Wednesday
January 28, 2015

Posts by DrBob222


Total # Posts: 46,601

Chemistry/Physics
Both are negative. Why did you subtract? dE = -2.43-5.00 = -7.43 kJ
November 20, 2014

Chemistry/Physics
Re-read your post. Did you omit calculating dE? 2430 J and it is exothermic. q = -2430J so dH is -2430 J. It does 5,000 J work so work - 5000 J. dE = q + w. dE will be in J. Convert to kJ and convert the other so to kJ.
November 20, 2014

Chemistry
Look up the bond energies for each bond. Then bond energies broken - bond energies formed = dH rxn Here is a tutorial + bond energies listed. http://www.kentchemistry.com/links/Kinetics/BondEnergy.htm
November 20, 2014

chemistry
You need to find the caps key on your computr and use it. While you are at it, find the period also. 2H2 + O2 ==> 2H2O When working with gases one can use volume interchangeably as moles. Therefore, If you have 250 mL H2 you will need 250 x 1 mol (O2/2 mols H2) = 125 mL O2 ...
November 20, 2014

college chem
Yes, that's what you do and when you get there there is nowhere else to go. You're subtracting kJ/mol from kJ/mol and the answer is in kJ/mol. (1*-285.8)-(-229.9+0) = ?
November 20, 2014

College Chem
This looks obvious to me. Exothermic reactions release heat so the heat to the calorimeter increases and the temperature goes up inside. T rises. Endothermic reactions absorb heat (where does the heat come from) so the calorimeter loses heat and the temperature goes down ...
November 20, 2014

chemistry
% w/w = (g solute/g solution)*100 = ? DON'T make the mistake that g solution is 250 g.
November 20, 2014

Chemistry
Isn't that all you need? Plot A vs C. You get concentrations from the volume of the stock solution diluted to x mL.
November 20, 2014

Chemistry
I would have measured the absorbance. Did you have a working curve; e.g., absorbance measurements of several solutions of different concentrations. They look like this. https://www.google.com/search?q=graph+beer%27s+law&client=firefox-a&hs=bnu&rls=org.mozilla:en-US:official&...
November 20, 2014

Chemistry
I'm confused by the language of "Can you tell if". Did you run any experiments specifically for this? If the dilute solutions follow Beer's Law and the stock solutions are not too concentrated, they may follow Beer's Law also. Generally, however, ...
November 20, 2014

Math Help
yes
November 20, 2014

int.science gtg in 8 min.
q = mass H2O x heat capacity
November 20, 2014

Int.Science
q1 = heat needed to raise T Cu from 83 C to it's melting point (you will look that up) is q1 = mass Cu x specific heat Cu x (Tfinal-Tinitial) where Tf is the melting point and Ti is 83C. q2 = heat needed to melt Cu at its melting point is q2 = mass Cu x heat fusion You ...
November 20, 2014

chem
I can't make sense of your post but here is how you work the problem. 1. Write and balanced equation. You've done that. 2. Convert grams HgO (whatever that is) to mols. mols = grams/molar mass 3. Using the coefficients in the balanced equation, convert mols HgO to mols...
November 20, 2014

chemistry
Is this at STP? If not what is the volume or total pressure? mols H2 = grams/molar mass mols SO2 = grams/molar mass total mols = mols H2 + mols SO2 XH2 = mols H2/total mols XSO2 = mols SO2/total mols pH2 = XH2*Ptotal XSO2 = XSO2*Ptotal
November 20, 2014

chemistry
1 ppm is 1 mg/L; therefore, 45 mg/L = 45 ppm.
November 20, 2014

chemistry
I don't know. Few of us have the solubility tables memorized. Do you have a graph or a table that lists the solubility at various temperatures. Post that and we can help.
November 20, 2014

Chemistry Numerical
You need k. k = 0.693/t1/2</su> Then ln(No/N) = kt Set No = 100 Set N = 100*1/8 =? k from above. Solve for t in days.
November 20, 2014

Chemistry
See your other post above.
November 20, 2014

chemistry
heat lost by hot 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 Substitute the numbers and solve for the only unknown in the equation.
November 20, 2014

Chemistry
Done
November 20, 2014

Chemistry
Use the dilution formula.
November 20, 2014

Chemistry
Sorry, but I don't see a 12:04 posting and no pH questions from you.
November 20, 2014

Chemistry
I'll give you some hints but it is up to you to put all of it together to make a coherent answer. For AuCl3 ==> Ag^3+ + 3Cl^- Write out Ksp expression and solve for (Cl^-). Plug in Ksp, (Au^3+) = 0.5M and calculate (Cl^-) Do the same for AgCl AgCl --> Ag^+ + Cl^- ...
November 19, 2014

Chemistry
Here is what I found. http://www.jiskha.com/display.cgi?id=1416382196
November 19, 2014

college chem
The electrode reactions are Zn ==> Zn^2+ + 2e Fe^2+ + 2e ==> Fe The anode is where oxidation occurs; therefore, Zn is the anode and it is the negative electrode. Electrons flow from Zn to Fe. Zn is more "active" than Fe.
November 19, 2014

Chemistry
HClO3 is a strong acid and ionizes 100%; therefore, (H3O^+) and (ClO3^-) = 0.28M.
November 19, 2014

Chemistry
You need to put some units on the numbers. Where does Qrxn come from? What kind of reaction?
November 19, 2014

Chemistry
C2H5NO2 + 2H2O → 2CO2 + NH3 + 3H2 a. That will be zero unless you want to count the vapor pressure of water at 27C. I assume that is to be neglected. b. You have 0.0125 mols glycine. You get 0.0125 mol NH3, 3*0.0125 mols H2 and 2*0.0125 mols CO2. c. PV = nRT d. PV = nRT
November 19, 2014

chemistry
(2*molar mass H2O)/(molar mass Na2CO3.2H2O)*100 =
November 19, 2014

chemistry
%Mg = (grams Mg/total mass)*100 = ? Note: total mass is 9.03 + 3.48 = ? %N = (3.48/total mass)*100 = ?
November 19, 2014

AP chemistry
In three words, not very easily. You don't need the O2. ZnCO3 + heat --> ZnO + CO2
November 19, 2014

chemistry
mols ethylene glycol = grams/molar mass = ? Then m = molality = mols/kg solvent (kg solvent is 0.0854 in the problem) The formula are similar for f.p. and b.p. f.p. delta T = Kf*m. Substitute and solve for delta T, then subtract from the normal f.p. of O C to find the new f.p...
November 19, 2014

chemistry
No. In trying to avoid confusing you because I didn't know what other compounds you had (and some may have been ionic), I ended up confusing you by including the i. Just forget the i or if yu want to keep it there, note that I TOLD you it was 1 for this problem. delta T = ...
November 19, 2014

chemistry
delta T = i*Kf*m and I assume all of the solutions you have are non-electrolytes and i = 1. You know Kf (the freezing point constant) and if you don't it will be in your text and it's all over the web. m is the molality and that's in the problem (in this case 1.40...
November 19, 2014

Chemistry
This is an excess of OH^-; therefore, (OH^-)= 1.96E-4M Then pOH = -log(OH^-) = ? Then pH + pOH = pKw = 14. You know pKw and pOH, solve for pH.
November 19, 2014

Chemistry
I don't think so. Count those zeros. I think you have one too many. Now you know what I don't like about those pesky things. 0.01 mmols/51 mL = 1.96E-4M 0.026L x 0.01M = 0.00026 mols NaOH 0.025L x 0.01M = 0.00025 mols HCl 0.00001 mols/0.051 L = 1.96E-4 Or you can let ...
November 19, 2014

Chemistry
Yes, I would say the missing information is important (at least to the problem). I like millimoles (you can change to mols if you like by dividing millimoles by 1000). I like millimoles because it keeps all of those zeros on the left from showing up. millimols NaOH = 26.0 x 0....
November 19, 2014

Chemistry
Mollie, I know all of the tutors on this site are good but we don't have ESP or any of those extraordinary powers. In other words, what solution are you talking about?
November 19, 2014

Science Technology
This site will provide more than you every wanted to know about the camera. http://en.wikipedia.org/wiki/History_of_the_camera
November 19, 2014

science
I believe O2 is generated by the electrolysis of H2O. I don't think it is generated as needed on a space suit but in the space ship, stored in canisters, and the canisters and contents are attached to the space suit with astronaut inside. If they don't wear a space ...
November 19, 2014

science
You've left me out in the wild; give me something to get my bearings
November 19, 2014

Chemistry: Sigma & Pi bonds
You know we don't do all of your homework. But we'll be happy to check your answers. For CH2Cl2 there are 2 sigma CH and CH bonds and 2 C-Cl bonds to make a total of 4 sigma bonds.
November 19, 2014

Chem
I think you're having trouble with the CH2O, too. I think by your formula and the way I do it that the formal charge on C is 0 and the formal charge on O is 0. (and the formal charge on each H is 0). Here is the dot structure. http://www.thegeoexchange.org/chemistry/...
November 19, 2014

chemistry
2Cu^2+ + 4I^- ==> 2CuI + I2 Then I2 + 2S2O3^2- ==> S4O6^2- + 2I^- mols S2O3^2- = M x L = ? Mols I2 = 1/2 that from the equation that 2 mol S2O3^2- = 1 mol I2. mols Cu = 2x mols I2 from the first equation = ? Then M Cu^2+ solution is mols Cu/volume = mols Cu/0.010 L
November 19, 2014

chemistry
mols Zn = grams/atomic mass = ? mols Zn = mols ZnCl2 since the equation shows 1 mols Zn produces 1 mol ZnCl2. Then M ZnCl2 = mols/L solution.
November 19, 2014

chemistry
(g solute/g solution)*100 = 3.1E-5 (0.225/g soln)*100 = 3.1E-5 Solve for g solution (in mL) and convert to L.
November 19, 2014

chemistry
Nope but close. 1 Al + 3N + 9O = ? NO3 = nitrate ion is not an atom. There are 4 atoms in 1 NO3^- and 3x that is 12 + the 1 Al = ?
November 19, 2014

Chemistry
I see a number of posts today. A couple of days ago I answered most if not all and suggested you give some indication of how to solve the problem or at least tell us what you don't understand or what you do understand about the problem. I will go through all of these again...
November 19, 2014

Chemistry
These usually are confusing for students; the secret is to follow the rules. 4Be is 1s2 2s2. For the 1s2 n = 1, s means l = 0 and there is 1electron with +1/2 and the other with -1/2. You simply go down the choices and pick any that has n = 1, l = 0, ml = 0 and ms = +/- 1/2. ...
November 19, 2014

Chemistry
How much heat do we need to raise T 112 g sample of water from 25 to 100? That's mass H2O x specific heat H2O x delta T = 112 x 4.184 x 75 = 35, 146 J. E 1 photon = hc/wavelenth E 1 photon = 6.626E-34*3E8^2/3.28E-3 E/photon x #photons = 3,146 J And solve for # photons.
November 19, 2014

chemistry
You've posted this before but provided no more information than the last time. I responded last time that I had no idea what version A is/was or what you're doing. We aren't clairvoyant and my crystal ball is hazy today. I can't help without knowing what the ...
November 19, 2014

chemistry
(P1V1/T1) = (P2V2/T2)
November 19, 2014

General
I don't know what you mean by "and field 100 bottles.." I suppose the answer is zero mm since mm is a length of measurement and not volume.
November 19, 2014

Chemistry
I agree with your thoughts about #1. For #2, I would look at it this way. If C combines with O2 we get C + O2 ==> CO2 and that is, indeed, a combination reaction. However, if we use a hydrocarbon, such as CH4 and combine that with O2 we get CH4 + O2 ==> CO2 + H2O and ...
November 19, 2014

chemistry
mass = volume x density You know mass and density, substitute and solve for volume in mL. Convert to L.
November 19, 2014

Chemistry
You have calculated the accuracy and not the % error. %error is 100-94.7 = 5.3% Or to do it another way, this is what you do. [(6.02-5.70)/6.02]*100 = same number above.
November 19, 2014

Chemistry
If you need further assistance you need to provide a few more details about exactly what you've done. To convert your answer of 352.49 J/g to kJ/mol, multiply that figures by the molar mass of H2O which gets you to J/mol and divide by 1000 to convert to kJ/mol. Your ...
November 19, 2014

chemistry
You may have made a typo but mass Mg from your data is -0.01 which is impossible. To answer your question, mols Mg = mass Mg/atomic mass mg. # molecules MgO (I wonder if you don't want mols of this, too) = (mass MgO/molar mass MgO)* 6.022E23 = ?
November 19, 2014

chemistry
dE = dm*c^2 E = (0.00650/1000)*3E8^2 =? The 1000 converts g to kg.
November 19, 2014

chemistry
volume H2O is 55.0 + 55.0 = 110.0 and that is 110.0g q = mass H2O x specific heat H2O x (Tfinal-Tinitial) mols Ba(OH)2 = M x L = approx 0.017 but you need a better number than that and that produces twice that number of mols or approx 0.034. Ba(OH)2 + 2HCl ==> BaCl2 + 2H2O ...
November 19, 2014

chemistry
mols H2SO4 in the solution is M x L = 0.400 x 0.1 = 0.04 grams = mols x molar mass = 0.04 x 98 = 3.92g
November 19, 2014

chemistry
Use dG = -RT*lnK and calculate dG. Then dG = dH - TdS.
November 19, 2014

chemistry
See your other post.
November 19, 2014

chemistry
Use delta T = K*m and solve for m. In dilute solutions the m and the M are very close to the same, then pi = osmotic pressure = MRT Post your work if you get stuck.
November 19, 2014

Chemistry
I didn't put units on my answers and I should have done so. They are (H3O^+) = 1.58E-7 M (OH^-) = 1.58E-7 M. And just to make things a little more interesting let me point out that this is a NEUTRAL solution with a pH = 6.80. We always talk about pH = 7.0 as being neutral ...
November 19, 2014

Chemistry
I saw only this one extra question whereas your post said "additional" questions. If there are more than this please just post everything again at the top of the page. You don't do anything with the 37C temperature. That's the temperature of the water at body...
November 19, 2014

Chemistry
I expect you meant H3O^+ and not H2O. (H3O^+)(OH^-) = Kw (H3O^+)=(OH^-) = sqrt Kw.
November 19, 2014

Chemistry
Some is correct. I don't know how your calculator works (calculators aren't standard) but I can tell you how to proceed. First, where did you get the pKa for formic acid? I looked it up on the web and found 3.77. You can get slightly different numbers depending upon ...
November 19, 2014

Chemistry
pKa = -log Ka
November 19, 2014

Chemistry
Use the Henderson-Hasselbalch equation. pH = pKa + log (base)/(acid) pH = ? pKa = look up (base) = 0.07 (acid) = 0.07
November 19, 2014

Chemistry
I can't confirm those dH formation values but I'll assume you have them right. In calculating, however, when you finish isn't that 3344.1 kJ that many kJ for the reaction as shown and not kJ/mol. It appears you used dHrxn = (n*dHformation products) - (n*dH ...
November 19, 2014

Chemistry
2. I've been thinking about this and I can't see a good reason why it isn't zero. It absorbs heat on the way up to vapor and releases heat on the way back down to the crystalline state. Think about that, is that right? If that is true it sure would save a lot of ...
November 18, 2014

Chemistry
3. How much heat (q) is needed to heat the water. q = mass H2O x specific heat H2O x (Tfinal-Tinitial) = approx 250,000 but that is an estimate as are all of the other numbers that follow. Look up the heat combustion for ethane, probably given in kJ/mol Convert that to the ...
November 18, 2014

Chemistry
1. mass H2O = 85 g + 85 g = ? (mass H2O x specific heat H2O x (Tfinal-Tinitial) + Ccal*(Tfinal-Tinitial) = 0 Substitute and solve for Tf.
November 18, 2014

chemistry
CH4 + 2O2 --> CO2 + 2H2O mols CH4 = 10.4 Using the coefficients in the balanced equation, convert mols CH4 to mols CO2. Then use PV = nRT and convert mols CO2 at the conditions listed to volume in L.
November 18, 2014

Chemistry
2. If 114.6 kJ are produced per 1 mol NO then 2*114.6 or 229.2 kJ will be produced for 2 mols(2*30 = 60 g). You may want to check this out in the problem because it always leaves one in doubt exactly what is meant. I'll assume the above is correct. Then 229.2 kJ x (1.16E4g...
November 18, 2014

Chemistry
1. q = Ccal*(Tfinal-Tinitial) for the rxn. q/0.1375 = q/gram q/gram x atomic mass mg = q/mol. Change those to kJ/g and kJ/mol
November 18, 2014

Chemistry
Don't hog the space. Post 1 question at a time. Show your work and/or explain what you don't understand about the problem. And don't change screen names;i.e., Vince and Tim.
November 18, 2014

Chemistry
The data you gave for #1 shows delta H as +483.6 and that makes it an endothermic reaction so there will be no heat given off. You may want to re-reread/re-type the question. 2. heat lost by Fe + heat gained by Au = 0 (mass Fe x specific heat Fe x (Tfinal-Tinitial) + (mass Au ...
November 18, 2014

chem
Always remember, think in terms of mols. How many mols do you want? That's gram/molar mass. Then M = mols/L solution. You know M and mol, solve for L and convert to mL.
November 18, 2014

Chemistry
After thinking about this for 5-6 hours I don't think it simplifies the problem and I don't think the process is the same.
November 18, 2014

Chemistry
I think that simplifies the problem somewhat but the process is the same.
November 18, 2014

Chemistry
I would do this. Since the SrSO4 is relatively more soluble than BaSO4 (Ksp about 3,500 times) I would calculate the solubility of SrSO4 in a saturated solution ignoring the solubility of BaSO4 and figure the small solubility of BaSO4 makes very little difference. So Ksp = (Sr...
November 18, 2014

chemistry
I would do it this way. pOH = -log(OH^-) = 8 pH + pOH = pKw = 14 pH = 14- pOH = 14-8 = 6 So what's wrong with 6? Your problem is that you failed to realize that since 7 is neutral where (H^+) = (OH^-) = 10^-7M each, so when OH^- is 1E-8 (NOTE THIS IS LESS THAN 10^-7) SO IT...
November 18, 2014

chemistry
mols EDTA = M x L = ? mols Ca = same since EDTA complexes 1:1 You titrated only 10 mL aliquot of a 250 mL sample so mols in the original sample is mols from above x 250/10 and that is mols Ca in the initial sample. grams Ca = mols Ca x atomic mass Ca and % in sample is (gCa/...
November 18, 2014

chem
H-C is a sigma bond C-Cl is a sigma bond C=O is a double bond which consists of 1 sigma bond and one pi bond. So I would count them as 3 sigma and 1 pi.
November 18, 2014

Chemistry Lab
I won't be much help. It may make a difference but it all depends upon what the instructor will be checking and how much work you completed the first day vs the next two days. If you heated the Mg ribbon and crucible the first day but continued heating the second day it ...
November 18, 2014

chemistry
Here is a site that may be interesting to you and may shed some light on your problem. http://en.wikipedia.org/wiki/Molar_ionization_energies_of_the_elements This shows ionization energy = 1312 kJ/mol H ATOMS.
November 18, 2014

chemistry
There may be an equation but if there is one I don't know it. I've always just counted them.
November 18, 2014

Chemistry
I don' see anything wrong with this. I calculated OH using 0.6 x 0.002/1.002 and came up with a slightly smaller number but it rounds to your value and if I square that and multiply by Ca (I used 0.850 x 1/1.002) = 0.848 but no matter how I slice it the answer always ...
November 18, 2014

Chemistry
Why don't you post your work and let us find the error.
November 18, 2014

chem
Reverse equation 1 and add to equation and that gives you 4PCl3 + 4Cl2 ==> 4PCl5 which is just 4x what yu want so divide everything by 4.
November 18, 2014

physical science
?cal = mass Al x specific heat Al x (Tfinal-Tinitial)
November 18, 2014

chemistry
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) is the heat absorbed. Then q/1.375 is heat absorbed per gram CaO and that x molar mass CaO is heat absorbed per mol CaO. I think those questions would follow although you didn't post them yet.
November 18, 2014

Chemistry 11
I've looked at your data again. %H2O appears to be done correctly. I disagree with the answer to the question about getting results similar to your neighbors results. If everyone is on the same page and everyone has good lab technique then the results should be comparable...
November 18, 2014

Chemistry 11
Bob Pursley answered this for you just a few minutes ago and you've hardly let the ink dry before re-posting as if you had heard nothing. No, the amount of material you took makes no difference in the formula nor the percent. If I have 5g out of 10 g sample, that's 50...
November 18, 2014

chem
mols Mg = mass Mg/atomic mass Mg. Wouldn't mass Mg be (mass xble+Mg) - (mass empty xble)? Now think about the MgO. Now that you see how easy it is to get mass Mg, wouldn't a very similar tactic obtain mass MgO.
November 18, 2014

chemistry
The problem tells you that you obtain 2511 kJ heat for burning 2*26 or 52 g C2H2. Instead of using 52g the problem is using 85g so the heat produced must be 2511 kJ x (85/52) = ?
November 18, 2014

chemistry
No table included. Take the sum of the B.E for bonds broken on the reactants side subtract the sum of the B. E, formed on the product side to calculate dHrxn
November 18, 2014

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