Saturday
May 28, 2016

Posts by DrBob222

Total # Posts: 52,440

oops-error--chemistry
equation 1 I wrote for you is not right. That should be mols HCl used for Mg(OH)2 + mols HCl used for Al(OH)3 = total mols HCl = M x L = ? Equation 1 should read, (X/2*mm Mg(OH)2) + (Y/3*mm Al(OH)3) = M HCl x L HCl Sorry about that.
May 4, 2016

chemistry
This is a two unknowns/two equation problem. The equations are solved simultaneously. Difficult to explain on a screen but I can show it in parts and leave most of the work to you. I should note that there is a piece of information missing; i.e., you don't have the mass of...
May 4, 2016

chemistry
You have not posted the entire problem.
May 4, 2016

chemistry
This is done the same way as the M2S3/MO2 problem. See that solution above.
May 4, 2016

Chemistry
That should work. Post your work and we can look for the error. Probably you just punched in a wrong number or hit the wrong key on your calculator.
May 4, 2016

chemistry
Where does Xmas come from? Look at your MO2/M2S3 problem above.
May 4, 2016

chemistry
Again, look at your MO2/M2S3 problem.
May 4, 2016

chemistry
This is another set of simultaneously solved equations but it is more complicated. You must separate, somehow, the Fe used to form FeO from the Fe used to form Fe2O3 and/or separate the oxygen used for form FeO and oxygen used to form Fe2O. Then solve stoichiometry problem to ...
May 4, 2016

chemistry
This is almost the same as the MO2/M2S3 problem but I would convert 12.13160 g Ag to the mass of AgBr. Post your work if you get stuck.
May 4, 2016

chemistry
Use PV = nRT
May 4, 2016

Chemistry
I will assume that 0.200m is a typo and you mean 0.200 M. mols NaOH = grams/molar mass = ? Then M NaOH = mols NaOH/L NaOH. YOu know mols and M, solvle for L and convert to mL.
May 4, 2016

chemistry
pH = -log(HCl)
May 4, 2016

chemistry
I assume that 250 mL is 250 mL of solution. mols urea = grams/molar mass = ? Then M = mols/L
May 4, 2016

chemistry
Is there a question here? Is there any absorbance data here? I don't see either.
May 3, 2016

Chemistry
6.0 x 455/2500 =
May 3, 2016

Chemistry
How may what of methane? liters? grams? tons? What is the temperature of the water from which the steam comes?
May 3, 2016

oops--typo--Chemistry
What is the molarity of the HCl? That's 1.2 g/cc x 1000cc x 0.65 x (1/36.5) = ? That line should read 1.2 g/cc x 1000 cc x 0.365 x (1/36.5) = ? It's the .65 that is not right. It should be 0.365 and that comes from the 36.5%
May 3, 2016

Chemistry
What is the molarity of the HCl? That's 1.2 g/cc x 1000cc x 0.65 x (1/36.5) = ? Then mols HCl = M x L = ? mols NaOH = mols HCl Then M NaOH = mols NaOH/L NaOH. You know mols NaOH and M NaOH, solve for L NaOH. Convert to mL if needed.
May 3, 2016

Chemistry
3Fe + 2O2 ==> Fe3O4 mols Fe = grams/atomic mass Fe = ? Using the coefficients in the balanced equation, convert mols Fe to mols Fe3O4. Now convert mols Fe3O4 to grams. g = mols x molar mass = ?
May 3, 2016

Chemistry
right
May 3, 2016

Chemistry
Ca(OH)2 ==> Ca^2+ + 2OH^- dGorxn = (dGo products)-(dGo reactants) Then dGo = -RTlnKsp
May 3, 2016

Chemistry
Look up (or calculate) the heat of combustion for methane. I'll call that Z. Then Z = mass H2O x specific heat H2O x (Tfinal-Tinitial) Keep mass and specific heat units consistent.
May 3, 2016

Chemistry
See your other post.
May 3, 2016

Chemistry
I don't either. With no grams listed, there is no way to know how much water is required to form a saturated solution.
May 2, 2016

Chemistry
You didn't list any quantities of pure water. From the problem 1 L would be a saturated solution but 1 L may not be listed as an answer and it may not be the smallest quantity.
May 2, 2016

Chemistry
A. You're right; Ba^2+ is reduced because it gains electrons. That makes it an oxidizing agent. B. So Co^2+ + 2e = Co and that's the same as A. C. Se. You have the choice of Se --> Se^4+ + 4e with a loss of electrons and that is oxidation or reducing agent. Or. Se...
May 2, 2016

Science
Science?
May 2, 2016

Chemistry
......2H2 + 2NO -> N2 + 2H2O I.....1.2..0.95....0......0 C.....-2x...-2x....x.....2x 5sec..1.02.0.95-2x..x....2x So 1.20-2x = 1.02 Solve for x and evaluate the other values you want to calculate.
May 2, 2016

chemistry
You had 25 g solution and obtained 4g anhydrous salt; therefore, you must have had 21 g H2O. So the saturated solution was 4g/21g solvent. Convert to g/100 mL. 4g salt x (100/21) = ?
May 2, 2016

Chemistry
It requires a little imagination and patience but it can be done. I don't think this will win a Pulitzer prize but here goes. Real chemicals often mesh, Eyes, noses, and other flesh, Digits, toes, and nails galore, On hands and feet that much more, Xenon free to the core.
May 2, 2016

chemistry
You must have meant BeCl2.
May 1, 2016

chem
HC2H3O2 = HAc ..........HAc + H2O--> H3O^+ + Ac^- I.......0.095...........0......0 C.........-x............x......x E.....0.095-x...........x......x Substitute the E line into the Ka expression for HAc and solve for x = (H3O^+).
May 1, 2016

chemistry
I don't get it?
May 1, 2016

Chemistry
You didn't like the answer I gave you earlier? What's wrong with it.
May 1, 2016

chemistry
What do you mean by a 10 DM vessel. Is that 10 dm^3 vessel. If I assume that then the (PCl5) = 0.1. If that isn't right correct accordingly. ......PCl5 ==> PCl3 + Cl2 I......0.1......0......0 C......-x.......x......x E......0.1-x....x......x Substitute the E line into ...
May 1, 2016

chemistry
dS = 100 x 4.2 x ln(T2/T1) Remember to use kelvin for T2 and T1.
May 1, 2016

Chemistry
I don't know the exact procedure you're using but if the original solution is in a volumetric flask you simply take an aliquot and go from there.
May 1, 2016

chemistry
See your other post. M = mols/L
May 1, 2016

chemistry
This is just a coefficient ratio. 0.00001 mol HSO3^- x (1 mol IO^-)/ mol HSO3^-) = 0.00001 x 1/3 = ?
May 1, 2016

science
(H^+) = sqrt(Kw/KaKb) and convert H^+ to pH. ammonium acetate = NH4Ac NH4Ac + HOH ==> NH4OH + HAc Plug in H^+ to Ka expression for HAc and calculate HAc.
May 1, 2016

Chemistry
How many mols H2CO3 are in 16 g? That's mols = grams/molar mass = ? Then there are 6.022E23 molecules in 1 mol.
May 1, 2016

Chemistry
Kayla, I answered this for you last night. You didn't like my answer?
May 1, 2016

Dissociation Equations Answer Check
Those look ok to me.
May 1, 2016

Chemistry
155 g x 0.14 = grams H2SO4. mols = grams/molar mass
May 1, 2016

Chemistry
a. 4 ppm = 4 mg/L. mols O2 = grams/molar mass and since that many mols are in 1 L that is the M. b. concn(M) = K*P You know M from part a and you K from the problem, solve for partial pressure of the gas.
April 30, 2016

chemistry
I haven't the slightest. You need to supply more information.
April 30, 2016

Chemistry
2H2O2 ==> 2H2O + O2 1 mol H2O2 will produce 1/2 mol O2 so 0.5 mol H2O2 will produce 0.5/2 = 0.25 mol O2.
April 30, 2016

chemistry
I looked up niobium hydride on Google and came up with NbH5; however, I think you have asked the wrong question. It appears to me that you aren't interested so much the oxidation state as in what product is formed. The electronic structure is [Kr]4d^4 5s^1. It would appear...
April 30, 2016

chem
I don't believe any of those numbers. First, I obtained 49.8 for the first instead of 50 g/L. You are allowed 3 significant figures; why not use all of them? Then 49.8 g/L x 30/1000 = ?
April 30, 2016

chemistry
KOH + HNO3 ==> KNO3 + H2O mols KOH = M x L = ? mols HNO3 = mols KOH (you know from the coefficients in the balanced equation; that is 1:1) M HNO3 = mols HNO3/L HNO3. You know M and mols, solve for L and convert to mL.
April 30, 2016

Chemistry
Look at the formula of C3H7OH. For every molecule of C3H7OH, you have 3 C atoms. Multiply by 8 to find the number of H atoms and multiply by 1 for find O atoms.
April 30, 2016

6th Grade Science - Connections Academy
http://education.seattlepi.com/five-examples-organisms-use-asexual-reproduction-5849.html
April 30, 2016

Chemistry
Do you know what the increase in CO2 concn is? Do you think the amount of precipitation has an effect on the pH. Do you know what that amount is?
April 30, 2016

Chem
dG = dH - TdS Set dG = 0, T is 200 and dH is -120. Solve for dS Then check it to see that temperatures below 200 have dG = negative and temperatures above 200 K have dG = positive.
April 30, 2016

chemistry
2 2Na ==> 2Na^+ + 2e S + 2e ==> S^=
April 29, 2016

Chemistry
mass = volume x density. You have the volume, look up the density, substitute and solve. I think the density of Fe is approx 8 g/cc but I don't remember all of that stuff that can be looked up.
April 29, 2016

Chemistry
I have copied the equation without the phases to save space. K2S + Co(NO3)2 = 2KNO3 + CoS mols Co(NO3)2 = M x L = ? Using the coefficients in the balanced equation, convert mols Col(NO3)2 to mols K2S. Since the coefficients are the same then mols Co(NO3)2 = mols K2S. Then M ...
April 29, 2016

Chemistry
85.5% H2SO4 w/w means 85 g H2SO4 + 15 g H2O or 85.5 g H2SO4/100 g solution. How many grams H2SO4 will be in 150 g of that solution? g H2SO4 = 85.5 x (150/100) = approx 128 but that's an estimate (but close). How many g H2SO4 will be 150 g of 45% solution. That will be 45 x...
April 29, 2016

Chemistry
https://www.google.com/search?q=lewis+structure+BrF3&ie=utf-8&oe=utf-8
April 29, 2016

Chemistry
That's ok but I believe you are allowed another significant figure. That 5 comes from the log. I obtained 5.34.
April 29, 2016

gum help asap
Of course not. Worst is the worst answer of all (unless that might fall to baddest). If you're referring to my response, I don't know what gum help has to do with that sentence. You are to type in the subject where you have gum help. I would think the subject might be ...
April 29, 2016

gum help asap
c but what does gum help have to do with this.
April 29, 2016

Chemistry-Dr.Bob222
HC2H3O2 = HAc NaC2H3O2 = NaAc millimols HAc = mL x M = approx 250 mmols NaAc = 250 0.150 mol NaOH = 150 mmol ......HAc + OH^- ==> Ac^- + H2O I.....250....0.......250....... add........150................. C...-150..-150.......+150 E.....100...0........400 Plug the E line ...
April 29, 2016

Chemistry
I assume you meant EXCESS HCl. Volume of H2 at what conditions of T and P.
April 29, 2016

Chemistry
If I read this correctly, there will be no A or B; all of it will be C.
April 29, 2016

Chemistry
Oxidation is the loss of electrons so those metals whose electrons are easily lost are oxidized the easiest. Those are the alkali metals (group I) of Li, Na, etc. Reduction is the gain of electrons. So F2 Cl2 etc gain electrons the easiest. (but those of course are not metals).
April 29, 2016

chemistry
You are correct with which is the anode and which is the cathode. The explanation of the difference in the two cells is easy. The DEFINITION of the anode is "The anode is where oxidation takes place." Which is the anode and which the cathode has nothing to do with ...
April 29, 2016

Chemistry
......ZnCO3 ==> Zn^2+ + CO3^2- But CO3^2- + H^+ = HCO3^- and HCO3^- + H^+ ==> H2CO3 Ksp = (Zn^2+)(CO3^2-) You can see from the equilibria that the H^+ increases the solubility of ZnCO3 because it adds to the CO3^2- to form H2CO3 (in two steps). You need to find the ...
April 28, 2016

help !! chemistry
I suggest you submit a question along with the appropriate information in standard form. I don't see a question here.
April 27, 2016

CHM130
With all gases one can use volume in L directly as mols as a short cut. 5.66 L CO2 x (9 O2/6 CO2) = ?
April 27, 2016

Chemistry
I would do this. Zn(OH)2 ==> Zn^2+ 2OH Ksp = ? Zn^2+ + 4OH^- ==> [Zn(OH)4]^2- Kf=? Add these two equations. Zn(OH)2 + 2OH^- ==> [Zn(OH)4]^2- .solid....+2x.........-x Krxn = Ksp*Kf = approx 6 Then [(x)/(2x)^2] = 6 Solve for x and convert to pH.
April 27, 2016

Chemistry
MgCl2 ==> Mg^2+ + 2Cl^- .........0.23....0.46 Substitute those number into the Ksp expression and solve for Ksp. part 2. Then Ksp = (Mg^2)(Cl^-)^2 Plug in 0.00537 for Mg^2+ and solve for Cl^-.
April 27, 2016

Chemistry
millimols F^- initially = mL x M = 50 x 0.5 = 25 At zero mL HCl added, the pH is determined by the hydrolysis of F^- ......F^- + HOH ==> HF + OH^- I....0.5M...........0.....0 C.....-x............x.....x E.....0.5-x.........x.....x Kb = approx E-11 = (x)(x)/(0.5-x) Solvr for...
April 27, 2016

Chemistry
CHNH2 + HCl ==> CH3NH3Cl b. Look up the delta Go values in your text/notes and solve for dGo for the rxn, then dGo = -RTlnK. Substitute and solve for K. c. ..CH3NH2 + HOH ==> CH3NH3^+ + OH^- I..0.030............0..........0 C.....-x............x..........x E..0.030-x...
April 27, 2016

Chemistry
Use (p1v1)=(p2v2)
April 27, 2016

Chemistry
We could help you better (and faster) if you didn't change screen names. Use (v1/t1) = (v2/t2) Remember to change the T to kelvin.
April 27, 2016

Chemistry
What is the mass of the 250 cc solution. That's 1.10 g/cc x 250 cc = 275 grams. 10% NaCl means 10g NaCl/100 g solution. So how much NaCl is in 250 cc? That's 10 g x 250/100 = 25 g. So 275 g total - 25 g salt = 250 g H2O. Check my work carefully.
April 27, 2016

Chemistry-Dr.Bob222
.......Ca(OH)2 ==> Ca^2+ + 2OH^- I.......solid.......0.......0 C.......solid.......x.......2x E.......solid......x.......2x (H^+) for pH 4 is 1E-4 Ksp = (Ca^2+)(OH^-) From the above, (Ca^2+) = x (OH^-) = x + 1E-4. Substitute and solve for x = solubility. Note: for the total...
April 27, 2016

Chemistry-Dr.Bob222
Why not simply calculate the solubililty of each.
April 27, 2016

Chem
2N2H4(g)+N2O4(g)→3N2(g)+4H2O(g) ...2......1.......3.......4 ...x......9.......y.......z (2)18.....9.......27......36 (3)....... (4)....... (5) x....3.6......y.......z ..2*3.6..3.6....3*3.6...4*3.6 I agree with 1 I did 2. Disagree with 3. I think that 14 should be 13.5 ...
April 27, 2016

Chemisty
NaOH + HCl ==> NaCl + H2O mols NaOH = M x L = ? mols HCL = mols NaOH since 1 mol NaOH reacts with 1 mols NaOH. Then M HCl = mols HCl/L HCl = ?
April 27, 2016

Chemistry
delta T = i*Kf*m i = van't Hoff factor = 1 for glucose i = 2 for NaCl. m is the same, Kf is the same;the only difference is i so ......is the lowest.
April 27, 2016

Chemistry
First calculate the molality. m = mols NaCl/kg solvent. 1.1% NaCl = 1.1 g NaCl/100 g solution. That is 1.1 g NaCl/98.9 g H2O) = 1.1 g NaCl/0.0989 kg H2O. mols NaCl in 1.1 g = 1.1/58.45 = 0.0188 mols NaCl = g NaCl/molar mass NaCl = about 1.1/58.45 = about 0.0188. Then m = mols ...
April 27, 2016

Chemistry
You want how many mols. That is M x L = 7.1 x 1.5 = ? mols = grams/molar mass. You know molar mass ethanol and mols, solve for grams.
April 27, 2016

Chemistry
But do you know why a,d,e are more soluble?
April 27, 2016

Chemistry
ade
April 27, 2016

chemistry
I can only guess as to the meaning of some of the symbols. Assuming my guesses are right, and I don't guarantee they are, I would go with B followed by A.
April 26, 2016

chemistry
See your post above.
April 26, 2016

Chemistry
........HCOOH ==> H^+ + HCOO^- I.........M.......0......0 C........-x.......x......x E........M-x......x......x The problem tells you the pH is 1.98. Use pH = -log(H^+) to find x. Then substitute into Ka expression for HCOOH (you will need to look up Ka for HCOOH) and solve...
April 26, 2016

chemistry
mols NaCl = M x L = ? grams NaCl = mols NaCl x molar mass NaaCl.
April 26, 2016

Chemistry
440 g Mg x (molar mass MgCl2/atomic mass Mg) = ? Another way to look at this is %Mg in MgCl2 is (atomic mass Mg/molar mass MgCl2) = approx 25.5% Then 0.255 x g MgCl2 = 440. Solve for g MgCl2.
April 26, 2016

Chemistry
See your other post above.
April 26, 2016

chemistry
avg speed = v = sqrt(8RT/pi*M) rrms = sqrt (3RT/M) Solve v equation for T Solve rms sped for T Set v = vrms = to each other. Substitute values and solve for T Post your work if you gets stuck.
April 26, 2016

chemistry
You need to re-read my explanation. If you want to convert 1.210 g to mols that's ok but more work than you need to do. I presume m1 is the mass of CH4 (that's my X) and m2 is the mass of C2H6(that's my Y). Your equation of m1/M1 + m2/M2 = 0.766 is right but ...
April 26, 2016

chemistry
I don't get a negative number. Post your work and I'll find the error.
April 26, 2016

chemistry
This is a problem with two unknowns; therefore, you need two equations and solve them simultaneously. Write and balance the combustion equations. CH4 + 2O2 ==> CO2 + 2H2O 2C2H6 + 7O2 ==> 4CO2 + 6H2O Let X = grams CH4 and Y = grams C2H6 ---------------------- eqn 1 is X...
April 26, 2016

Chemistry
HNO3 + KOH ==> KNO3 + H2O millimols HNO3 = M x mL = approx 200 but this is just an estimate. mmols KOH - 0.57 x 400 = approx 228. So HNO3 is the limiting reagent; all of the HNO3 will be used and 28 mmols KOH will be left over. So the pH will be determined by the excess KOH...
April 26, 2016

Chemistry
You don't? 1.20-2x = 1.02 -2x = 1.02-1.20 = -0.18 2x = 0.18 x = 0.18/2 = 0.09 So N2 = x = 0.09 H2O is 2x = 2*0.09 = 0.18 N2 left is 0.95-2x = ?
April 26, 2016

Chemistry
.......2H2 + 2NO->N2 + 2H2O. I.....1.20..0.95..0.....0 +5sec.-2x....-2x..x....2x final.1.20-2x.0.95-2x..x...2x The problem tells you that 1.20-2x = 1.02. Solve for x and evaluate the other values. Post your work if you get stuck.
April 26, 2016

Chemistry
dE = dH + w dH = +100 J w = -50 J
April 26, 2016

  1. Pages:
  2. <<Prev
  3. 1
  4. 2
  5. 3
  6. 4
  7. 5
  8. 6
  9. 7
  10. 8
  11. 9
  12. 10
  13. 11
  14. 12
  15. 13
  16. 14
  17. 15
  18. Next>>