Tuesday
January 27, 2015

Posts by DrBob222


Total # Posts: 46,592

chemistry
You get better service if you use the same screen name. Look up teflon on google.
November 7, 2014

chemistry
I would choose c.
November 7, 2014

Chem please help
right
November 7, 2014

Chem please help
I just answered your earlier post.
November 7, 2014

chemistry
M1V1 = M2V2 15.0*17.6 = 0.495*mL
November 7, 2014

Science
mols = M x L = ?
November 7, 2014

Chemistry due soon
Nope. If the reaction as written is -56 kJ/mol then it would be -112 kJ for 2 mols and rhe REVERSE of that is +112 kJ/2 mol.
November 7, 2014

Chemistry
Actually, I don't know and perhaps another tutor can add to this but I think the H bonding in solids is quasi-permanent and there is not the transitions one sees in liquids. I say this because H bonds adds to the rigidity of polymers such as nylons and polyamides and if ...
November 7, 2014

chemistry due soon
Why not add arrows for clarity? That's what they're for. Have I put them in at the right place? Only the author knows. A. C(s) + 2S(s)==> CS2(l) + 89.4 kJ B. C(s) + 2S(s) + 89.4 kJ ==>CS2(l) C. C(s) + 2S(s) + 89.4 kJ ==> CS2(l) + 89.4 kJ D. C(s) + 2S(s) ==>...
November 7, 2014

Chem
K = (NH3)(HCl). Substitute and calculate K.
November 7, 2014

chemistry
b,c
November 7, 2014

Chemistry
1s2 2s2 2p3
November 7, 2014

chemistry
It has a double bond so is an alkene.
November 7, 2014

science help!!!!!!!!!
I would choose D also.
November 6, 2014

Chemistry due soon
You're right. It's a or b but it is not kinetic energy. http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Enthalpy
November 6, 2014

physical science
6,500 J x 0.12 = ? B. 1 J/s = 1 watt.
November 6, 2014

chemistry
I believe David used the wrong symbol (among other issues); i.e., He is zero electron affinity (but isn't in the list but Ne is zero also) and I is the largest. So from lowest to highest it is Ne<Na<Bi<Te<I The problem ask to rank from most positive to most ...
November 6, 2014

Chemistry
I believe you have them grouped right except for N^3- but you don't delineate between the groups very well. I think you should put them in "boxes" so to speak, something like this just to separate them. Ne Mg2+ S2- Ti4+ K+ He Li+ Be2+ I think N3- should go with ...
November 6, 2014

Chemistry
You got it. Great! Good work.
November 6, 2014

CHEMISTRY
You need to post ALL of the problem.
November 6, 2014

Chemistry
What's the density of the solution? mols C6H12O6 = grams/molar mass = ? volume solution = mass/density Then M = mols/L solution. Or did you mean molality =m If so then m = mols solute/kg solution = mols from above/0.560 = ?
November 6, 2014

chemistry
http://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/N2O5-lewis-structure.html
November 6, 2014

chemistry
Ptotal = pO2 + pH2 + pN2
November 6, 2014

chemistry
Something is wrong with the post. You're titrating NH3 with HNO3 but you want the pH before any KOH is added. Where did the KOH come from? I assume you want the pH of a solution of 0.1M NH3. At the beginning, before any HNO3 has been added ...........NH3 + HOH ==> NH4...
November 6, 2014

Chemistry
Try this. n initial = PV/RT Solve for n initial. Then redo at the lower pressure and solve for n final. The delta n = n final- n initial = mols O2 used. Then grams = mols x molar mass.
November 6, 2014

Chemistry
http://pubs.acs.org/doi/abs/10.1021/je60034a008
November 6, 2014

Organice Chem
Why not try it and see. I've not done it but I think the it may be too soluble in acetone.
November 6, 2014

Chemistry
I think the answer is A. I would choose heat, also, but T measures heat. And choosing heat means ONLY heat would be evidence but CHANGE in T is what counts; i.e., it could get colder or hotter and be evidence that a change in energy had occurred. So temperature I think is the ...
November 6, 2014

chemistry
It's a math symbol for divide. 4/2 = 2 15/5 = 3
November 6, 2014

chemistry
Yes as long as the copper you left wet doesn't oxidize or otherwise react with something OR if the wet stuff left is not pure H2O. Water from the faucet usually contains Ca and Mg salts and when the liquid evaporates it leaves a very small residue in a beaker. It isn't...
November 5, 2014

Chemistry
A ppm is 1g solute/10^6 g solution. You want 1.40 ppm and you want 5.40E6 g of that solution; therefore, (x g solute/5.40E6g soln)*10^6 = 1.40 x = 1.40*5.40 = 7.56 g Cl2. Then 7.56 g Cl2/5.40E6 g solution = 1.40 ppm
November 5, 2014

Chemistry
delta T = i*Kf*m So if Kf is he same and m is the same you can forget about K and m. Differences will be due only to i. i is the van't Hoff factor and i = # particles on dissociation; e.g., LiNO3 will be 2 and that will have the highest freezing point of the three listed. ...
November 5, 2014

Chemistry Physics
That's 1.3E-5 WHAT/C. meters, cm, ? Substitute into the following: delta L = alpha*delta T*L You can read more about it here. http://en.wikipedia.org/wiki/Thermal_expansion
November 5, 2014

chemistry
A single bond from overlapping s orbitals.
November 5, 2014

chemistry
By the way, you don't list the length of the cell.
November 5, 2014

chemistry
You need to proof what you post. How do you transfer a 1000 mL aliquot OUT of a 200.0 mL volumetric flask INTO a 100.00 mL volumetric flask containing......? I would like to see this done.
November 5, 2014

Chemistry
dO2/dT
November 5, 2014

Chemistry
mols Cr2O7^2- = M x L = ? Convert to mols C2H5OH by ?mols Cr2O7^2- x (1 mol C2H5OH/2 mol Cr2O6^2-) = ?mols Cr2O7^2- x 1/2 = ? Then grams ethanol = mols ethanol x molar mass ethanol %w/w = (mass ethanol/mass sample)*100 = ? mass ethanol from above. mass sample is 30.50 g from ...
November 5, 2014

Chemistry
2NO(g) + O2(g) --> 2NO2(g) When all gases are listed one can take a shortcut and use volume directly as mols. volume NO = 150 mL Convert NO to O2 150 mL NO x (1 mol O2/2 mols NO) = 150 x 1/2 = 75 mL O2 required.
November 5, 2014

Chemistry
I believe this is a first order with respect to H2O2. rate = k(H2O2)^1
November 5, 2014

chemistry
You have a typo. 3Cl2 +6KOH->KClO3 +5KCl + 3H2O z mols KOH x (1 mol KClO3/5 mols KOH) = ? mols KClO3
November 5, 2014

chemistry
You have a typo. 3Cl2 +6KOH->KClO3 +5KCl + 3H2O z mols KOH x (1 mol KClO3/5 mols KOH) = ? mols KClO3
November 5, 2014

chemistry
right!
November 5, 2014

chemistry
determine z as in the previous question for the reaction CH4 + O2 --> CO2 + H2O. z kJ/mol CH4 x # mols CH4 = 3287 kJ. Solve for # mols CH4. Then use PV = nRT and solve for V at the conditions listed. # mols CH4
November 5, 2014

chemistry
What's Table 9.4? 2CH3OH(g)+ 3O2(g) ==> 2CO2(g) + 4H2O(g) Look in you text/notes and find delta H formation. dHrxn = (n*dHformation products) - (m*dHformation reactants) = some number which I'll call z. Then z x (1.2/2*molar mass CH3OH)
November 5, 2014

Chemistry Physics
[mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute all of the numbers; mass cool H2O @ 20C is the only unknonwn.
November 5, 2014

organic biochemistry
mols = M x L = ?
November 5, 2014

chemistry
My best guess is that this forms KNH2 + H2 in a reaction analogous to K + HOH --> KOH + H2 However, I can't confirm that anywhere.
November 5, 2014

Chemistry, Check 1 easy question!!!
I guess ionic bonds would be pleased to be called iconic.
November 5, 2014

chemistry
So the number of moles on the left side would have to equal the number of moles on the right side of the equation?For the same element yes And does this apply for the total number of moles on one side, or just the number of moles of one element on one side? I don't know ...
November 5, 2014

chemistry
As long as you're careful what you are equating I would say yes. 4Cu + O2 ==> 2Cu2O So you have 4 mols Cu atoms on the left and 4 mols Cu atoms on the right. BUT you have 4 mols Cu atoms on the left and only 2 mols Cu2O on the right. Also note that you mols are not ...
November 5, 2014

Chemistry
Dial up each and take a look at adding or subtracting electrons to make the ion isoelectronic with a noble gas. don't think Zr^4+ will do it. http://www.webelements.com/
November 5, 2014

chemistry
What is column f? What experiment are you doing?
November 5, 2014

chemistry
Use the ideal gas law and solve for P. Use the van der Waals equation and solve for P. Compare the two. %deviation = [(delta P)/(Pvan der waals]*(100 = ?
November 5, 2014

chemistry
http://dev.nsta.org/ssc/pdf/v4-TS_51.pdf
November 5, 2014

Chemistry please help
Bob Pursley did this for you yesterday.
November 5, 2014

Chem 130
The easy. I just didn't go through the mopls steps. 976 tons ZnCO3 x 0.78 = approx 750 tons ZnCO3 750 tons ZnCO3 x (atomic mass Zn/molar mass ZnCO3) = ? tons Zn metal.
November 4, 2014

Chemistry
c = fw. f = freq in Hz w = wavelength in m c = speed of light in m/s
November 4, 2014

Chemistry
If you had typed in what you did I could have found the error by now. I don't know what you did wrong either. How do you know 22 is wrong? What are you using for specific heat Cu.
November 4, 2014

Chemistry
The mistake you're making is trying to manipulate the numbers to get answer; i.e., guessing. Reason it out. You know mol = grams/molar mass. mol = grams/molar mass 0.85 mol = 23.8/molar mass molar mass = ? approx 28 Therefore X2 = 28 and X must be 28/2 = ? You should have ...
November 4, 2014

Chem 130
%H2O = [(5*molar mass H2O/molar mass CuSO4.5H2O)]*100 ?
November 4, 2014

chem 2
..........CH3NH2 + HOH --> CH3NH3^+ + OH^- I..........0.025............0..........0 C...........-x..............x..........x E........0.025-x............x..........x You need the Kb for CH3NH2. You can find that in your text/notes/web. Substitute the E line into Kb ...
November 4, 2014

Chemistry
This is a regular stoichiometry problem. All of these are worked with the following steps. 1. Write and balance the equation. You have that. 2. Convert grams of what you have (in this case B2O3) to mols. mols = grams B2O3/molar mass B2O3 = ? I'm estimating mols = approx 0....
November 4, 2014

chemistry
mmols HCl added initially = mL x M = 5.00 mmols NaOH to neutralize = 4.21 x 1.2 = 5.05. Check you numbers. mmols NaOH > mmols HCl which means antacid tablet did not neutralize any of the HCl.
November 4, 2014

chemistry
https://www.google.com/search?q=lewis+structure+N2O5&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a&channel=sb
November 4, 2014

Chemistry
We can't draw structures on this forum. Do what I do; i.e., Google lewis structure ......and fll in the blank
November 4, 2014

Chemistry
Use the Henderson-Hasselbalch equation and substitute millimols for base and millimols for acid. Solve for pH.
November 4, 2014

Science
The cattle egret does that but other egrets do not.
November 4, 2014

Chemistry pretty urgent!!!!!!!
http://faculty.ncc.edu/LinkClick.aspx?fileticket=L2sV5SDzIhE%3D&tabid=1871
November 4, 2014

chem
pO2 = XO2*Ptotal pO2 = 0.21*760 = approx 160 mm Then pressure remaining is 760-approx160 = approx?
November 4, 2014

chem
4Fe + 3O2 --> 2Fe2O3 mols O2 = 80 mL x (1 mol/22,400 mL) = ? Using the coefficients in the balanced equation, convert mols O2 to mols Fe. Then convert mols Fe to grams. g = mols x atomic mass Fe.
November 4, 2014

Chemistry
I see Bob Pursley ignored the difference in temperature. Is that 23.3 and 23.8 a typo?
November 4, 2014

Chemistry
Is that 50 mL total? How much HCl and how much NaOH? What's the molarity of HCl and NaOH? You need to know mols H2O formed if you are to calculate dH/mol.
November 4, 2014

chemistry
millimols base = 1,500*0.17 = 255 mmols acid = 1,500*0.32 = 480 add 54 mmols HCl .........Bz^- + H^+ ==> HBz I.......255......0.......480 add.............54............... C.......-54....-54......+54 E.......201.....0.......534 Plug these E line values into the HH equation ...
November 4, 2014

Chemistry
Is this what you're looking for? Universal indicator--mixture of two. http://www.all-about-ph.com/ph-indicators.html
November 4, 2014

chemistry
You have two "addeds" in your post. I assume you want to know how much NaOH was ADDED FROM the buret TO the titration solution and not how much was added to the buret initially. 24.8-1.2 = ?
November 4, 2014

Chemistry
CO3^2- + Cu^2+ ==> CuCO3 millimols CuSO4 = mL x M = 23.7*0.167 = approx 4 but that's an estimate so you need to redo all of these calculations. So you will need approx 4 mmols NaCO3 and M Na2CO3 = mmols/mL or mL = mmols/M = approx 4/0.122 = ? mL. g CuCO3 = mols CuCO3 x ...
November 4, 2014

science
If you will look up those values and post them here we may be able to help. Include infor on HgSCN+. Why? My table values may not agree (probably will not) so we can't get the same answer. Show what you know to do also.
November 4, 2014

organic chemistry laboratory
Here is my answer of two days ago. Did you see it? http://www.jiskha.com/display.cgi?id=1414966391
November 4, 2014

Science
How many mols do you want? That's mols = M x L = ? Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
November 4, 2014

To Chandler---chemistry
Remember you get faster service if you don't post piggy back. Go to the top of the page and post your own question all by itself. heat lost by steam + heat lost by 1.3g liquid H2O + heat gained by 31 degree H2O = 0 [mass steam x -heat vap] + [mass steam to H2O x specific ...
November 4, 2014

chemistry
If there is a question here I don't see it.
November 4, 2014

chemistry
Look on the web for drawings of how to do this. Most I have seen ascarite or drierite. We can make drawings on the forum. Textbooks also have drawings showing how to do it.
November 3, 2014

Chemistry Lab
I suppose we are to assume that the volumes are additive. M1V1 = M2V2 0.150*1.5 = M2*(3.5+1.5) M2 = 0.045M Another way. You start with 1.5 mL and end up with 5.0 mL (if the solutions are additive) so 0.150M x (1.5/5.0) = 0.045M
November 3, 2014

chemistry
dE = q + w q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial) to get from 25 C to 100C. q2 = mass H2O x heat vaporization to get from liquid water at 100 to steam at 100 q3 = mass steam x specific heat steam x (Tfinal-Tinitial) to get from steam at 100 C to steam at 111 C. ...
November 3, 2014

big time oops----chemistry
The first part of my response is ok. It's the work that isn't. As you've probably already noticed I used mols for volume and that's a no, no. Work = is pdV with the proper sign. pdV = dn RT. Substitute R, T and delta n (delta n is 34 mols) and solve for pdV ...
November 3, 2014

chemistry
I would do this. dE = q+w dE = -2252 kJ + w. Modifying PV = nRT and remembering that pdV = work, then pdV = dnRT and dn is 11 + 23 = 34 mols. pdV = 34*0.08205*298 = ? L*atm. Convert to J by multiplying by 101.325 and convert that to kJ. What't the sign of work in this case...
November 3, 2014

Chemistry
I worked this below. I'll try to find it and give you a link. http://www.jiskha.com/display.cgi?id=1415056292 Any questions?
November 3, 2014

Chemistry
I worked this below. I'll try to find it and give you a link.
November 3, 2014

chemsitry
delta g lost = 0.1 %error = (0.1/1.3)*100 = ?
November 3, 2014

chemistry
This is black magic, not chemistry. Stick to the subject.
November 3, 2014

Organic Chemistry
Frankly, I don't know why it redissolves upon adding too much HCl but why not recalculate the c1v1 = c2v2 and see how much NaOH to add to neutralize that extra HCl you added. That should get you back to the starting point or at least where you had the ppt form.
November 3, 2014

Organic Chemistry
I wish I could help but there is nothing in your post to tell us what experiment you were doing. Sorry. My tea leaves aren't working tonight.
November 3, 2014

chemistry
q = Ccal*dT = 4.84 kJ*(27.1-23.7) = ? kJ heat combustion is -?kJ/0.009298 = -?kJ/mol
November 3, 2014

Chemistry
mL1 x M1 = mL2 x M2 Substitute. mL1 x 0.355M = 19.9 mL x 15.0M Solve for mL1
November 3, 2014

Chemistry
% w/w = mass solute/mass solution. (X/369)*100 = 3.04 Solve for X = grams NaCl.
November 3, 2014

Chemistry
Take 100 g sample which gives you 7.252 g Na 75.084 g U 17.664 g O ------------ convert to mols by mols = grams/atomic mass. 7.252/23 = approx 0.315 75.084/238 = approx 0.315 17.664/16 = approx 1.10 For the empirical formula you want to find the ratio of these elements to each...
November 3, 2014

chemsitry
I don't know but if all of the Ag in AgNO3 is converted to Ag2CrO4, then 1 mol 2AgNO3 = 1 mol Ag2CrO4; therefore, 1 mol AgNO3 will produce 1/2 mol Ag2CrO4. You have mols AgNO3 = 146/molar mass AgNO3 = approx 0.86 but that's an estimate. Then 0.86 mols AgNO3 = 0.43 mols...
November 3, 2014

To Maria---chemistry
It is better to post your own question instead of riding piggy back on another post. I've already answered your question above.
November 3, 2014

chemistry
heat lost by water + heat gained by calorimeter = 0 [(mass H2O x specific heat H2O x (Tfinal-Tinitial)] + [(Ccal*(Tfinal-Tinitial)] = 0
November 3, 2014

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