Saturday
April 18, 2015

Posts by DrBob222


Total # Posts: 48,372

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

To Mathmate--Chemistry

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

chemistry
The problem is worded in a confusing manner. I think the weight loss represents the decomposed part. % = (0.384/4.90)*100 = ? If you want the percent remaining of the original it is [(4.90-0.384)/4.90]*100 = ?
March 31, 2015

Chemistry
It appears to me you have accounted for only a part of the heat. q = [mass x specific heat x (Tfinal-Tintial)] + [Ccal*(Tfinal-Tinitial)] The first term takes care of the heat gain in the water and the second term takes care of heating the calorimter.
March 31, 2015

Chemistry
I don't think other pain relievers would affect the results IF they were not acidic/basic and therefore, not tirated with acid or base titration. I would be concerned with acid relievers as in Tums and that kind of thing because those can be titrated with acid.
March 31, 2015

Chemistry
Cu + 2AgNO3 ==> Cu(NO3)2 + 2Ag mols Cu = grams/molar mass mols AgNO3 = M x L = ? How much Cu is needed? That's mols AgNO3 x 1/2 = ? Do you have that much?
March 30, 2015

Chemistry
C12H22O11 + 12CO2 + 11H2O mols CO2 = grams/molar mass = ? mols sugar = mols CO2 x (1/12) = ? grams sugar = mols x molar mass 4.00-g sugar = g salt % salt = (g salt/4.00)*100 = ?
March 30, 2015

chemistry
specific heat = (J/c)*g
March 30, 2015

Chemistry
No, no ICE chart needed. Just remember this. Ka*Kb = Kw = 1E-14
March 30, 2015

chemistry
Calculate the pH with what? You didn't provide any information to determine pH of the solution before the HCl is added.
March 30, 2015

chemistry
Few people have the solubility tables memorized for all of the salts. If you will provide the solubility data we can help.
March 30, 2015

chemistry
CH4 + 2O2 ==> CO2 + 2H2O You may use a shortcut when dealing with gases and use volume as if they were mols directly. Therefore, we will use 2.65 L CH4. Convert to L O2. 2.65 L CH4 x (2 mols O2/1 mols CH4) = 2.65 x 2/1 = ? L O2. Then convert ?L O2 at STP to L O2 at the ...
March 30, 2015

Chem
I hope you revisited the ethanol problem After your last post I found two errors; one yours and one mine.
March 30, 2015

chemistry help
I thought I worked this problem for you a couple of days ago. No? Use the HH equation, use x for millimols NH4Cl and use millimols for NH3. Convert x to grams NH4CL. You must first convert Kb to pKb and then to pKa.
March 30, 2015

AP Chemistry
For #2. NaOH + HNO3 ==> NaNO3 + H2O mols HNO3 = M x L = ? mols NaOH = mols HNO3(from the coefficients in the balanced equation.) M NaOH = mols NaOH/L NaOH For #1. Assume some volume of the 5.0 M HNO3. Calculate the volume NaOH needed to titrate half of it. Calculate the NO3...
March 29, 2015

chemistry
11 m or 11 M? What is 29" concentration?
March 29, 2015

chemistry
mols Al = grams/molar mass = ? mols KOH = M x L = ? Use the coefficients in the balanced equation to convert mols Al to mols of the product. Do the same for the mols KOH to mols of the product. It is likely that the two values for mols product will not agree; the correct value...
March 29, 2015

Chemistry
2NaOH + H2SO4 ==> Na2SO4 + 2H2O How many mols H2SO4 do you have? That's M x L = > How many mols NaOH do you need to neutralize that? That's mols NaOH = 2x mols H2SO4 from the coefficients in the balanced equation. At that point you have a pH = 7 (neutral solution...
March 29, 2015

chemistry
H2O(s) ==> H2O(l) dG = dHfus - TdS(fus) dG = 0 since the system 9s at equilibrium dH fusion you know from the tables. I think is is about 334 J/g. T is 273.15, solve for dS.
March 29, 2015

chemistry
I THINK all you have to do here is to calculate the pH of the 0.02 M NaOH but just for good measure I would soolve for the contribution from the Na salt. If it is large enough add it to that from hydrolysis of the Na salt.
March 29, 2015

probability
I haven't looked at all of this long line of posts but the last three say correct and no need to respond. So why are you posting and cluttering up the board if you don't need the help.
March 28, 2015

chemistry
millimols NH4^= = 3 x mL x M = ? millimols Na^+ = 2 x mL x M = ? Total millimols cations = ? Concn cations in M = millimols/total mL.
March 28, 2015

science
Thank you for posting your thoughts. I think you have painted us with too broad a brush. I have found that rude is one of those words that started creeping in about two or three years ago. It is used most often in cases where the tutor asks for clarification or asks for the ...
March 28, 2015

chemistry
Use the Henderson-Hasselbalch equation. You will need to convert Ka to pKa. pyruvic is the acid, of course, and the sodium salt is the base.
March 27, 2015

Chemistry 1002
dSf rxn = (n*dSf products) - (n*dSf reactants) Look up the S values in tables in your text/notes, substitute, and solve for dS rxn
March 27, 2015

chemistry
A few comments. Your math is hard to follow. 4.35 certainly is not = 3.89 I would count 1.28E-4 as the same answer as 1.27E-4 but maybe your prof is tougher than that. I think the reason your answer came out slightly larger (1.28 and not 1.27) is because you took the numbers ...
March 27, 2015

chemistry
You did something wrong. I got the answer. Post your work and I'll find the error
March 27, 2015

chemistry
Use the Henderson-Hasselbalch equation and solve for pKa. Then remember that pKa + pKb = pKw = 14. Solve for pKb. Then pKb = -logKb and solve for Kb. No, you're SUPPOSED to solve for M base and M acid but I never do because that's more work and the answer comes out the...
March 27, 2015

chemistry
2KOH + H2SO4 ==> K2SO4 + 2H2O mols KOH = 25 x 0.4 = 10 millimols. That will neutralize 1/2 that or 5 millimols H2SO4. How many millimols H2SO4 do you have?
March 27, 2015

biochemistry
9 mg = 0.009 g (0.009g/5 mL)*100 = ?%w/v
March 27, 2015

Chemistry
pH = pKa + log (base)/(acid) Let x = mL base then 100-x = mL acid millimols base = mL x M millimols acid = mL x M Substitute into the HH equation and solve for x = mlL base and 100-x = mL acid. That's it.
March 27, 2015

Chemistry
NaOH + HCl ==> NaCl + H2O mols HCl = M x L = ? Since the equation shows 1 mol NaOH = 1 mol HCl then mols NaOH = mols HCl. Then M NaOH = mols NaOH/L NaOH. You know mols NaOH and you know L NaOH. The only complicating factor is that the buret was not full for the titration. ...
March 27, 2015

Chemistry
That means you put on your thinking cap, read up on PCl3 and PCl5 (I suggest you Google one or both and read up on that on Wikipedia). Then I would go the route of starting with a known amount of P4 and calculating how much Cl2 it takes to form PCl3 or pCl5. You work all of ...
March 27, 2015

Chemistry
You're talking about writing just short of a thesis here and few of us will do that. If you can break it down, explain exactly what you don't understand(in detail) about these one at a time, perhaps someone can give you some helpful tips on how to complete the ...
March 27, 2015

chem
How many mols do you need? That's mols = M x L = ? How many grams is that? That's g = mols x molar mass = ?
March 27, 2015

Chemistry
C2H5OH + 3O2 ==> 2CO2 + 3H2O delta Hcomb = -296.6 kJ. So you have 296.6 kJ released for 10 g; you want to know how much is released for 1 mol (46.07g); therefore, dH = -296.6 x 46.07/10 = ?
March 27, 2015

biddies
I learn something every day. I never used flux in my life except for magnetic flux in chemistry and physics. I didn't know it had other meanings.
March 27, 2015

Science
I wonder what generally very low means? Here is a site that may be useful. The salinity is 35 ppt (about 3.5%)
March 27, 2015

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