Saturday
May 23, 2015

# Posts by DrBob222

Total # Posts: 49,010

ap chem
96,485 Coulombs can plate out approx 209/3 grams Bi. You need a better answer than that estimate. How many coulombs do you have? That's C = amperes x seconds = approx 5.60 x 23.8 x 60 s/min = ? 209/3 x ?coulombs/96,485 = g plated out.
May 22, 2015

AP Chem ---
k = 0.693/t1/2 Solve for k and substitute into the below. ln(No/N) = kt No = 100 N = 25 k from above Solve for t in days.
May 22, 2015

AP Chem ---
May 22, 2015

AP Chem ---
zXw ==> 11Na24 + 2He4 z = 11 + 2 = ? (13 and that is Al) w = 24 + 4 = ? So the equation is 13Al28 ==> 11Na24 + 2He4
May 22, 2015

Chemistry
24Cr51 ==> 25Mn51 + -1eo So that must be a beta decay.
May 22, 2015

Chemistry
Pb is +2 on the left and +2 on the right O is zero on the left and -2 on the right. S is -2 on the left and +4 on the right.
May 22, 2015

Chemistry
Oxidation is the loss of electrons. Cu is +2 on the left and zero on the right. H2 is zero on the left and +1 on the right.
May 22, 2015

is the ph= o for the H+
10^-0.249 = 0.56 so from these numbers the pH = 0.56.
May 21, 2015

is the ph= o for the H+
If you will show your work I can check it.
May 21, 2015

CHEMISTRY ( I NEED HELP AGAIN QUICK )
The first question is a limiting reagent problem. You work it the same way as the previous LR problem. The second one is not a LR problem but a simple stoichiometry problem. So instead of converting two reactants to mols of product and choosing which one to use you have only ...
May 21, 2015

chem 100
a. .........PbI2 ==> Pb^2+ + 2I^- I........solid.....0.......0 C.........-x.......x.......2x E........solid.....x.......2x Sustitute the E line into Ksp expression and solve for x = (Pb^2+) b. You need Kc for the reaction and you can get that from the sum of the two ...
May 21, 2015

Chem 100
http://www.chemistry.wustl.edu/~edudev/LabTutorials/naming_coord_comp.html
May 21, 2015

Chemistry (I NEED HELP REAL QUICK)
This is a limiting reagent (LR) problem and you know that because amounts for BOTH reactants are given. I work these the long way. BaCl2 + MgSO4 ==> MgCl2 + BaSO4 mols BaCl2 = M x L = ? mols MgSO4 = M x L = ? Use the coefficients in the balanced equation to convert mols ...
May 21, 2015

chem
You don't have a workable equation.
May 21, 2015

Chemistry
NF3 is trigonal pyramidal; therefore, it is not symmetrical in three dimensions. NF4^+ is tetrahedral and is not polar because the N-F polar bonds balance each other. That is not true for NF3.
May 21, 2015

science
I would choose a,b,d with a note that heat energy is the only end product of the transfer. The others are USING electrical energy to make the transfer from an electromagnetic wave in space that contains the video and audio signals into light and sound.
May 21, 2015

science
I don't think much of this question. It's worse than the one above it. The answer is it depends upon where you place it. If you place it on the stove top burner it is mostly conduction. If you place it in the oven it is mostly convection but with some radiation.
May 21, 2015

Chemistry
It takes 96,485 coulombs to deposit 65.4g/2 = estimated 33 g Zn You want 81.5 g so that will require how many coulombs? That's 96,485 x (81.5/33) = ? C. Then C = amperes x seconds. You know C and amperes, solve for seconds. Remember that 33 g is an estimate so you need to ...
May 21, 2015

Chemistry
The Al is not right. Al atoms balance. H atoms balance O atoms balance. Charge does not balance. You have 6- on the left and zero on the right. You can fix that by placing the 3e on the right. The CrO4^2- half reaction is balanced. The total equation is not balanced. Al is ok ...
May 21, 2015

Chemistry
H2 gas is released. You can test it by collected some of the gas in a test tube (it is lighter than air so it floats UP) and hold a lighted match at the mouth of the test tube. A slight explosion shows the presence of H2 gas. Zn + 2HCl ==> H2 + ZnCl2.
May 20, 2015

college chemistry
You have some number = (H^+)^14 You can do it two ways. The easier way I think is to key in the number, punch the x^y button on your calculator, and key in 0.07142 (that's 1/14), then hit the enter or = button depending upon how you get the answer out of the calculator. ...
May 20, 2015

college chemistry
You fill in all of those numbers. (H^+) is the only unknown in the entire equation (I know I said calculate Q but of course you can't do that). After you've solve for H^+ you convert to pH. The easy way to do that is to start out with Ecell = EoCell - 0.05916/n)*log Q ...
May 20, 2015

college chemistry
Ecell = EoCell - 0.05916/n)*log Q Ecell is given. Calculate Eocell the usual way n = 6 Q = (Al^3+)^2(Cr^3+)^2/(Cr2O7^2-)(H^+)^14 and you plug in the concentrations given in the problem to find Q.
May 20, 2015

college chemistry
You need to look up the reduction potentials and explain what you don't understand. Do you know what an oxidizing agent is? reducing agent?
May 20, 2015

ap chemistry
Refer to the problem just below that I worked for you (the Cu one).
May 20, 2015

college chemistry
Look up the reduction potentials. The most positive reactions is #1.
May 20, 2015

college chemistry
Coulombs = amperes x seconds = ? Let's estimate that at 3000 C. We know 63.54/2 = estimated 32 grams Cu will be deposited by 96,485 C. Therefore you will have deposited 32 x 3000/96,485 = ?
May 20, 2015

college chemistry
I don't know what you're asking. For b, you want to know anode and cathode for which reaction? The spontaneous or non-spontaneous cell. For the cell given, look up each half reaction and calculate the Ecell. If it is +, the reaction proceeds as written and the anode is...
May 20, 2015

Chemistry
Write and balance the three steps as equations. Convert 118 kg isobutylbenzene to mols. mols = grams/molar mass = ? Use the coefficients in the balanced equation to convert mols isobutylbenzene to mols ibuprofen. Multiply by 0.24, then convert mols of product to grams. g = ...
May 20, 2015

Chemistry
Rf value = (distance by unknown/distance by solvent) Substitute and solve for distance by the 0.81 material and the 0.21 material, then subtract one from the other to find the distance between them.
May 20, 2015

chemistry
What kind of a course is this? What are your information sources? text, notes, internet, other? I'm asking these questions because I don't understand why these answers don't stand out in your resource material.
May 20, 2015

Chemistry
0 = qcal + qcombustion 0 = Ccal*(delta T) + qcombustion 0 = 7854*delta T + qcomb Solve for qcomb Then q/mol = (qabove/1.76)*molar mass quinone
May 20, 2015

chemistry
Another. You didn't tell us how much metal(III) chloride OR how much HgI2 was formed. We beed one of them.
May 20, 2015

chemistry
The lower case i is not correct. It should be, for example, mercury(II) chloride. What does metal(iii) mean? Is that an M^3+ ion. And where does the I of HgI2 come from? I think your post needs editing and re-posting.
May 20, 2015

cell notation
What you have is ok under most circumstances. The molarity of the Zn ion and H ion usually are shown. Usually we write (g) after H2. Finally, usually we add a comma and Pt since H2 gas is non-conductive. If these were 1 M concentrations I would do it this way. Zn|Zn^2+(1M)||H...
May 20, 2015

physical science
Because the column of air is not as high (nor as dense) which means it has less weight.
May 20, 2015

Chemistry
delta T = i*Kf*m We need i, Kf, m. i = 2 for NaCl Kf we know for water is 1.86 c/m 1.25 kg ice eventually becomes 1.25 kg water. So we need m. m = mols/kg solvent. We have kg solvent so we need mols. mols NaCl = grams/molar mass. You know grams and molar mass, solve for mols ...
May 20, 2015

Chem
For #1. 0.01 M HCl. So H^+ = 0.01 M and since (H^+)(OH^-) = 1E-14, then (OH^-) = 1E-14/0.01 = 1E-12 M For #2. (NaOH) = 0.001 M so (OH^-) = 0.001 M Then (H^+)(OH^-) = 1E-14 and (H^+) = 1E-14/0.001 = 1E-11 M I'm confused by your question. If you know (H^+) OR (OH^-) you ...
May 20, 2015

Chem
(H^+) from HCl = (HCl) since HCl is 100% ionized. Then (H^+)(OH^-) = Kw = 1E-14 That allows you to calculate OH^-, the pH = -log(H^+). For NaOH, it is 100% ionized; therefore, OH^- = (NaOH). Then solve for H^+ and pH. Ba(OH)2 is a strong electrolyte; therefore, [Ba(OH)2] = 0....
May 20, 2015

Chemistry
Answered in your previous post. I would correct you on the carbon dioxide. That carbon(2) oxide technically means carbon(II) oxide and that means carbon with a valence of 2; however, CO2 has carbon with a valence of 4. Carbon dioxide should be called carbon dioxide and should ...
May 20, 2015

Chemistry
You CAN form two products. K2CO3 forms first. Continued passage of CO2 produces KHCO3.
May 20, 2015

Chemistry
See my responses above for CO2 vs CO.
May 20, 2015

Chemistry
You could try this yourself and see what happens but I believe the solution cools which makes it endothermic.
May 20, 2015

Chemistry
The burning candle is extinguished (because there is no oxygen to sustain the flame).
May 20, 2015

CHEMISTRY
1.836 g/mL x 50 mL = ?
May 20, 2015

chemistry
2 m is 2 mols/kg solvent 2 mols NaOH = 2*40 = 80 g NaOH Total mass solution is 1000 + 80 = 1080 g. mass = volume x density. You know the density and mass, solve for volume, then M = mols/L
May 20, 2015

physical science
momentum = mass in kg x speed in m/s. You will need to change 12 mph to m/s. 1 mph = 0.447 m/s.
May 19, 2015

physical science
You must provide much of this informaton from your text/notes/other sources. Convert 5 lbs Au to grams. There are 453.6 g in 1 lb. Look up the melting point of Au, in degrees C and convert that to kelvins. That is done with K = 273 + C. Look up the specific heat Au. Look up ...
May 19, 2015

Chemistry
mols BaCl2 = grams/molar mass = ? Use the coefficients in the balancd equation to convert mols BaCl2 t mols BaSO4. Then convert mols BaSO4 to grams. g BaSO4 = mols BaSO4 x molar mass BaSO4.
May 19, 2015

college chemistry
My crystal ball is a little fuzzy tonight. What did you do? What was the initial concentration? What was the new concentration? What was the other electrode? Tell us what you did?
May 19, 2015

chemistry
.........BaF2 ==> Ba^2+ + 2F^- I.......solid.....0........0 C........-x.......x........2x E.......solid.....x........2x Ksp = (Ba^2+)(F^-)^2 You know (Ba^2+) = 2E-3M Solve for F^- ..........ScF3 ==> Sc^2+ + 3F^- I.........solid.....0.......0 C..........-x.......x..........
May 19, 2015

collage chemistry
You must be over thinking this problem. What's the volume of the box? That's 4m x 2m x 3m = 24 m^3 So the concn in the box must be 1500 mg/24 m^3 = ?
May 19, 2015

Chemistry
I agree that CH4 is the LR but here is how you prove it. Convert 2.05 mols CH4 to mols needed for O2. 2.05 mols CH4 x (2 mols O2/1 mol CH4) = 4.10 mols O2 needed. Do you have that much? Yes. So CH4 is the limiting reagent.
May 19, 2015

caley
See my response above.
May 19, 2015

To Kibito
Yes, those are the ones. I may be able t find them with the time.
May 19, 2015

To Kibito
I posted a response to your last two queries and now I can't find your original question nor my answers. Did you get everything taken care of on that H2S problem? My final answer was x = 0.00167 but also talked about how to solve the cubic equation using a web calculator ...
May 19, 2015

science
Ca(OH)2 + CO2 ==> CaCO3 + H2O mols CaCO3 formed = grams/molar mass = estimated 0.2 mol. NaHCO3 + HCl ==> NaCl + H2O + CO2 Looking at the equations, you had 0.2 mol CaCO3 formed, and that was produced by 0.2 mol CO2. From the lower equation, for form 0.2 mol CO2 you need ...
May 19, 2015

Science
lead(II) nitrate The "old" name is plumbous nitrate
May 19, 2015

science
I looked up the definition on two web based dictionaries and didn't like them. Here is a definition by Wikipedia. http://en.wikipedia.org/wiki/Mixture
May 19, 2015

Science
You want a landform map.
May 19, 2015

Science (Chemistry)
Yes, that's all you do. And since I made that error, the estimate I gave you last night of 0.0017 for x will be slightly lower. I have recalculated and I obtained 0.001671 and although that's too many significant figures you almost must use all of them when you check ...
May 19, 2015

Science (Chemistry)
No. You're right and I made an error. I multiplied but didn't double it as I should have done. That means that my estimated answer I gave you last night is slightly off. Thanks for letting me know.
May 19, 2015

Science
Why don't you repost and star those answers at the question. This scrolling up and down a half dozen times for each question takes too long. I agree with answers to 1 and 2.
May 19, 2015

science
q = mass ice x heat fusion
May 19, 2015

Chemistry
Here is a very good site for redox in general including balancing equations. Let me know if you have any questions but explain clearly what you don't understand. http://www.chemteam.info/Redox/Redox.html
May 19, 2015

Chemistry
You're right but the - you have for 400-mL is not needed.
May 19, 2015

Chemistry
1. (V1/T1) = (V2/T2) What throws students on this kind of problem is that they don't give an initial volume and you must use that and what that volume changes to. The easy way to do these is to assume an initial volume for V1 (any convenient number will do), then take 1/2 ...
May 19, 2015

chemistry
1. The cell potential would drop to zero since the circuit has been broken. 2. Yes, Sn will displace Cu^2+. Sn(s) + Cu^2+ ==> Sn(2^+) + Cu(s) Here is the activity series. https://www.google.com/search?q=activity+series&tbm=isch&imgil=COkl6kSFzOOkZM%253A%253BO97ZY4ZQ8eeUCM%...
May 19, 2015

college chemistry
Zn + Cu^2+ ==> Cu + Zn^2+ 2. I assume when you say standard Zn electrode you mean (Zn^2+) = 1 M. Zn|Zn^2+(1M)||Cu^2+(0.001M)|Cu 3. Ecell = Eocell - (0.05916/n)*log Q n = 2 for the transfer of 2 electrons. Q = (Zn^2+)(Cu)/(Cu^2+)(Zn) Fo that (Zn^2+) = 1M; (Cu)=(Zn) = 1; (Cu^...
May 19, 2015

Science (Chemistry)
When you reach that point you have not copied the complete equation. What you have is K = 4.2E-6 = 4X^3/(0.07-2x)^2. When I expand the denominator I obtained 4.2E-6 = 4X^3/0.0049-0.14x + 4X^2 To proceed you "cross multiply" and get 4.2E-6 * (0.0049 - 0.14x + 4X^2) = ...
May 19, 2015

chemistry
When I was in school and I got a problem like this I changed up the problem a little to see if I couldn't figure it out. So let's change the problem. We have 250 g sample that is 50% water. How much water is there. We know almost instantly that is 125 g H2O available. ...
May 19, 2015

science
NaHCO3 + CH3COOH=>CH3COONa + H2O + CO2 8.4........20g.........x......y....4.4 So you start with 20+8.4 = 28.4 You lose 4.4 which should leave you with 24.0 grams. (In case you're interested, that's the CH3COONa(that's x = 8.2 g) + H2O(that's y=1.8 g) + ...
May 19, 2015

Chemistry
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O millimols H2SO4 = 21 x 0.54 = approx 11 but you need a better answer than that. mmols NaOH = 39 x 0.12 = estimated 5. Which is the limiting reagent. 5 mmols NaOH will require 2.5 mmols H2SO4. You have that much; therefore, NaOH is the ...
May 19, 2015

chemistry
How many equivalents do you want? That's L x N = equivalents = 60 x 0.5 = 30 How many grams is that? That's # equivalents x equivalent weight = 30 x 40 = ? grams if it were pure NaOH. But it is only 88% pure so you need ? grams/0.88 = ? The reasoning goes like this. I ...
May 19, 2015

100^12/100^8 = 100^12-8 = 100^? When you move the 100^8 from the denominator to the numerator, you change the sign of the exponent and 12-8 = 4.
May 19, 2015

Chemistry
Ah, I see. You figures it out while I was typing. Good work. Note that for a redox equation to balance it must balance three things. 1. atoms must balance (yours did) 2. charge must balance (yours did) 3. e lost must = e gained. (yours didn't. Yours had 5e gained and 2 ...
May 18, 2015

Chemistry
I think your MnO4^- is right for acid solution. Br2 is also. Perhaps it the electrons. Let's check. MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O 2Br- ==> Br2 + 2e So multiply eqn 1 by 2 and eqn 2 by 5. Apparently you used some other numbers. This is what you should get. 2MnO4...
May 18, 2015

Chemistry
It may appear to be correct but it isn't. The atoms balance and the charge balances BUT there are two things wrong. 1. A balanced equation never has electrons on either side amd you have 3e on the left side. 2. The electrons lost do not balance the electrons gained. How ...
May 18, 2015

Chemistry
Note the correct spelling of celsius. The easy way to do these is to assume an initial volume, any number that is convenient will do. Then the new volume is 0.04 x that initial volume. {P1V1/T1) = (P2V2/T2) Remember T must be in kelvin.
May 18, 2015

Science (Chemistry)
First, I think you made a typo. I assume you meant to write H2S ==> 2H2 + S2 Second, you need to explain to me how you obtained a,b,c without any numbers. You need numbers to calculate a reaction quotient. Why didn't you provide your answer and what you did. That way I ...
May 18, 2015

Chemistry
q = mass Fe x specific heat Fe x (Tfinal-Tinitial) Substitute and solve for Tfinal
May 18, 2015

Chemistry
Do you have anything (a graph, table, etc) that tells you (or allows you to obtain) the solubility of KNO3 at 70 C and 45 C?
May 18, 2015

science
This is similar to the problem I worked for you two days ago with the 100 g iron producing rust. http://www.jiskha.com/display.cgi?id=1431803547
May 18, 2015

science
Balance the equation. Then mols AgNO3 = grams/molar mass = ? Use the coefficients in the balanced equation to convert mols AgNO3 to mols NO2 and do the same to convert to mols O2. Add mols NO2 + mols O2 Convert total mols of the gas to volume in liters by L gas = mols gas x (...
May 18, 2015

chemistry
I like to do these in millimoles. Although technically they are to be done with concentration in molarity, mmols works just as well and I think it is easier. You have two equations. Let a = acid and b = base. eqn 1 is a + b = 500 x 0.75 mmols. eqn 2 is 7.5 = pKa2 + log (b/a) ...
May 18, 2015

Biochemistry
I have no idea what this question means. My best suggestion is that if you understand the question that you use the same kind of process I used for that phosphate buffer I did last night.
May 18, 2015

Chimistry
heat lost by metal + heat gained by water = 0 [mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial) = 0 Solve for specific heat metal Note that you have calculated the specific heat capacity of the metal. The problems asks ...
May 18, 2015

Biochemistry
I didn't answer this last yesterday when you first posted because the post was screwed up. It looks ok now. You want 100 mL of 200 mM phosphate buffer. That is 100 mL x 0.2 M = 20 millimols phosphate. So you know base(b) + acid (a) = 20. That is equation 1. You obtain ...
May 17, 2015

Chemistry-Decay
You didn't type in the initial rate of decay but I THINK it is 15 dps. Check that out and use the correct value. Also you didn't include the half life but I think that is 5730 years. Again, check that out and use the correct value. First, determine k. k = 0.693/t1/2 ...
May 17, 2015

Analytic Chemistry
I think you need to redo the ICE table and I don't understand what (or why) you did with the NaI. The (AgNO3) for the first addition is 0.0260 x (15.00/15.50) = 0.02516 M. The NaI is reduced likewise as 0.0130 x (0.5/15.5) = 0.0004194 and you should check these numbers and...
May 17, 2015

Science 8
Light travels with different speeds in different materials. It travels at 3E8 m/s in a vacuum but slower in air and slower still in glass. Because it travels with different speeds it is refracted when moving from one medium to another. I think you can get at least two ...
May 17, 2015

physics
momentum before = momentum after 4*10 = 4*5 + 10*x
May 17, 2015

chemistry
dG = dH - TdS Substitute and solve for dG. Then dG = -RTlnK Substitute and solve for K.
May 17, 2015

College Chemistry
I don't mind helping you through it but tell me what you know about this. Just working the problem won't help you. I assume part a means the equation for the SPONTANEOUS REACTION or is it the reaction for the way the cell is set up (and that could be spontaneous). So ...
May 17, 2015

M = mols/L (that's the definition--start with that). Then mols = grams/molar mass = ? and I'll estimate that at approx 25/40 = about 0.5 mol (remember this is only close so yo need to redo the math to obtain a more accurate answer). M = mols/L. You know mols and you ...
May 17, 2015

Chemistry - Equilibrium
I tried to find some graphs; here is the best I could do.http://en.wikipedia.org/wiki/Reaction_coordinate You're right with your statement about catalysts and I agree that the last two change. I am ambivalent about c because of the wording. Adding a catalyst certainly ...
May 17, 2015

Chemistry
The approximate concentration of the HCl solution that you have on hand is 0.5M. Therefore, if you transfer 10.00 mL of that HCl solution to an Erlenmeyer flask, the concentration of the HCl still is approximately 0.5M. Since HCl is a strong acid and ionizes 100%, the ...
May 17, 2015

Math (Solving Equations)
I think what "you do from here" is to expand and factor the terms; however, can you subtract 2x/3 from X^2/6. Those are not like terms. I don't believe you can do that, at least not legally. Therefore, I don't agree with the "getting here" part. ...
May 17, 2015

Chemistry
Look up the redox potential of each and note that the problem SHOULD have specified acid or basic conditions.
May 17, 2015

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