Saturday
August 1, 2015

Posts by DrBob222


Total # Posts: 49,606

chemistry
How would you calculate pH when Kw is 1E-14 at 25C? This is done the same way; but don't expect the pH to be 7.00 which is the whole point of the problem.
April 5, 2015

science
Ptotal = pH2 + pH2O Ptotal is what you read as the room pressure. pH2O is the vapor pressure of H2O which you can obtain from tables in your book at the corresponding temperature.
April 5, 2015

Chemistry
Of course. It will change the calculated volume and since you are measuring the volume of the titrant + the volume of the bubble, it will be larger. Since this is a molarity of the base, you will calculate M = mols/L The indicator will change when mols acid you're ...
April 4, 2015

chemistry
work = -pdV = -p(Vfinal - Vinitial) Convert 100 w to watt*sec = joules and that will be q dE = q+w
April 4, 2015

chemistry
UF4 + 2Ca ==> 2CaF2 + U mols UF4 initially = grams/molar mass = 292,000/314 = approx 930 but you need to do this more accurately. All of the following calculations are estimates; recalculate those also. mols Ca = grams/atomic mass = 47,000/40.1 = estimated 1180 Using the ...
April 4, 2015

chemistry
This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants. This is a standard LR problem. What do you not understand?
April 4, 2015

chemistry
It's the Arrhenius equation. Substitute the numbers I gave in the earlier response and solve for k2.
April 4, 2015

chemistry
I don't understand. What does "but that get with this????" mean?
April 4, 2015

chemistry
Use the Arrhenius equation. Choose any concentration you wish but the easy number is 1. k1 is 1/4 k2 = ? T2 is 811 C converted to K. T1 is 748 C converted to K. Ea is 150,000 Post your work if you get stuck.
April 4, 2015

Chemistry
Double displacement is right. The equation is wrong. Your error in the equation is that you have tried to balance the equations by making the products fit the balanced numbers you need. It doesn't work that way. Here is what you do. The product will be K and NO3 for one ...
April 4, 2015

chemistry
C + O2 --> CO2 20g at 90% purity = 18 g pure C. Can you take it from there?
April 4, 2015

Chem 12
It would be much easier to help you if you didn't change screen names. See your post above on K (where you didn't post the equation). Mathmate answered that.
April 4, 2015

Chemistry
If you will post how you obtained the 152 mL I shall be happy to check it for you.
April 3, 2015

Chemistry
mols NaCl = 50/molar mass = ?M which is approx 0.9 but you need a more accurate answer than that for this as well as the other calculations. M of the NaCl solution is about 0.9/0.5 = about 1.8 M. Then mL1 x M1 = mL2 x M2 mL1 x 1.8 = 1300 x 0.1M You're right. Remember the 1...
April 3, 2015

chemistry ap
I don't think so. mols Al = grams/atomic mass = ?
April 3, 2015

chemistry ap
After working those other problems, what about this do you not understand. Explain in detail. And while you're at it, balanced the equation.
April 3, 2015

chemistry
Start by balancing the equation. Convert 1.09L H2O vapor to mols. Convert mols H2O to iron(III) hydroxide using the coefficients in the balanced equation, the convert that to grams. I can help you through it if you get stuck but explain in detail what you don't understand.
April 3, 2015

chemistry ap
See the problem below.
April 3, 2015

chemistry ap
When using gases one can take a shortcut and use volume as if it were mols. 734 L NO2 x (1 mol NO/3 molls NO2) = 734 x 1/3 = ?
April 3, 2015

chemistry
PV = nRT and solve for n = mols gas. Then n = grams/molar mass. You know grams and n, solve for molar mass.
April 3, 2015

chemistry ap
See your problems above. Same procedure for all gas problems.
April 3, 2015

chemistry
Here is one just like it. http://www.jiskha.com/display.cgi?id=1428076442
April 3, 2015

chemistry
There are 6.02E23 molecules of NO2 in a mole of NO2. Since NO2 is a covalently bonded molecules that is the number of particles. For N. You have 1 atom N for each molecule NO2; therefore, you have that many atoms N. For O. You have 2 atoms O for each molecule NO2; theefore, ...
April 3, 2015

Chemistry
Try -nRTln(V2/V1)
April 3, 2015

chemistry
mols HBr = 0.025 from your work mols KOH = 0.015 from yur work. .......HBr + KOH ==> KBr + H2O I....0.025..0.015.....0.....0 C...-0.015.-0.015....0.015..0.015 E....0.010....0......0.015 HBr is in excess. (HBr)= mols/L = 0.010/(0.06+0.05) = approx 0.09 Then pH = -log(H...
April 3, 2015

chemistry
millimols HPO4^- = mL x M = approx 14 but you need to use a better number than that so recalculate all of these. mmols HCl added = 50 x 0.275 = approx 14. .......HPO4^2- + H^+ ==> H2PO4^- I......14........0........0 add .............14............ C.....-14.......-14......+...
April 3, 2015

chemistry
PV = nRT What's the trouble. Plug in P, V, R, T and solve for n. Remember to convert C to Kelvin
April 3, 2015

chemistry
AgCl ==> Ag^+ + Cl^- Ksp = (Ag^+)(Cl^-) Ag^+ + 2NH3 ==> [Ag(NH3)2]^+ Kf = --- Add the equations to get this: AgCl(s) + 2NH3 ==> Cl^- + [Ag(NH3)2]^+ So Keq = Ksp*Kf Set up an ICE chart and solve for (NH3). I think a quadratic will be necessary. Post your work if you ...
April 3, 2015

chemistry
Set up an ICE chart and use the Henderson-Hasselbalch equation.
April 3, 2015

chemistry
Use the Henderson-Hasselbalch equation to solve for pH of the solution. Use NH4Cl as the acid and NH3 as the base. Convert the pH to (OH^-) and set up Ksp with Mg(OH)2 and see if Qsp is greater than Ksp for Mg(OH)2. In the regular qual scheme of things, Mg ions are in group V ...
April 3, 2015

Chem pls!!!!!
To the first part, use #2 rxn as an example. #2 is a single replacement reaction. For #2, the field is wide open for an answer. I would try ppt but other answers will fit there too. And I might note that I don't like the wording. A solid would be formed from CuCl but not ...
April 2, 2015

chemistry
2HClO4 + Ca(OH)2 ==> 2H2O + Ca(ClO4)2 mols Ca(OH)2 = M x L = ? mols HClO4 = 2x that. M HClO4 = mols HClO4/L HClO4. You have mols and M, solve for L.
April 2, 2015

chemistry
See your post below under the screen name of Branson. That should tell you how to work the problem. By the way, you make it difficult to help if you change screen names.
April 2, 2015

chemistry
IF you are using the equation of N2 + 3H2 ==> 2NH3 (if not this won't work) mols NH3 = grams/molar mass mols H2 = mols NH3 x (3/2) = ? Then g H2 = mols H2 x molar mass H2 = ?
April 2, 2015

Chemistry
Shouldn't you have used the salt for the concentration of (A^-) and not x which is the same as (H^+). From your work, Ka=[x][x]/[9.03 x 10^-3 M] would be Ka = (x)(x+salt)/(HA) where salt is 3.87E-3 M.
April 2, 2015

Chemistry
58.8 C is the boiling point of Br2. At the boiling point you have equilibrium between the liquid phase and vapor phase and dG = dH - TdS. dG = 0 because it is equlibrium. dH vap-- look that up in tables. You know T, solve for dS. If you want to read more about this look up ...
April 2, 2015

Life science,agric science,geography,history,maths
Talk to your advisor/counselor. S/he can look at what you have and what you've done well in and make recommendations for the future.
April 2, 2015

Chemistry
This is a limiting reagent (LR) problem; you know that because amounts are given for BOTH reactants. This is worked just like two simple stoichiometry problems. CH4 + 2O2 - cO2 + 2H2O mols CH4 = grams/molar mass = ? mols O2 = grams/molar mass = ? Using the coefficients in the ...
April 2, 2015

Chemistry
(rateH2/rateun) = sqrt(M un/M H2) Substitute ad solve for rate H2. un is unknown which I assume is He.
April 2, 2015

Chemistry
HA + BOH ==> BA + H2O HCl + NaOH ==> NaCl + H2O
April 2, 2015

chemistry
There is something inherently wrong with these equations. That's the major reason I didn't answer when you had this posted a few days ago. You simply can't get H2 from 2 H^+ and you can't get Br2 from 2Br^- IF you could get two H^+ to combine (and I doubt that ...
April 2, 2015

Chemistry
It is not a weak base. A buffer consists of A. a weak acid and its salt OR B. a weak base and its salt. Mg(OH)2 does not meet either criteria. It is not a weak base and there is no salt.
April 2, 2015

Chemistry
Note the correct spelling of celsius. pH + pOH = pKw = 14 Solve for pOH. Then pOH = -log(OH^-)
April 2, 2015

Chemistry
4FeCl2(aq) + 3O2(g) - 2Fe2O3(s) + 4Cl2(g) mols O2 = 8.71E21 molecules x (1 mol/6.02E23 molecules) = ? Using the coefficients in the balanced equation, convert mols O2 to mols Fe2O3 Then convert mols Fe2O3 to g Fe2O3 with g = mol Fe2O3 x molar mass Fe2O3.
April 2, 2015

chemistry
Reduction procedure for what? to what? with what? What are you doing? My crystal ball is hazy today.
April 2, 2015

chemistry
What what the original procedure?
April 2, 2015

GeneralChemistry
You worked all of those stoichiometry problems (even the limiting reagent problem) beautifully that I checked last night. This is the same thing with just a slight twist at the end. 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g) mols H2 = grams/molar mass = ? Using the ...
April 2, 2015

chemistry
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2 mols HCl added initially = M x L = approx 0.0025 but you need a more accurate answer than that estimate. That HCl is more than enough to dissolve all of the tablet and have some HCl left over. mols NaOH needed to neutralize the excess HCl...
April 2, 2015

chemistry
I assume you mean the standard at 298 K. Look up the heat of formation. That will be in a table in your text and you probably can find it on the web.
April 2, 2015

science
I agree with what you've said but without knowing the particulars of the experiment I can't help you. Are you titrating I2 with thiosulfate? If so you measured the KIO3 precisely. The KI is an excess reagent and the amount of the excess is not that important and the ...
April 1, 2015

chemistry
I may have stared at the sun too long today but why isn't dG = 0 when the reaction has reached equilibrium? You can calculate those concentrations if you wish AND you can calculate Kp at 25C with dGo = -RTlnK. But I don't think any of that is necessary since we know dG...
April 1, 2015

Chemistry
2HNO3 + Ba(OH)2 ==> Ba(NO3)2 + 2H2O Look at the coefficients in the balanced equation. mols HNO3 = 2 x mols Ba(OH)2
April 1, 2015

chemistry
HBr + NaO ==> NaBr + H2O This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. mols HBr = grams/molar mass = ? mols NaOH = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols HBr to mols H2O. ...
April 1, 2015

Chemistry
You meant Na2CO3 for NaCO3. 2Fe^3+ + 3Na2CO3 ==> Fe2(CO3)3 + 6Na^2+
April 1, 2015

chemistry
I think you need to back up before going forward. (HA) = 0.01 (H^+) = (A^-) = 0.01 x 0.037 = 3.7E-4 ............HA ==> H^+ + A^- I.........0.01.....0.....0 C........-3.7E-4..3.7E-4..3.7E-4 E.......you finish Substitute the E line into the Ka expression and solve for Ka. ...
April 1, 2015

Science
I'm not a meteorologist but I would look at A.
April 1, 2015

Chemistry
How can we help you with this assignment?
April 1, 2015

chemistry help
Tell us what you understand about this problem and what you don't understand. I can help you through that part. And please explain in detail what you don't understand.
April 1, 2015

chemistry help
See your post above.
April 1, 2015

chemistry help
I answered this earlier, at least partially. See your other posts and help me understand what you don't. Explain in detail.
April 1, 2015

Chemistry
The molecular equation is AgNO3 + Na2S ==> Ag2S + NaNO3 You balance. The net ionic equation is Ag^+ + S^2- ==> Ag2S(s) You balance.
April 1, 2015

Chemistry
Is this % w/w or % w/v? I'll assume % w/v % w/v = (grams solute/50 mL)*100 Solve for grams.
April 1, 2015

chem
The process looks ok to me but you have too many significant figures in the answer. I would round that to 161 grams.
April 1, 2015

chem
This looks ok but you have too many s.f.. I would round that answer to 665 g.
April 1, 2015

chem
I agree with your answer. As for the 19/2, I think many profs will not allow it but I don't see anything wrong with it.
April 1, 2015

chemistry
Devron is right; volumes are not additive but at these concentrations the difference is strictly theoretical. I don't think you would ever be able to measure it.
April 1, 2015

chemistry
mols = grams/molar mass M = mols/L
April 1, 2015

chemistry
mols = M x L = ? Then grams = mols x molar mass
April 1, 2015

Chemistry
You can convert mols of one material in the equation to any other material in your equation by simply using the coefficients in the balanced equation. It works every time. Follow this. You have 1.75 mols H2O2. How many mols O2 will it give? That's 1.75 mols H2O2 x (1 mol ...
April 1, 2015

Chmeistry
I don't understand. grams/L; grams/100 mL, mols/L,
April 1, 2015

Chemistry
Note the correct spelling of celsius. Few of us have the solubility tables of all of the known salts memorized; if you can give us the solubility at 50 C we can tell you. Probably you have a graph in your text/notes that should tell you this and it's a matter of reading it...
April 1, 2015

chemistry
What about this problem troubles you. This is pretty standard; however, the buret reading may be a problem. 29.58 mL is final reading -0.23 mL is initial reading. ---------------- 29.35 ml is titration volume.
April 1, 2015

chemistry
Can you tell me what the main points troubling you? Please explain in detail.
April 1, 2015

chemistry
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) Tfinal = 98.5 C Tinitial = 23.4 C
April 1, 2015

chemistry
Do you know the density. 1000 x density(g/mL) x 0.69 x (1/63) = ?M = approx 14M but you need that more accurate than that. Then mL1 x M1 = mL2 x M2 1000 x 2.5M = mL2 x approx 14 Solve for mL2, add that amount to a 1L volumetric flask, make to the mark, stopper, share thoroughly.
April 1, 2015

chemistry
Use the HH equation. pH = pKa + log (base)/(acid)
March 31, 2015

chemistry
Do you know it is a weak acid? .......HA ==> H^+ + A^- I....,0.5.....0.....0 C......-x.....x.....x E.....0.5-x...x.....x You know pH so you know x = (H^+) and A^-. Plug these into the Ka xpression and solve for Ka after evaluating 0.5-x.
March 31, 2015

Chemistry physics
I worked this below. You should be able to find it. If not I can try and locate it. It is filed under a chemistry banner, not physics.
March 31, 2015

CHE
I would think that since Cl2 is added as a bactericide for many public water supplies that the Cl2 might be a source (from its reaction with H2O and any bacteria) of chloride ion.
March 31, 2015

chemistry
mols NaOH = M x L = ? Using the coefficients in the balanced equation, convert mols NaOH to mols H2SO4. That should be mols H2SO4 = 1/2 mols H2SO4. Then M H2SO4 = mols H2SO4/L H2SO4. Solve for L H2SO4 and convert to mL.
March 31, 2015

chemistry
The reaction is H2SO4 + 2NaOH ==> Na2SO4 =+ 2H2O; BUT you didn't post the molarity of the H2SO4 and that is needed to do the problem.
March 31, 2015

Chemistry
pH = 1.0; (H^+) = 0.1 So you added 56 mL of ?M to 85 mL H2O and the result was 0.1M. ?M x (56/(56+85) = 0.1 Solve for ?M This assumes, of course, that the volumes are additive. Technically they are not but is dilutions that is usually considered to be so. Or if you want to do ...
March 31, 2015

Chemistry DrBob please answer
If it consumes 5.10L, does it matter what the initial V and final V actually are. The difference will be 5.10 L and I would use 5.10 as V1 and V2 as zero (since it was consumed). Of course you could always use any other set of numbers such as 35.10 and 30.0 L or 125.10 and 120.0.
March 31, 2015

Chemistry
See my response above.
March 31, 2015

science help.. pls help me dr.bob22 or damon
I believe the answer is covalent bond. I think the question is about the bonging between the H atoms and the O atom of a water molecule. Although water forms hydrogen bonds, and many of them, those bonds are intermolecular and not intramolecular. I believe the question is all ...
March 31, 2015

science help.. pls help me dr.bob22 or damon
And can you imaging that if the discovery of electrons didn't do enough damage that the discovery of protons and neutrons would do it in. Having said all of that, let's not condemn the theory without recognizing that with just a slight word change here and there the ...
March 31, 2015

science help check
very good. yes.
March 31, 2015

Chemistry
dG = dH - TdS At equilibrium dG = 0 and the melting point you have equilibrium. Substitute heat fusion for dH and T and solve for dS. Watch the signs.
March 31, 2015

Chemistry
dE = Ccal*dT and that is kJ (if you use Ccal in kJ) for 1.578/110.7 mol xample. Convert that to kJ/1 mol.
March 31, 2015

science help
Yes. It is a transition metal. It is is row 6 and column 4 so that makes it group 4 and period 6.
March 31, 2015

Chemistry
v2 is zero V1 is 5.10L
March 31, 2015

Chemistry
dE = q+w dq = 0 w = -p*dV = -p*(V2-V1)
March 31, 2015

Chemistry
dE = q + w You have heat given in the problem. w is -pdV Substitute and solve for dE
March 31, 2015

chemsistry
Sarah, a stoichiometry problem is a stoichiometry problem. See your last post with hexane. Post what you know to do and explain in detail what you don't understand next.
March 31, 2015

chemistry
See your hexane/CO2 post below. Same kind of problem.
March 31, 2015

chemistry
2C6H14 + 19O2 ==> 12CO2 + 14H2O Use the coefficients in the balanced equation to convert mols C6H14 to mols CO2 Then convert mols CO2 to grams. g = mols x molar mass.
March 31, 2015

Science
..........NH4^+ H2O ==> NH3 + H3O^+ I.......0.250............0.....0 C.........-x.............x.....x E.......0250-x...........x.....x Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.250-x) Solve for x = (NH3)
March 31, 2015

Physical science
q1 = heat needed to raise T from -8.5 to zero c. q1 = mass ice x specific heat ice x (Tfinal-Tinitial) q2 = heat needed to melt ice at Zero C and convert to liquid H2O. q2 = mass ice x heat fusion q3 = heat needed to raise T of water from zero C to 100 C q3 = mass H2O x ...
March 31, 2015

science
Do you want 0.5m or 0.5M?
March 31, 2015

Chemistry
dHrxn = (n*dHf products) - (n*dHf reactants)
March 31, 2015

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