Sunday
March 29, 2015

Posts by drbob222


Total # Posts: 47,990

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

suggestion
Some have put a space (sometimes more than one space) in the address and it sometimes posts that way. Then a tutor posts that URL, deletes the spaces, and the site is SUPPOSED to come up. My experience is that sometimes that works and sometimes it doesn't.
March 27, 2015

Chemistry
I don't know what you want. If the woman uses 96 kJ and the efficiency is 35%, she actually used 96/0.35 or about 274 kJ for the 1 km. (you can do that more accurately) If the car gets 8.7 km/L, then to go 1 km you will use 1/8.7 = ? L gasoline. That x 1000 = mL gasoline ...
March 27, 2015

chemistry
NaOH + HCl ==> NaCl + H2O mols HCl = M x L = ? mols NaOH = mols HCl (since the reaction shows 1 mol HCl = 1 mol NaOH Then M NaOH = mols NaOH/L NaOH. The problem is confusing because it appears there is no unknown; ignore the "approximately 0.1 M NaOH". That's ...
March 27, 2015

chemistry
I guess the last name is Mcilvaine but I couldn't find the other names. However, I found this reference to Journal Biol Chemistry in 1921. Here it is with chemical abstracts number. J Biol Chem 1921, 49, 183. Chem Abs 16-729). You can look up that information at that ...
March 27, 2015

Chemistry
Answer to what?
March 27, 2015

Chemistry
You say equation but you have four equations listed. I don't see a question.
March 27, 2015

Chemistry help
8O16. 1s2 2s2 2p4 2 + 2 + 4 = 8. So there are 4 p electrons.
March 27, 2015

CHEMISTRY
This is a limiting reagent problem and is just like the KClO3 problem EXCEPT you work it twice and take the smaller value. See how many mol product is formed from mols of first reagent. See how many mol product is formed from mols of the other reagent. In limiting reagent ...
March 26, 2015

To CASSIE
I found an error in that ethanol(whiskey) problem. I posted a correction at the original question. If you can't find it let me know and I'll give you a link to it. At this time it's on page 5 or 6 but that will change with time.
March 26, 2015

CHEMISTRY
Balance the equation. You have that. mols KClO3 = grams/molar mass = ? Use the coefficients in the balanced equation to convert mols KClO3 to mols O2. That's mols KClO3 x (3 mols O2/2 mols KClO3) = mols KClO3 x 3/2 = ? Now convert mols O2 to grams. grams = mols x molar ...
March 26, 2015

chemistry
Calculate mL NaOH needed to reach the equivalence point. mLacid x Macid = mLNaOH x MNaOH Let HBz = benzoic acid then HBz + NaOH ==> NaBz + H2O The pH at the equivalence point is determined by the hydrolysis of the salt, NaBz. The (NaBz) = molsNaBz/L = ?. I will call that z...
March 26, 2015

chemistry
Reverse equation 1 and add to the reverse of equation 2 and add equation 3 as is. Remember to change the sign of dH when reversing an equation. Post your work if you have questions.
March 26, 2015

chemistry
I've shown you how to do these above. Try your hand at this. There is little I can do except do it for you and I don't want to do that. If you still have trouble explain in detail what your problem is and I'll try to help you through it.
March 26, 2015

chemistry
Haven't you seen this before? dHrxn = (n*dHf products) - (n*dHf reactants) Post your work if you get stuck.
March 26, 2015

chemistry
q = [mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [Ccal*(Tfinal-Tintiial)] Substitute and solve for q, then convert to kJ.
March 26, 2015

chemistry
q = mass x specific heat x (Tfinal-Tinitial)
March 26, 2015

Science
Do you have choices. I think several words might go there.
March 26, 2015

chemistry
8.3 mol H2S used. Use the coefficients in the balanced equation to convert mols H2S to mols H2O. That's 8.3 mol H2S x (2 mols H2O/2 mols H2S) = 8.3 x 2/2 = 8.3 mols H2O formed. Convert that to grams. grams H2O = mols H2O x molar mass H2O = estimated 149 g (but you need to ...
March 26, 2015

To Ty---chemistry
Did you read Bob Pursley's message? He is telling you that you didn't write the "above reaction". We can't read your mind and you wrote the problem part but not all of the question. Write the equation and we may be able to help.
March 26, 2015

chemistry
PV = nRT n = total mols if you want total pressure. n = mols of each if you want partial pressure of each gas by itself.
March 26, 2015

to tt---chemestry
If you want to post a question, please go to the top of the page and click on "Post a new Question".
March 26, 2015

chemestry
Explain what you don't understand about this and I can help you through it. Or if you have it worked out I'll be glad to check your answers.
March 26, 2015

Chemistry
What mass does that 7.0 mL vinegar have? You need the mass or the density of the vinegar to calculate that. NaOH + CH3COOH --> CH3COONa + H2O mols NaOH = M x L = ? mols CH3OOH = mols NaOH since the equation tells you 1 mol NaOH = 1 mol CH3COOH. Then grams CH3COOH = mols ...
March 26, 2015

chem
You posted a long string of questions a few days ago and I responded that I would be glad to help if you would tell me what you didn't understand. You didn't respond. On this long list I will do the same but I'm not interested in doing a lot of typing so you can ...
March 26, 2015

chemistry
Ptotal = pH2 + pN2 + pO2 + pCH4. Substitute and solve for pCH4 in atm and convert to kPa.
March 26, 2015

Chemistry
Use PV = nRT and solve for n = number of mols. Convert that to grams if needed; i.e., mols = grams/atomic mass.
March 26, 2015

chemistry
You need to look on the boxes or on the internet and find % caffeine in each bag and/or each serving of tea. I think you may come up largely empty handed because the FDA does not require that to be listed. I went through the same kind of thing for decaffeinated coffee and ...
March 26, 2015

Chemistry
2Zn + O2 ==> 2ZnO mols O2 = grams/molar mass = ? Using the coefficients in the balanced equation convert mols O2 to mols ZnO. That will be just mols ZnO = mols O2 x 2. Then convert mols ZnO to grams ZnO with grams = mols x molar mass.
March 26, 2015

Science
If you will show your work on this and the questions you've posted below and explain exactly what you don't understand I will be glad to critique what you have written and give you some helpful hints.
March 26, 2015

practical chemistry
Please tell me what you don't understand about this. Do you realize you're asking someone to write a book on qualitative analysis? That's what book stores are for.
March 26, 2015

AP Chemistry
This question can't be answered without know the pH of the solution. If you know the pH of the solution, substitute it into the Henderson-Hasselbalch equation and solve for the ratio of base to conjugate acid.
March 26, 2015

oops--Chemistry
Note that 12.67 is mL and must be converted to L before multilying by M to get mols.
March 25, 2015

Chemistry
I don't follow much of what you've done. That doesn't mean it isn't right. Apparently your database thinks you haven't corrected for the blank and I don't see that in your work either. Do you know how to do that? From your post I surmise that the data ...
March 25, 2015

Chemistry
I'm leaving this one for you. After the last two you should be able to do large chunks of this on your own. If you have trouble post the balanced equations and explain in detail what you don't understand about the problem/next step. I can help you through it but one ...
March 25, 2015

Chemistry
I've gone over this many times and found nothing UNTIL I checked the formula. I found two errors. The first is mols Cr2O7^2- x 3 (not divided by 3) = mols ethanol (and you should have caught that too). When I first responded I balanced the equation with dichromate + C2H5OH...
March 25, 2015

Chemistry
No. But large chunks are ok. Here is your work. moles for dichromate initially. That's 0.024M x 0.05L = 0.0012 mol Fe^2+ used in the back titration 0.1255M x 0.019L = 0.0023845. Convert that to mols Cr2O7^2- (that's 1/6 the iron(II) So 3.974 x 10^-4 and subtract from ...
March 25, 2015

Chemistry
3C2H5OH + Cr2O7^2- ==> 2Cr^3 + 3CH3COOH Again, I've not worried about the rest of the stuff here. In the back titration part. 6Fe^2+ + Cr2O7^2- ==> 6Fe^3+ + 2Cr^3+ How many moles did you have for dichromate initially. That's M x L = ? How many mol Fe^2+ used in ...
March 25, 2015

Chem
mols KMnO4 is M x L = 0.02805 x 0.1 =? You dissolved the sample in 50 mL, the titration with KMnO4 was 28.05 mL. Correct that step and follow through with the rest of the problem The set up looks ok for the rest of it. I ran through it and obtained approx 0.35 g Fe/g sample.
March 25, 2015

Chem
I don't like the Fe2 business. Probably you mean Fe^2+ and even though that is a perfect way of writing it the Fe2 bit makes one think iron is diatomic and it isn't. 5Fe^2+ + KMnO4 ==> 6Fe^3+ + Mn^2+ (I know the equation is balanced but the redox part is and we aren...
March 25, 2015

CHE 107
SA ==> ASA You want 90/0.95 = 94.7g ASA in order to produce 90 g at 95% yield. mols ASA = 94.7/molar mass ASA mols SA = mols ASA since the rxn is 1:1. g SA = mols SA x molar mass SA.
March 25, 2015

chemistry
The nice thing about balanced equations is that they allow you to convert mols of anything to mols of any other material in the equation. Use the coefficients. Here is how you do a. I'll let you do b. mols N2 = 1.75 x (2 mols N2/2 mols NH4NO3) = 1.75 x 2/2 = 1.75 mols N2. ...
March 25, 2015

Chemistry
In essence you will place the same number of mols that occupy 1.81 m^3 into a volume of 0.0155 m^3 so the pressure will increase by the inverse of the volumes; i.e., 1 atm (the initial P) x (1.81/0.0155) = ? atm
March 25, 2015

chemistry
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. 4Cr + 3O2 ==> 2Cr2O3 mols Cr = grams/molar mass = ? mols O2 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols Cr to mols Cr2O3. Do the ...
March 25, 2015

college chemistry
It most certainly does. I worked it and it produced exactly that number. If you will post your work I will find the error but I expect I can tell you without looking at your work. You probably wrote (0.1)(2x). It should be (0.1)(2x)^2 and x = 4.44E-5M big as life.
March 25, 2015

chemistry
NaOH + HCl ==> NaCl + H2O mols NaOH = M x L = ? mols HCl = mols NaOH since the ratio in the equation is 1:1. Then M HCl = mols HCl/L HCl. YOu knowL and mols, solve for M
March 25, 2015

chemistry
Set up K expression, substitute the equilibrium mixture of N2 and H2 and solve for NH3.
March 25, 2015

chemistry
Follow up on your later post but the answer really is 4.44E-5M. You just didn't set it up right. Probably didn't square F^- OR didn't include the 2 which squared is 4.
March 25, 2015

chemistry
This is a Ksp problem with a common ion add on. Two equilibria. Sr(NO3)2 is a strong electrolyte and ionizes 100% in solution. ..........Sr(NO3)2 ==> Sr^2+ + 2NO3^- I..........0.100........0........0 C.........-0.100.......0.100...0.200 E...........0..........0.100...0.200 ...
March 25, 2015

Physical Science
This is a limiting reagent (LR) problem and I know that because amounts are given for BOTH reactants. I solve these the long way--there is a shorter way but I think it's harder to explain, especially by typing. 2Bi + 3F2 ==> 2BiF3 mols Bi = grams/molar mass = ? mols F2...
March 25, 2015

Chemistry CP
Because of intermolecular forces, H2O is not a simple H2O molecule but a cluster much like (H2O)n where n represents a small whole number. The last time I read something on this n was something like 6 at room temperature but was 1 as steam at 100 C. So the molar mass of H2O is...
March 25, 2015

chemistry cp
If you mean increasing boiling points from top to bottom, yes.
March 25, 2015

chemistry cp
What's your main problem with this. You do it with intermolecular forces, H bonding, molar mass, etc.
March 25, 2015

chemistry
Technically, H bonding occurs between H and O, N, F atoms. Look at H bonding of HF.
March 25, 2015

Chemistry
You need to supply all of the question.
March 25, 2015

chemistry
Is that m/10 or M/10. I assume you know there is a difference. Also you must assume the volumes are additive. Technically they are not. 0.1M x (400/600) = ?
March 25, 2015

Chemistry
Convert pH 2.3 to (H^+) with pH = -log(H^+) Then convert 9 cups H2O to mL and I'll call that ?mL Then (H^+) x (volume original lemon juice/? mL in 9 cups) = ? = (new H^+ in mols/L. Convert that back to pH.
March 25, 2015

chemistry
3 mols x 24 mol/dm^3 = ?
March 25, 2015

Physical science
I didn't look up the values but my best guess is, from slowest to highest: water, methylated spirits, ethanol, nail polish remover.
March 25, 2015

Physical science
Look up the vapor pressure of each at whatever temperature is of intrest and compare. The highest vp evaporates fastest.
March 25, 2015

chemistry
Tartaric acid has two ionizable H^+ but k1 is very close to k2; I assume the 0.50 mols reduced both and that's the equivalence point to which you refer. H2T + 2NaOH ==> Na2T + 2H2O 0.5 mol NaOH x (1 mol H2T/2 mols NaOH) = 0.5 x 1/2 = 0.25 mols H2T
March 25, 2015

Chemistry
You can get pH from HH equation. From that calculate pOH. I assume you can get K easily. Then for H3PO4 calculate the fraction in that from from H3PO4 = (H^+)^3/D where D is (H^+)^3 + k1(H^+)^2 + k1k2(H^+) + k1k2k3
March 25, 2015

chemistry
ln(No/N) = kt
March 25, 2015

chemistry
http://www.chemguide.co.uk/physical/basicrates/orders.html
March 25, 2015

chemistry
mols H3PO4 = 2.00 # molecules H3PO4 = mols x 6.02E23 # atoms O is 4x that.
March 25, 2015

math
That's right.
March 24, 2015

math
If it is 31 F now and it drops by 15 F, then the new T will be 31-15 = ?
March 24, 2015

chemistry
Use the HH equation. I would start with any convenient number for the initial concentration of HA. Any number will work but I would choose 1 to make it easy. So if we start with 1M HA it will be 0.25 at 75% titrated and the salt (the base) will be what? Plug those values into ...
March 24, 2015

Chemistry
I would write the reaction as Zn|Zn^2+(0.4M)||Zn^2+(1.45M)|Zn Zn ==> Zn^2+(0.40) + 2e at the anode Zn^2+(1.45) + 2e ==> Zn at the cathode ----------------------------- Zn + Zn^2+(1.45M) ==> Zn + Zn^2+(0.4) Then Ecell = Eocell - [0.05916/2]*log Q where Q is 0.40/1.45. ...
March 24, 2015

math
b is correct for #1.
March 24, 2015

chemistry
You remember that 1 mol of a gas at STP occupies 22.4 L.
March 24, 2015

chemistry
You can convert any mol in this equation to any other mol in this equation by using the coefficients in the balanced equation. For example: How many mols NaN3 are necessary to obtain 10 mols Na? 10 mols Na x (2 mols NaN3/2 mols Na) = 10 x 2/2 = 10. The number on the top of the...
March 24, 2015

Chemistry
2C4H10 + 13O2 ==> 8CO2 + 10H2O This is a limiting reagent (LR) problem. mols C4H10 = grams/molar mass = ? mols O2 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols C4H10 to mols CO2.' Do the same for mols O2 to mols CO2. It is likely ...
March 24, 2015

Chemistry
right
March 24, 2015

Science
If that is 0.3 ohm apiece, then total R is 0.1. Then E = IR = 8*0.1 = ? Is that right?
March 24, 2015

Chemistry
Why don't you post this under a physics banner? This is more a physics question than a chemistry question. I would answer that F = m*a. Since m is the same and a is the same (it isn't moving) then F must be the same. But I would feel better if you got your answer from ...
March 24, 2015

Science
Look up life cycle of mosquito on Google. Each stage will be there. Or see if it's in your text (and don't tell me you don't have a text).
March 24, 2015

Analytical Chemistry
I found it. http://www.jiskha.com/display.cgi?id=1425612744
March 24, 2015

Analytical Chemistry
Didn't I do one of those "list all species of H3PO4" for you recently? I don't want to write all of that stuff again. Look at your notes (you do take notes here I would think). What can you do on your own?
March 24, 2015

chemistry
Look up Ksp for PbCl2 and solve for solubility PbCl2 @ 25 C. (You will get solubility at 25C because that temperature is used for determining Ksp). Then convert M to mols/L, then to grams/L. If the solubility is >1.0 g no ppt; if < 1.0 g, yes a crystallization will occur.
March 24, 2015

Chemistry
I don't think so. dHrxn = (n*dHf products) - (n*dHf reactants) 4*-973.49 = (8*dHfCO2 + 10*dHf H2O) - (4*dHf glycine) Solve for dHf glycine
March 24, 2015

chemistry
I've been burned on question like this in the past and vowed to leave them unanswered. However, it appears to me that c is the way to go. I'm assuming when they say "less work" that is compared to the work available if the piston were to fully compress the ...
March 24, 2015

Chemistry
No, I don't get that answer. I only went through it once so I could have made an error. Here is what I did. The reverse of twice equation 3 + the reverse of equation 2 + 3x equation 1 and that gave me 6FeO + 6CO ==> 6Fe + 6CO2 and -101 kJ/mol. That divided by 6 = about...
March 24, 2015

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