Wednesday
August 27, 2014

Posts by DrBob222


Total # Posts: 43,386

chemistry
C6H12O6 ==> 2C2H5OH + 2CO2 mols glucose = grams/molar mass = estimated 0.5 Using the coefficients in the balanced equation, convert mols glucose to mols ethanol. That is 0.5 x (2 mols ethanol/1 mol glucose) = 0.5 x 2 = 1 mol ethanol produced. Then g ethanol = mols x molar ...
July 11, 2014

science
mols H2SO4 = grams/molar mass Then M = mols H2SO4/L solution. Note: 100 mL = 0.100 L.
July 11, 2014

electric circuit
Probably 6 cells in series to give 12 v, then 6 sets of those (36 cells total) in parallel.
July 11, 2014

Clincal
I don't see a question here.
July 11, 2014

chemistry
Please read and rephrase. It makes no sense as is.
July 11, 2014

chemistry
See your other post above.
July 11, 2014

Chemistry
mL1 x M1 = mL2 x M2 mL1 x 17.8 = 250 x 2.00M Add some distilled H2O to the 250 mL volumetric flask, then pipet the value of mL1 of the 17.8M H2SO4 into the 250 mL flask adding concentrated H2SO4 cautiously. Add distilled H2O to the mark and mix thoroughly.
July 10, 2014

chemistry
1 in^3 = 16.38 cc 13.4 in^3 to cc = ? 1 lb = 453.59 grams 5.14 lb to grams = ? Then d = mass/volume
July 10, 2014

CHEMISTRY
How much voltage do you need? How much can you get from these two materials? Look up the Eo values and calculate.
July 10, 2014

CHEMISTRY
Here is a simplified activity series table. A metal will displace any metal ION BELOW it in the activity series; i.e., Al will displace Cu^2+ but Cu will not displace Al^3+. 2Al + 3Cu^2+ ==> 3Cu(s) + 2Al^3+ Cu^2+ + Al^3+ ==> no reaction. http://www.cod.edu/people/faculty...
July 10, 2014

CHEMISTRY
Look up the standard reduction potentials for I2, Br2, and Al^3+. Then set them up on a sheet of paper something like this, along with H2. ..........................Eo vs SHE Br2 + 2e --> 2Br^-..........1.07 I2 + 2e --> 2I^- ...........0.54 2H^+ + 2e --> H2...
July 10, 2014

CHEMISTRY
See the activity series question.
July 10, 2014

CHEMISTRY
By definition, reduction occurs at the cathode.
July 10, 2014

CHEMISTRY
You obtain the electrolysis products of H2O and not Al.
July 10, 2014

CHEMISTRY
It's an arbitrary standard and an arbitrary number.
July 10, 2014

CHEMISTRY
I wouldn't say they contained H2O but I agree they are moist. Compared to a wet cell; however, relatively speaking the dry cells are dry.
July 10, 2014

CHEMISTRY
You repeated my line but didn't answer if this was a made up problem or not. I don't like that answer until I have more information about the question.
July 10, 2014

CHEMISTRY
I wonder if this is a made up problem? Sn^2+ + 2e ==> Sn E = -0.14 Cr --> Cr^3+ + 3e E = ? ---------------------- Ecell = -0.60 then Cr must be -0.46 written as an oxidation so as a reduction it would be +0.46 but that isn't what it gives in the tables.
July 10, 2014

chemistry
What numbers do you have?
July 10, 2014

Chemistry
Some picky points. n N2 I obtained 0.007279 so I rounded to 0.00728. For n O2 I obtained 0.00436 I obtained 0.0029 for mols N2 just as you did; however, it is done by 0.0036 x 2/3 and not 3/2. Finally,(this is not so picky) the problem asks for mL so you should convert your 0....
July 9, 2014

Chemistry
almost. n = 0.0079? Why did you throw away the 1 on 0.00791. You have three places in yur other numbers; you should keep three for the answers. Then if I multiply 0.00791 x 16 I get 0.1265678 which rounded to 3 s.f. is 0.127grams CH4.
July 9, 2014

Chemistry
Change -15F to degrees C. That's approximately -26 Then delta T = Kf*m approx 26 = 1.86*m solve for molality. m = mols/kg solvent = mols/1kg You have m and kg, solve for mols. Then mol = g/molar mass. You have molar mass and mols, solve for g
July 9, 2014

CHEMISTRY
H2 gas at a pressure of 1 atmosphere bubbled over "black platinum" in a 1 M solution of H^+ is arbitrarily set at zero. Black platinum is platinum with a fine coating of extremely small particles (granules) of Pt. That surface apparently gives better reproducibility ...
July 9, 2014

CHEMISTRY
3 M+1...M+2...M+3 Then the two half reactions are M^2+ ==> M^+3 + e M^2+ + e ==> M^1+ then add the two half rxns --------------- 2M^2+ ==> M^3+ + M^+
July 9, 2014

CHEMISTRY
Just think of an example of an element in a compound with the element at its max oxidation state. For example take SO3 where S is +6. It CAN'T be oxidized any further (if this is the max) to go to S7 or S8 or higher) but it can go down to something like SO2 or S or H2S. So...
July 9, 2014

CHEMISTRY
You don't think so or you don't know so. Did you check it? I checked it and all atoms appear to balance. I even found a site in which someone did a lot of work to do it with half reactions. Here it is if you are interested. The numbers match those from another site ...
July 9, 2014

CHEMISTRY
It's not simple but here is what I found on the internet. I don't vouch for its authenticity. 10[Cr(N2H4CO)6]4[Cr(CN)6]3 + 1176KMnO4 + 1399H2SO4 = 35K2Cr2O7 + 420CO2 + 660KNO3 + 1176MnSO4 + 223K2SO4 + 1879H2O
July 9, 2014

CHEMISTRY
I'm confused by your use of the terms reactant/products side. From the original question I thought that was a compound and if it is a compound there is no reason why you can't have an element exhibiting two different oxidation states. In coordination compounds that is ...
July 9, 2014

CHEMISTRY
I would start this way. This is a zero charged compound so the cation must be +3 and the anion part must be -4. We know CN is -1. I always go with +1 for H and -2 for O unless I know better. A value of +4 for C is always a good choice (unless it interferes with my +1 for H ...
July 9, 2014

CHEMISTRY
right. 3KIO3 + 2KMnO4 + H2O ==> 3KIO4 + 2K^+ + 2OH^- + 2MnO2 but that is not the usual way of doing it. Usually we have an ionic equation or a molecular equation but a mixed is not wrong (unless your prof says to write a molecular equation in which case you will have points...
July 9, 2014

CHEMISTRY
Actually, there is nothing wrong with having a "mixed" equation in which you have both molecules and ions. However, in your case you added 5K^+ on the left and you added only 3K^+ on the right. So in changing from the ionic to the molecular equation your molecular ...
July 9, 2014

CHEMISTRY
I think the answer is no but you just didn't finish. You added 5K^+ on the left. 3 of those on the right go with 3KIO4 and the other two go with 2KOH
July 9, 2014

CHEMISTRY
Here is a step by step of what I do for half reactions. This is the LONG way of doing it but I use it for two reasons: 1. It requires the student to know oxidations states. 2. The same procedure is used for acid OR basic solution. And I'm going to omit the spectator ions; ...
July 9, 2014

chemistry
You don't give the density of the solution so I have no way of knowing what the final volume. You also don't give any units for the 190^3 for water. If we assume units are cc If we assume the volumes are additive (and you can make that assumption since the volume of ...
July 9, 2014

chemistry
You must have a reaction in mind. You can't get PbCl2 from MgCl2 without some reaction.
July 9, 2014

chemistry
HNO3 + KOH ==. KNO3 + H2O mols HNO3 = M x L = 2.5 x 0.1 = 0.25 mols KOH = 2.55 x 0.1 = 0.255 Since the reaction is a 1:1 ratio, the HNO3 is the limiting reagent and 0.25 mols H2O will be produced. g H2O = mols H2O x molar mass H2O = ?
July 8, 2014

pharmacy calculations
I answered this for you earlier.
July 8, 2014

ph
I'm not a pharmacist but if that means take 250 mg every 8 hours for 7 days, the total dose is 250 mg/dose x 3 doses/24 hrs x 24 hours/1 day) x 7 days = ?
July 8, 2014

questions
So you want 15 tablets for 1/2 the month (30/2 = 15 days) and another 15/2 day supply so that is 15 + 7.5. You dispense 15 + 8 = 23 tablets unless you want to cut one of the tablets into two. I don't know if you are allowed to cut tablets or not. If so then you want 22.5 ...
July 8, 2014

Chemic
mL1 x %1 = mL2 x %2 mL1 x 98% = 50 mL x 5% Solve for mL 1. Add that amount to 50 mL volumetric flask and make to the mark with distilled H2O.
July 8, 2014

CHEMISTRY
You're right. I turned it around. Sorry about. Just reverse everything and you will have that half reacted. For KPb(OH)3 the cation is K^+ and the anion is Pb(OH)3^- So I would write PbO2 ==> Pb(OH)3^- I assume this is a basic solution. Pb is +4 on the left and +2 on ...
July 8, 2014

CHEMISTRY
In acid solution.? 2H^+ + ClO^- + 2e ==> Cl^- + H2O In basic solution? H2O + ClO^- + 2e --> Cl^- + 2OH^- The way to learn this is to do them. Looking at what someone else has done won't get it. If you will share what your problem(s) is/are in doing these perhaps I ...
July 8, 2014

Chemistry
Yes, and if your prof is picky about the number of significant figures you whould round that to 514 mL. You are limited to 3 from the 325 to start.
July 8, 2014

Chemistry
When it's an all gas sample and the T and P don't change, you may use a shortcut and use volume directly as if they were mols. 2C6H14(g) + 19O2(g) -> 12CO2(g) + 14H2O(g) 325 mL CO2 x (19 mols O2/12 mols CO2) = 325 mL x (19/12) = mL O2 consumed.
July 8, 2014

Chemistry
Use PV =nRT and convert NH3 and O2 (each) to mols = n. Find the limiting regent Convert to mols N2 formed, the use PV = nRT and convert to mL at the conditions listed.
July 8, 2014

Chemistry
I don't think you did anything wrong. v1 = 683 mL p1 = 0.792 atm t1 = 19C = 273+19 = 292K v2 = 733 mL p2 = ? in atm. t2 = 76C = 273+76 = 349K (0.792*683/292) = (p2*733/349) If I used 292.15 and 349.15 the answer is 0.8819 which rounds to 0.882 atm. Usually when something ...
July 8, 2014

Chemistry
(P1V1/T1) = (P2V2/T2)
July 8, 2014

Chemistry
Since they threw that T as 27.2 then I would add to 273.2 to convert to K (I missed that above) so I would use 300.4 and not 300.2. That gives you n = number of mols. The problem asks for grams methane. g methane = mols x molar mass CH4 = mols methane x 16 = ? Note: Check your...
July 8, 2014

Chemistry
P in atm (741/760) = ? atm V in L. 200 mL = 0.200 L T in kelvin. K = 273 + 27.2 = ? With all of those, R = 0.08205.
July 8, 2014

Chemistry
If P is changed to atm [(741/760) = ?] then R is 0.08205 L*atm/mol*K
July 8, 2014

Chemistry
Use PV = nRT and solve for n = number of mols CH4. Then n = grams/molar mass. You have molar mass and n, solve for grams.
July 8, 2014

Chemistry
Ptotal = pN2 + pO2 + pCO2 You have only one unkinown; solve for that. However, note that the units are not the same. Since you want pCO2 in torr (mm Hg), I suggest you change the others to mm Hg.
July 8, 2014

Chemistry
density = mass/volume
July 8, 2014

chemistry
I have no idea which experiment you are conducting or what this problem is about.
July 8, 2014

CHEMISTRY
Here is a simplified set of oxidation state rules. http://www.chemteam.info/Redox/Redox-Rules.html You must understand that oxidation rules are arbitrary rules mostly for bookkeeping. When H is with a metal, such as CaH2, I always give it an oxidation state of -1 because it&#...
July 8, 2014

CHEMISTRY
I'll play your game. What about them? You add H^+ if acid and OH^- if basic along with H2O in the appropriate amounts.
July 8, 2014

CHEMISTRY
You look at the activity series. Here is one on the web. http://www.cod.edu/people/faculty/jarman/richenda/1551_hons_materials/Activity%20series.htm Any metal will displace any ion BELOW it in the activity series. So Al(s) + Ni^2+(aq) ==> Ni(s) + Al^3+(aq) then balance it; ...
July 8, 2014

science
q1 = heat removed to lower T from 10C to zero C. q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial) q1 = 200g x 4.184 x (10) = ? q2 = heat removed to freeze the H2O (change from liquid to solid phase). q2 = mass H2O x heat fusion H2O. q2 = 200 x heat fusion = ? Look up heat...
July 8, 2014

CHEMISTRY
...........AB2 ==> A^2+ + 2B^- I..........solid...0.......0 C..........solid...s.......2s E..........solid...s.......2s Ksp = (A^2+)(B^-)^2 Ksp = (S)(2S)^2 Ksp = 4S^3
July 8, 2014

CHEMISTRY
Ca5(PO4)3F--> 5Ca^2+ + 3PO4^3- + F^- I..solid.......0.........0.......0 C..solid.......5x......3x........x E..solid.......5x......3x........s Ksp = (Ca^2+)^5*(PO4^3-)^3*(F^-) 1E-60 = (5x)^5*(3x)^3*(x) Solve for x = solubility in mols/L.
July 8, 2014

chemistry
You use the water as a way of knowing how much heat was lost or gained. So the salt changed T from zero c to -1.0 C. Since it cooled the water you know the reaction is endothermic; i.e. it took heat away from the water so the rxn required heat and that is endothermic. Rxns ...
July 7, 2014

chemistry
salt + H2O ==> solution - heat or salt + H2O + heat ==> solution. endothermic for #1 and #3. Exothermic for #2. Therefore, for the salt, q is + for 1 and 3 and - for #2. Do you want q/g? You don't have the molar mass so you can't get q/mol. qH2O = mass H2O x ...
July 7, 2014

chemistry
Will this get you started. S on the left is +4; on the rght is +6 Mn on the left is +7 and on the right is +2. I think you failed to add the charge to MnO4. It should be MnO4^-.
July 7, 2014

chemistry
And how would you like for us to help you? The directions are there, the formulas are there, the only thing left is for you to substitute the numbers into the formula to obtain the answers. We shall be happy to check your work.
July 7, 2014

chemistry
This must be a question associated with an experiment you performed. What experiment did you conduct? What was your experience?
July 7, 2014

chemistry
Frankly I don't know what you are asking? The left hand side (Mg and HCl) are the reactants. The right hand side (H2 and MgCl2) are the products. HCl is hydrochloric acid Mg is magnesium H2 is hydrogen MgCl2 is magnesium chloride.
July 7, 2014

chemistry
I've shortened the equation. HAc is acetic acid, RCH2OH is the alcohol and RCOOR is the ester. ..........HAc + RCH2OH --> RCOOR + H2O I.........1.0....1.0........0.......0 C..........-x.....x.........x.......x E.........1.0-x..1.0-x......x.......x Substitute the E line ...
July 7, 2014

chemistry
q = mass concrete x specific heat concrete x (Tfinal-Tinitial)
July 7, 2014

CHEMISTRY
AgCl --> Ag^+ + Cl^- Ksp = (Ag^+)(Cl^-) = about1.8E-10 Compare Qsp with Ksp. In soution (and I'm assuming that adding the salts to 200 mL H2O is the same as producing 200 mL solution). Ag^+ = 1.7/170 = about 0.01 mols. You should go through these calculations yourself; ...
July 7, 2014

CHEMISTRY
This is done the same way as the AgCl problem. Compare Qsp with Ksp.
July 7, 2014

chemistry
See above.
July 7, 2014

Science
15% w/v means 15 g solute in 100 mL solution. [(x g solute)/250 mL] = 0.15 x g solute = 250 x 0.15 = 37.5g HCl Therefore, (37.5g HCl)/(mL HCl + mL H2O) = 0.15 Since mL HCl + mL H2O = 250, you can solve for mL H2O if you know the volume occupied by 37.5g HCl. There is no date ...
July 7, 2014

chemistry
I don't see a question here.
July 7, 2014

Science
no question.
July 7, 2014

Science
See your methyl alcohol problem.
July 7, 2014

Science
% v/v means (volume solute/total volume)*100 = ? = (80/154)*100 = ?
July 7, 2014

Chemistry
........HCHO2 --> H^+ + CHO2^- I........0.2......0......0 C.......-0.006..0.006...0.006 E......0.2-0.006..0.006..0.006 Ka = (H^+)(CHO2^-)(HCHO2) Ka = (0.006)(0.006)/(0.2-0.006) Ka = ? pH = -log(H^+) = -log(0.006) = ? HCHO2 --> H^+ + CHO2^- Le Chatelier's Principle ...
July 7, 2014

Science
% w/w means (grams solute/grams solution)*100 = ? [(15g NaCl/(70+15)]*100 = % NaCl {(70g H2O/(70+15)]*100 = % H2O
July 7, 2014

Science
See your NaCl problem. Same thing.
July 7, 2014

general chem
Don't forget to multiply by 100 to convert to %.
July 7, 2014

chemistry solution preparations
To begin, I don't think this is possible using standard volumetric equipment. I'll explain below. Let's first determine the N of the 37% stuff. The density of 37% HCl is 1.19g/mL so the N is 1000 mL x 1.19 g/mL x 0.37 x (1/36.5) = about 12N for the 37% HCl. How ...
July 7, 2014

CHEMISTRY
Here is the first one in detail (for CO2). 1) 2CO + O2 <==> 2CO2 + 167 kJ and I will rewrite the problem to more clearly show how the heat is treated. Basically the heat is made to look just like an added product (if an exothermic rxn) or a reactant (if an endothermic ...
July 7, 2014

CHEMISTRY
See you last post for that detail explaination I gave for CO2
July 7, 2014

chemistry, biology
I believe you can use the Henderson-Hasselbalch equation for this pH = pKa + log (base)/(acid) You have the zwitterion at the isoelectric point and you are taking away the H^+ from the NH3^+ group. Acid + OH ==> base + H2O etc. You have 480 x 0.02M = 9.6 millimols to start...
July 6, 2014

science
q = mass iron x specific heat iron x (Tfinal-Tinitial) q = 114
July 6, 2014

chemistry
Use PV = nRT and sole for n = number of mols. You must convert 1.00 MPa to kPa.
July 6, 2014

chemistry
I would use PV = nRT. You know n, R, T, and P. Solve for V. Or you may use the other bit of information given that 1 mol at STP occupies 22.4L. So you know that at STP you will have 14 x 22.4L, then use (p1v1/t1) = (p2v2/t2)
July 6, 2014

chemistry
(P1V1/T1) = (P2V2/T2)
July 6, 2014

chemistry
Use PV = nRT and solve for n = number of mols. You must convert cubic meters to L. Then n = grams/molar mass. You know molar mass and n, solve for grams.
July 6, 2014

Chemistry
0.0900M = 0.0900 mols/L solution. g = mols x atomic mass = 0.0900*19 = 1.71g. mass solution = volume x density = 1000 mL x 1.0 g/mL = 1000 grams. mass % = (g solute/grams soln)*100 = ? Convert to mg/L and that is ppm
July 6, 2014

investing
The interest of 6% compounded quarterly is 6/4 = 1.5% each quarter. There are 20 quarters in 5 years. (1.015)20*4,000 = ? = final value of thd $4,000. Interest earned is ?-4000 = ?
July 6, 2014

chemistry step by step for me thank a lot
1 mol Ca(OH)2 removes 1 mol Ca(HCO3)2; therefore, to remove 100 ppm harness (as CaCO3) will require (100 x 74/100) = 74 ppm Ca(OH)2. (Note: To convert harness in ppm CaCO3 to Ca(OH)2, use the coefficients in the balanced equation. That is a 1:1 mole ratio in which 1 mol Ca(OH)...
July 6, 2014

College Algebra I
I don't see a question here.
July 6, 2014

chemistry step by step thank a lot
Is that m or M (molality or molarity) M is molarity m is molality Did you make a typo? Is that question for molarity of the H^+. That is 0.888M (not m). Total molarity, however, is much more than 0.888M.
July 6, 2014

Chemistry
pi = MRT pi = 869/760 = ? M solve for this R = 0.08205 L*atm/mol*k T = 273 + 21 = ? M gives you part b. Then M = mols/L solution, you know L solution and M, solve for mols and that is part c. Part a is mol = grams/molar mass. You know grams and mols, solve for molar mass.
July 6, 2014

Soil Chemistry
I don't get it. There are no carbon-halogen bonds in any of those molecules.
July 6, 2014

chemistry
wavelength = h/mv Convert 1A to m.
July 6, 2014

CHEMISTRY
dG = dH - TdS 0 = 8000 - 298(dS)
July 6, 2014

CHEMISTRY
q = mass In x specific heat In x (Tfinal-Tintial) q = 23J mass = atomic mass In solve for sp.h.
July 6, 2014

chemistry
CO2 is the gas evolved. It is colorless and odorless but will extinguish a burning flame.
July 6, 2014

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