Saturday
May 28, 2016

Posts by DrBob222

Total # Posts: 52,440

Chemistry
Answered above.
April 2, 2016

chemistry
Ca3(PO4)2 + what?
April 1, 2016

Chemistry
RbOH + HBr ==> RbBr + H2O mols HBr = M x L = ? mols RbOH = mols HBr M RbOH = mols HBr/L HBr = ?
April 1, 2016

Chemistry
ZnCO3 ==> ZnO + CO2 ZnO + 2HCl ==> ZnCl2 + H2O mols ZnCO3 = grams/molar mass = ? 1 mol ZnCO3 gives 1 mol ZnO 1 mol ZnO produces 1 mol ZnCl2. grams ZnCl2 = mols ZnCl2 x molar mass ZnCl3.
April 1, 2016

Chemistry
10.5 g Zn heated to constant mass may form ZnO but not ZnCO3.
April 1, 2016

Chemistry
M.O. changes color approx 4 (a little < 4.0) and H2C2O4 and NaoH don't change at that point.
April 1, 2016

Chemistry
Fe2O3 ==> 2Fe so you see there are 2 mols Fe in 2 mol Fe2O3. To get 1100 kg Fe will require 1100 kg Fe x (1 mol Fe2O3/2 mols Fe) = about 550 kg Fe2O3. That is only 84% Fe2O3 in the ore itself; therefore, 550/0.84 = ? kg of the ore.
March 31, 2016

Chemistry
Ksp = (Ba^2+)(F^-)^2 You know (F^-) = 1.5E-2 M You know (Ba^2+) = 1/2 the F^- Plug those into Ksp expression and solve for Ksp.
March 31, 2016

Chemistry
I assume you must have some connection to the two equations.
March 31, 2016

chemistry
(p1v1/t1) = (p2v2/t2) Remember T must be in kelvin
March 30, 2016

Chemistry
2 electrons for the two H atoms, +8 electrons for the O atom = 10 for one H2O molecules. That x 4 = 40 for the 4 H2O molecules. That means M^2+ must be 50-40= 10. If M^2+ is 10, the neutral metal M must be 12. What has an atomic number of 12?
March 30, 2016

Chemistry
mols Mg = grams/atomic mass = ? Using the coefficients in the balanced equation, convert mols Mg to mols H2. You can see it's a 1:1 ratio; therefore, mols Mg = mols H2 produced. Then convert mols H2 to grams. g H2 = mols H2 x molar mass H2 = ? This is the theoretical yield...
March 30, 2016

Chemistry
A slight twist on limiting reagent (LR). CaCl2 _+ Na2SO4 ==> CaSO4 + 2NaCl mols CaCl2 = grams/molar mass = ? mols Na2SO4 = grams/molar mass = ? Using the coefficients in the balanced equationk convert mols Cacl2 to mols CaSO4. Do the same and convert mols Na2SO4 to mols ...
March 30, 2016

Chemistry
Use PV = nRT and the conditions listed to determine the volume of F2. Then specific mass = density = molar mass/volume in L = ?
March 30, 2016

chemistry concentrations
0.171 ppm in initial 64 oz bottle. Of course the concn is the same no matter what kind of bottle you put it in. Remove 10.0 mL and dilute to 40 mL in another bottle. The concn in the other bottle is 0.171 x (10/40) = ? ppm.
March 30, 2016

Chemistry
HCl(aq) + MnO2(s) → 2 H2O(l) + MnCl2(s) + Cl2(g). mols MnO2 = grams/molar mass = ? mols Cl2 produced = mols Cl2 produced since the reaction of 1 mol MnO2 produces 1 mol Cl2. Then grams Cl2 = mols Cl2 x molar mass Cl2 = ?
March 30, 2016

Chemistry
You have 0.07 mols (NH4)2SO4. There are 6.02E23 molecules in 1 mol (NH4)2SO4 so there are 6.02E23 x 0.07 = ? molecules (NH4)2SO4 in that 0.07 mols. Since there are two NH4^+ in 1 molecule of (NH4)2SO4, there must be twice the number of NH4^+ ions as there are (NH4)2SO4 molecules.
March 30, 2016

chemistry
..........PbI2 ==> Pb^2+ + 2I^- I.........solid....0........0 C.........solid....x........2x E.........solid....x........2x ......CaI2 ==> Ca^2+ + 2I^- I...0.0123.....0........0 C..-0.0123....0.0123..2*0.0123 E.......0.....0.0123...0.0246 Ksp - (Pb^2+)(I^-)^2 (Pb^2+) = x...
March 30, 2016

Chemistry
mols = grams/molar mass = ?
March 30, 2016

chemistry
For many solids the polar nature of the solid with that of water makes them more or less soluble. Remember like dissolves like and even organic compounds with not much polarity contain OH groups and that is similar to H2O. For gases dissolved in water, there is no such ...
March 30, 2016

chemistry
CuSO4 is a polar molecule as is water. Many organic solvents are non-polar.
March 30, 2016

Chemistry
Alex, you got me. I don't know. It looks ok to me. USUALLY it is a matter of an incorrect number of significant figures but this looks ok to me. The problem asks for pH and you didn't finish the calculation. Perhaps that is the problem.
March 30, 2016

chemistry
I assume you mean acetic acid solution. acetic acid = HAc HAc ==> H^+ + Ac^- H2O ==> H^+ + OH^- Charge balance: (H^+) = (Ac^-) + (OH^-)
March 30, 2016

chemistry
It makes it difficult with no arrow. I have rewritten the equation and moved the dH so you can see it is an endothermic reaction. 2NOBr(g)+ heat =>2NO(g) + Br2(g) Remember Le Chatelier's Principle says that when a system at equilibrium is disturbed, it will shift so as ...
March 30, 2016

Chemistry
I don't think so. Are you guessing? Post your work and I'll find the error.
March 29, 2016

Chemistry
Post your work and let me check it.
March 29, 2016

Chemistry
Mg(OH)2 + 2H^+ ==> 2H2O + Mg^2+ mmols Mg(OH)2 = mg/molar mass = approx 250/58.3 = about 4 but that's only an estimate. mmols H^+ = twice that approx. Again, an estimate. CaCO3 + 2H^+ ==> 2H2O + Ca^2+ mmols CaCO3 = 300/100 = about 3 mmols H^+ = twice that. Add mmols H...
March 29, 2016

Chemistry
Sarah, I showed you in detail how to do the last one. Just follow that procedure. The only difference in the two problems is that the other one was a strong acid and the salt of a weak acid. This one is a strong base (CsOH) and the salt of a weak acid (H2S or technically HS^-).
March 29, 2016

Chemistry
I don't think so. Where did you get 43.6 mL? That isn't anywhere in the problem. mols CuSO4.5H2O = 0.096/249.7 = approx 3.8E-4 M = mols/L = 3.8E-4 mols/0.5L = approx 7.7E-4 M for the solution in the first 500 mL flask. Then you dilute 10.8 mL to 500 so 7.7E-4M x (10.8/...
March 29, 2016

Chemistry
millimols HCl = 0.4 x 50 = 20 millimols NaF = 85 x 9.4 = 34 .......F^- + H^+ ==> HF I.....34.....0.......0 add.........20............ C....-20...-20.......+20 E.....14....0........20 Ka for the rxn = (HF)/(H^+)(F^-) Doesn't that look like 1/Ka of HF to you. You could ...
March 29, 2016

oops--Chemistry
I hit the wrong key. The answer should be a little less than 3.
March 29, 2016

Chemistry
pH + pOH = pKw = 14. You know pH, solve for pOH. Then pOH = -log(OH^-) Substitute and solve for OH. You should get a little less than 2 but that's just an estimate.
March 29, 2016

Chemistry
2H2 + O2 ==> 2H2O Use the coefficients to calculate this. 27.4 mols H2 x (1 mol O2/2 mols H2) = 27.4 x 1/2 = ?
March 29, 2016

Chemistry
Do you have choices? Ga^3+ would be smaller but so would many other ions.
March 29, 2016

chemistry
NaC2H3O2 = NaAc acetic acid = HC2H3O2 = HAc NaAc diluted = 0.293 x (190/750) = approx 0.074 but you need to confirm this and all other calculations that follow. I estimate and round. .......Ac^- + HOH ==> HAc + OH^- I....0.074............0......0 C.......-x............x...
March 29, 2016

AP Chemistry
I answered this at your other post above (with anonymous screen name).
March 29, 2016

chemistry
This is a limiting reagent (LR) problem; you know that because amounts are given for both reactants. 3Mg(s) + N2(g) --> Mg3N2(s) The long long way to do this. mols Mg = 8 from the problem. mols N2 = 2 from the problem. Using the coefficients in the balanced equation, ...
March 29, 2016

Chemistry- Dr.Bob222
No point in my weighing in!
March 28, 2016

Chemistry
CaCO3(s)-> CaO(s) + CO2(g) Convert 71.42 g CaCO3 to mols. mols = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols CaCO3 to mols CaO. Now convert mols CaO to grams. g CaO = mols CaO x molar mass CaO. This is the theoretical yield (TY). The ...
March 28, 2016

Chemistry
No. Your error is you convert mols NaOH to grams NaOH then used GRAMS for mols in M = mols/L but you can't do that. You should convert 2 g to mols KHP, then to mols NaOH, then mols NaOH/M = L NaOH
March 28, 2016

chemistry
8.5E2 cal x (1 Cal/1000 cal) = ? 0.45 kJ = 450 J 450 J x (1 cal/4.184 J) = ?
March 28, 2016

Chemistry
Ben, I did this far below. http://www.jiskha.com/display.cgi?id=1459199032
March 28, 2016

Chemistry
You explanation still doesn't make sense to me. reducing oxide? number of this oxidation reaction?
March 28, 2016

@ Korina--Chemistry
You need to make it clear as to what you want. Also it helps if you explain exactly what you don't understand.
March 28, 2016

Chemistry
see http://www.jiskha.com/display.cgi?id=1459199032
March 28, 2016

Chemistry
Ptotal = pH2O + pgas Ptotal = 100.2 kpa pH2O--look up vapor pressure H2O at 30 C (in kPa or convert to kPa or convert 100.2 kPa to mm Hg and use pH2O in mm Hg) Substitute and solve for pgas.
March 28, 2016

Chemistry
Look up Ksp for Mg(OH)2. .........Mg(OH)2 --> Mg^2+ + 2OH^- I........solid.......0........0 C........solid.......x........2x E........solid.......x........2x Substitute the E line into the Ksp expression for Mg(OH)2 and solve for x = solubility in mols/L. Convert to g. g = ...
March 28, 2016

Chemistry
Buffered at what pH. pH low the solubility is increased significantly. pH high, solubility is decreased significantly.
March 28, 2016

Chemistry
I have no idea what you're talking about. Valley?
March 28, 2016

Chmistry
What's your problem in doing this. The problem tells you the products they want. a) 2C2H2 + O2 ==> 4C + 2H2O b) 2C2H2 + 5O2 ==> 4CO2 + 2H2O c and d are done the same way.
March 28, 2016

Chemistry
First determine the M of the concentrated (the 37 %) HCl solution. That is 1000 x 1.19 x 0.37 x (1/36.45) = approx 12 M but you need to do it more exactly. Then use mL1 x M1 = mL1 x M2 mL1 x 12M = 1000 mL x 2.5M Solve for mL1 which is mL of the 37% stuff.
March 28, 2016

Chemistry
2Mg + O2 ==> 2MgO mols Mg = grams/atomic mass = 12.01/24.3 = ? mols O2 = grams/molar mass = 6.56/32 = ? Using the coefficients in the balanced equation, convert mols Mg to mols MgO produced. That's #1. #2. Convert mols O2 to mols MgO. #3. In limiting reagent problems ...
March 28, 2016

Chemistry
How many moles do you need? That's mols = M x L = ? Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
March 28, 2016

Chemistry
I'm not familiar with the experiment you are doing but I would go with the second equation with a +. Again, I'm not sure of the - sign in front of the second equation since I don't know how it's being measured.
March 28, 2016

Chemistry
(0.2*10) + (0.8*11) = mass B. To be accurate you should have the masses of 10B and 11B given and plug those values in instead of 10 and 11.
March 28, 2016

chem
Use molar mass and H bonding to determine. The higher molar mass (more van der Walls forces) the higher the b.p. The more H bonding (polarity) the higher the b.p. Butane, for example, would be the lowest b.p. You can look these up on google and see if you have them right.
March 28, 2016

chemistry
acetic acid is HC2H3O2 molar mass 60 Oxalic acid is H2C2O4 molar mass 90. Oxalic acid is both more London forces as well as more H bonding.
March 28, 2016

Chemistry
N = #H or OH*M or M = N/#H or OH For H3PO4. 6N/3 = 2M For Ca(OH)2. 4N/2 = 2 etc. M =
March 28, 2016

Chemistry
(p1v1/t1) = (p2v2/t2) Remember T must be in kelvin. Remember to use the same units for p.
March 28, 2016

Chemistry
This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants. 2Fe + 3O2 ==> Fe2O3 mols Fe = grams/atomic mass = ? mols O2 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols Fe to mols Fe2O3. Do the ...
March 28, 2016

Chemistry
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants. Your first step is ok. Using the coefficients in the balanced equation convert mols CaCO3 to mols CO2. Do the same and convert mols HI to mols CO2. It is likely that these two ...
March 28, 2016

Equilibrium Consentration
H2 + Cl2 ==> 2HCl Kc = (HCl)^2/(H2)(Cl2) You know Kc, (H2) and (Cl2). The only unknown is (HCl). Substitute and solve for that.
March 28, 2016

Chemistry
CH3NH3Br ==> CH3NH3^+ + Br^- Then the CH3NH3^+ is hydrolyzed ..CH3NH3^+ + H2O ==> CH3NH2 + H3O^+ I..0.01...............0........0 C...-x................x........x E.0.01-x..............x........x Ka for CH3NH3^+ = (Kw/Kb for CH3NH2) = (x)(x)/(0.01-x). Solve for x = (...
March 28, 2016

chemistry12_2016-4
See your Mg/HCl post below. Follow the directions. Post your work if you get stuck.
March 27, 2016

chemistry12_2016-3
This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants. The following steps will work almost any LR problem. 1. Write and balance the equation. Mg + 2HCl ==> MgCl2 + H2 2. Convert reactants to mols a. mols HCl = M x L = 0.5 x 0....
March 27, 2016

chemistry12_2016-2
M = mols/dm^3 = mols/0.1 mols = grams/molar mass = 14.0/approx 56 = ? Substitute into M = mols/L and solve for M.
March 27, 2016

Chemistry
A poorly worded question. What you want to know is how much H2SO4 was used, not how much xs was used. Technically, none of the excess was used. Did your problem provide an equation. Check and see that it looks like this. Ca5(PO4)3F + 5H2SO4 ==> 5CaSO4 + 3H3PO4 + HF If my ...
March 27, 2016

chemistry12_2016-1
RTP is Room Temperature & Pressure. Room T is considered 298 K.
March 27, 2016

chemistry12_2016-1
Another limiting reagent problem Follow the directions in your posts above. Post your work if you get stuck and I can help you through it.
March 27, 2016

Chemistry
Thank you for the equation. See your post above and change the equation I proposed to the proper one. Post your work there if you run into trouble and I can help you through it. For this problem, mols Ca5(PO4)3F = grams/molar mass = 5/504.3 = approx 0.006 but you need a more ...
March 27, 2016

Chemistry
You want me to write a term paper for you?... No thanks.
March 27, 2016

Chemistry
Look up Graham's Law. The rate is proportional to the square root of the molar mass.
March 27, 2016

Chemistry
ptotal = pH2 + pH2O ptotal = 757 pH2O = 17.5 pH2 = ? Substitute and solve.
March 27, 2016

Chemistry
I don't understand anything you wrote. Are you working this problem or another one? Plug values into that equation I gave you and solve. What you have posted is gibberish.
March 27, 2016

Chemistry
(p1v1/t1) = (p2v2/t2) Remember T must be in kelvin.
March 27, 2016

chemistry
Use the H-H equation. Technically one should use concentrations but I like to work in millimols = mL x M. And since both numerator and denominator for M = millimols/mL and mL is the same (because it's the same solution) then using millimoles gives the same answer. mmols ...
March 27, 2016

chemistry
Do you have any other information? You need, at least, the heat of neutralization for H^+ + OH^- ==> H2O. I don't remember but think it's about 55 kJ/mol. HCl + NaOH ==> NaCl + H2O You have mols HCl = M x L = 0.1 x 0.1 = 0.01 mols. So approx 55 kJ/mol x 0.01 mol...
March 27, 2016

chem
Use PV = nRT and solve for n = number of mols air in the lungs. Then n x 0.21 = mols oxygen.
March 27, 2016

Chemistry
I should point out that m and M are not the same. I assume that is a typo. It isn't easy to convert m to M with knowing the density.
March 27, 2016

CHEM QUESTION!
A very easy way to do these ppm problem in water solution is to remember that 1 ppm = 1 mg/L or 1 ug/mL. So 0.6 ug/mL = 0.6 ppm.
March 26, 2016

Chemistry
I think the correct solution is to recognize this as a buffer problem use the Hendersib-Hasselbalch equation. pH = pK2 + log (base)/(acid) pH = 1.88 + log (0.574/0.2) = ? pH approx 2.3 OR you can work it as a common ion; i.e., HSO4^- ==> H^+ + SO4^- k2 = ? K2SO4 ==> 2K...
March 26, 2016

Chemistry
The first half equation is not balanced. Cl on the left is 7+ and on the right is 1- so delta e is 8e on the left. You will notice the charges don't balance in your half equation. 1+ on the left and 1- on the right.
March 26, 2016

Chemistry
Em, I answered this only 10 minutes after you posted it first. Give us a break.
March 26, 2016

Chemistry
.....H2 + Cl2 ==> 2HCl E...0.26..0.087....x Kc = (HCl)^2/(H2)(Cl2) You know Kc, (H2), and (Cl2), substitute and solve for HCl.
March 26, 2016

~ CHEMISTRY ~
The first two lines ask a question. If P goes up, V goes down. I don't know what you want for the rest of the post.
March 25, 2016

Chemistry
2H2 + O2 ==> 2H2O 1.249 g + 9.99 g = ? The law of conservation of mass says that you get out what you put in.
March 25, 2016

AP Chemistry
I think I answered this for you yesterday. Use the Henderson-Hasselbalch equation.
March 25, 2016

chemistry
Here is the way to do these. The reaction is this. ....M^2+ + 4CN^- ==> [M(CN)4]^2- I..0.170...1.04.......0 C.-0.170..-4*0.170 ...0.170 E...0.....0.36.......0.170 You can see that with a K of a bouat 10^16 that the reaction is essentially complete to the right. Now you ...
March 25, 2016

chemistry
I looked up the density of a 37% solution at 20 degrees C = 1.18 Then 1000 x 1.18 x 0.37 x (1/36.45) = 12.3 M for the HCl. Then mL1 x M1 = Ml2 x M2 mL1 x 12.3 = 1000 mL x 0.2 assuming you want to prepare 1000 mL of the solution. So you take mL1 you calculate, add to a 1000 mL ...
March 25, 2016

Chemistry
mL1 x M1 = mL2 x M2 mL1 x 12 = 2000 mL x 3.0 M Solve for mL1
March 25, 2016

Chemistry
There is 1 mol N in 1 mol HNO3. 1 mol N = 14 g IF you mean nitrogen gas as in N2, then there are 7 g N2 since there is 1/2 mol N2 in 1 mol N.
March 25, 2016

chemistry
To clarify, do you want 0.2m or 0.2M?
March 25, 2016

Chemistry
Using the 6.0 M stock. mL1 x M1 = mL2 x M2 mL1 x 6.0 = 250 mL x 0.01 Solve for mL 1, add that to a 250 mL volumetric flask and make to the mark with DI water. 2.5 M stock is done the same way.
March 25, 2016

chemestry
Use the coefficients to do this. 0.3 mol N2 x (2 mols NH3/1 mol N2) = 0.3 x 2/1 = ?
March 24, 2016

chemestry
See your other post and note the correct spelling of chemistry.
March 24, 2016

Chemistry
2 mols NH3 formed. 1 mol N2 and 3 mols H2 are represented. That is 6.02E23 molecules of N2 and 3*6.02E23 molecules of H2.
March 24, 2016

Chemistry
I don't see a question here.
March 24, 2016

Chemistry
I don't see a question here.
March 24, 2016

chemistry
........NiCO3 ==> Ni^2+ + CO3^2- I.......solid.....0........0 C.......solid.....x........x E.......solid.....x........x Substitute the E line into the Ksp expression and solve for x = (NiCO3) in mols/L. Then grams = mols x molar mass = ? Compare the solubility with the 5 mg...
March 24, 2016

chemistry
.......CaCrO4 ==> Ca^2+ + CrO4^2- I......solid.......0.......0 C......solid.......x.......x E......solid.......x.......x ........Ca(NO3)2 ==> Ca^2+ + NO3^2- I........0.25M........0.......0 C.......-0.25.......0.25......0.25 E..........0........0.25......0.25 Note that ...
March 24, 2016

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