2016

Chemistry
Why don't you look these up on Google. Better for you to do and for me to do it.
April 5, 2016

Chemistry
See your other post above.
April 5, 2016

Chemistry
What reaction?
April 5, 2016

CHEM
What's wrong with what I gave you last night? The directions are there; I made 1 L of solution so adjust to 100 mL for this and you have it. I'll be glad to help you through it but I don't want to do the work for you.
April 5, 2016

Chemistry
2HCl + Ca(OH)2 ==> 2H2O + CaCl2 mols HCl = M x L = ? mols Ca(ON)2 = 2 x mols HCl (look at the coefficients in the balanced equation). Then M Ca(OH)2 = mols Ca(OH)2/L Ca(OH)2. You know M and mols, solve for L.Convert to mL if needed.
April 5, 2016

CHEM
Usually you know the pH but you don't show that. You know pKa for NH3 to be about 9.26 but use the number in your text/notes. If you want the buffer to be between 0.01 and 0.5, why not choose say 0.3. If base + acid = 0.3, then base = 0.1 and acid must be 0.2 so pH = pKa...
April 5, 2016

Chem
(p1v1/t1) = (p2v2/t2) Remember t must be in kelvin.
April 5, 2016

chemistry
Calculate X for A(which I'm calling a) and X for B (which I'm calling b). Poa and Pob are Poa and Pob respectively. pa + pb = 40 mm pa = Xa*Poa Knowing pa you calculate pb Then pb = Xb*Pob You know pb and Xb, solve for pob. Post your work if you get stuck and I can ...
April 5, 2016

Chemistry
This is a little hard to show in writing but here goes. ......M^2+ + 4CN^- ==> [M(CN)4]^2- I..0.150....0.870.......0 C.-0.150...-0.600....+0.150 E.....0.....0.270.....0.150 The first equilibrium shows the formation of the complex on the right. With such a large Kf of over ...
April 5, 2016

Chemistry
Pb(OH)2 ==> Pb^2+ + 2OH^- Al(OH)3 ==> Al^3+ + 3OH^- Calculate OH^- needed for a 0.21M Pb^2+ solution to ppt Pb(OH)2, Calculate OH^- needed for a 0.41M Al^3+ solution to ppt Al(OH)3. You will want the (OH^-) to be less than you calculate to prevent pptn of Pb(OH)2. ...
April 5, 2016

Chemistry
........PbSO4 ==> Pb^2+ + SO4^2- ........SrSO4 ==> Sr^2+ + SO4^2- You know Ksp for each and you know metal is 0.039. I like to solve for OH^- needed to ppt each. Using my values for Ksp (which won't be the same as yours probably since texts differ), SO4^2- for Pb = 4...
April 5, 2016

Chemistry
http://chemistry.elmhurst.edu/vchembook/568denaturation.html
April 5, 2016

Chem
100.0 g CO2 x (molar mass O2/molar mass CO2) = ? g O2 in 100.0 g CO2.
April 5, 2016

Chemistry
Add eqn 1 to eqn 3 to the reverse of eqn 2. Add the dH values; when you reverse the rxn change the sign of dH for that rxn. Post your work if you get stuck.
April 5, 2016

Chemistry
Probably several but you can't beat this one for simplicity. Any other method that will test the acidity/basicity will work.
April 5, 2016

Chemistry
Add water, test with litmus paper.
April 5, 2016

Chemistry
Kw = 1E-14 = (H^+)(OH^-) You know Kw and (H^+). Solve for (OH^-).
April 4, 2016

Chemistry
q = 1387.4 kJ x (12.5 g/26 g) = ? kJ liberated
April 4, 2016

Chemistry
I'm not sure this is the question you intended to ask but here is the answer. mols AgNO3 = grams AgNO3/molar mass AgNO3.
April 4, 2016

Chemistry-Dr.Bob222
When you post a problem like this please include your work and/or what you used for constants; otherwise we can't get the same answer you obtained. I obtained an answer of 0.005 M using k2 of 5.6E-11 so your answer probably is correct.
April 4, 2016

Chemistry
Your post makes no sense to me as written. I understand what is being asked but I don't understand how the solution was made.
April 4, 2016

Chemistry
Do you have an equation to use? I don't see how you can get HNO3 out of N2 and O2.
April 4, 2016

Chem
The question tells you how to do it. SnO(s) +2HF(aq) ==> SnF2 + H2O
April 4, 2016

Chemistry
mols Ag2SO4 = grams/molar mass - ? and that is the M since it is in 1 L. I estimate that to be 0.015 M ........Ag2SO4 <--> 2Ag^+ + SO4^2 I.......solid........0.......0 C.......solid........2x......x E.......solid........2x......x Substitute the E line into the Ksp ...
April 4, 2016

Chemistry- Dr.Bob222
I'm not sure how to answer this question. Technically there is no such compound as NH4OH and I hope your prof is teaching you that. Based on that being zero, then NH3 > NH4^+ = OH^- > NH4OH = 0 What this should tell you is that the the NH3 gas dissolves in the ...
April 3, 2016

Chemistry
See your previous post. However, aren't these definitions in your text/notes?
April 3, 2016

Chemistry
BF3. Can you draw the lewis dot structure for this. Look on the web and google Lewis dot structure BF3.It's close to this. F:B:F :F: wotj extra dots around the F atoms. Note the BF3 does not obey the rule of eight; i.e., it only six electrons around B in BF3. Therefore, ...
April 3, 2016

Chemistry
I think you have made a typo. That surely is a Ka of 3.5E-8. ........HClO ==> H^+ + ClO^- I......0.45M.....0......0 C......-x........x......x E.....0.45-x.....x......x Substitute the E line into the Ka expression and solve for x = (H^+) = (ClO^-) Convert H^+ to pH
April 3, 2016

chemistry
What's 1 com? What were your observations? All you gotta do is look, unless, of course, you're doing a dry lab.
April 3, 2016

Chemistry - Dr.Bob 222
Can't you calculate that> pH = -log(H^+) 0 = -log(H^+) (H^+)= 10^-pH = 1.0M Now you do 14.
April 3, 2016

chem
(COOH)2.2H2O + 2NaOH ==> (COONa)2 + 4H2O mols NaOH = M x L = ? mols (COOH)2 = 2 x mols NaOH from the coefficients in the balanced equation. grams (COOH)2.2H2O = mols x molar mass. This is the theoretical yield (TY) %(COOH)2.2H2O in the sample is (TY/mass sample)*100 = ?
April 3, 2016

Chemistry
1. Balance the equation. That's done. 2. Separate into ions with the following rules. a. If weak acid or weak base or weak electrolyte (such as H2O), keep as molecule. b. If solid, keep as molecule. c. If gas, keep as molecule d. All others show as ions. This step gives ...
April 3, 2016

Chemistry
Yes, d is the correct answer.
April 3, 2016

Chemistry
C'mon, delta T = i*Kb*M i - 2 ]Kb = 1.52 M = 1 M Then delta T = 2*1.52*1 = ? dT = 3.04 for a. The others are done the same way. Go back and look at my response to your question on the 0.08M monoprotic acid. If you will explain what you don't understand perhaps I can ...
April 3, 2016

Chemistry
You can calculate each if your wish or take a short cut. To calculate, delta T = i*Kb*m i is the van't Hoff factor and is 2 for all cases in this question. You know Kb is 1.52. The shortcut is just i*M since Kb is the same for all of them. The higher i*m the higher delta T...
April 3, 2016

Chemistry
Since solubility is proportional to the pressure, wouldn't that be just half the initial P.
April 3, 2016

chemistry
2C8H18 + 25O2 ==> 16CO2 + 18H2O mols C8H18 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols C8H18 to mols CO2. Now convert mols CO2 to g. grams CO2 = mols CO2 x molar mass CO2.
April 3, 2016

Chemistry
1000 mL x 1.02 g/mL x 0.0013 = approx 1.4 grams Ca(OH)2 but that's just a close estimate. Then mols = grams/molar mass Ca(OH)2 and since that is mols/L that is the M and = approx 0.02. Again, that's an estimate. .......Ca(OH)2 ==> Ca^2+ + 2OH^- I......0.02.........0...
April 3, 2016

Chemistry
Look it up in the tables, dHo formation C2H2.
April 3, 2016

Chemistry
That doesn't help me. What are you confused about. In detail. Show your work as far as you can get and explain what you don't understand about the next step.
April 3, 2016

Chemistry
Weak monoprotic acid is HA ..........HA --> H^+ + A^- I.......0.08.....0.....0 C........-x......x.....x E......0.08-x....x.....x Substitute the E line into the Ka expression. The problem tells you that x = 1E-4. You may need to solve the quadratic; i.e., you may not be able...
April 3, 2016

Chemistry
Yes grams are and no grams are not. You calculated grams in part 1 by using PV = nRT, then since n = grams/molar mass, you substituted molar mass of 4 for He and calculated grams. For part 2, you used PV = nRT again and if you wanted to use grams, then mols = n = grams/molar ...
April 3, 2016

Chemistry
n = 1.25 and 5.025 g He is correct for #1. #2. PV = nRT P is 0.565 atm T is 6.0. Convert to kelvin first. n from above is 1.25 R you know Solve for V and compare it with the manufaturer's value of not bursting up to 47.0 L
April 3, 2016

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