Monday
September 1, 2014

Posts by DrBob222


Total # Posts: 43,434

Science(chemistry)
law of constant proportions.
May 21, 2014

Chemistry
The limiting reagent problems are really just two ordinary stoichiometry problems solved more or less together. Here is how it is done. The first step is to write and balance the equation which you have supplied. CaCl2 + Na2CO3 ==> CaCO3 + 2NaCl Step2a. Convert g CaCl2 to ...
May 21, 2014

Science
http://en.wikipedia.org/wiki/Prepuce http://books.google.com/books?id=5XogAQAAMAAJ&pg=PA19&lpg=PA19&dq=define:preouce&source=bl&ots=CI40WcXkl_&sig=5WQoN85Ehia0eynrI-LJfSKFvpQ&hl=en&sa=X&ei=uRN8U-eKKtHNsQSyqYDAAQ&ved=0CDwQ6AEwBA#v=onepage&q=define%3Apreouce&f=false
May 20, 2014

Science
http://en.wikipedia.org/wiki/Prepuce
May 20, 2014

Chemistry
This isn't so much a normality problem as it is a dilution (concentration) problem. If you take 4 ml and dilute to 20 you have diluted it 5 times, right? Then another 50 times in the 1:50. So the total dilution is 5*50 = 250 times. So if you have 1N solution and it's ...
May 20, 2014

Chemistry
Solubility is 1g aspirin in 300 mL. So how much aspirin is lost with a 10 mL rinse and I use the term rinse lightly which I will comment on later. 1000 mg aspirin x 10mL/300 mL = 33.33 for the first rinse. 33.33 for the second rinse. 33.33 for the third rinse. Total is 100 mg ...
May 20, 2014

chemistry
I would name that 3,4-dimethyl-2-pentene.
May 20, 2014

AP chemistry
I don't like to do it this way because students ALWAYS get confusted so I'll put it together for you. heat lost by Al is mass Al x specific heat Al x (Tfinal-Tintial) heat gained by H2O is mass H2O x specific heat H2O x (Tfinal-Tinitial) So set heat lost + heat gained...
May 20, 2014

AP chemistry
heat lost by aluminum + heat gained by water = 0 heat lost by Al = mass Al x specific heat Al x (Tfinal-Tinitial) heat gained by water = mass H2O x specific heat H2O x (Tfinal-Tinitial) Substitute into the heat lost + heat gained = 0 and solve for Ti.
May 20, 2014

Chemistry
I don't know where you obtained the 85 C? m = 312, right c = unknown, right The problem tells you delta T is 15, where did the 85 come from?
May 20, 2014

Chemistry
I hesitate to answer all of these for you for that way you aren't learning how to do them yourself. I will get you started. For the others show what work you've done, how you think the problem should be solved. and/or exactly what you don't understand about the ...
May 20, 2014

Chemistry
Do you have suggestions for these?
May 20, 2014

Chemistry
.........Ag + ....xN2 --> AgxNy I......344.588.....0........0 add.............15.906......... C.....-344.588.....-x......359.503 So you end up with 359.503 You started with...344.588 How much N2 was added? Then N2 remaining = start-end = ?
May 19, 2014

Chemistry
If we look at this the reverse direction. H2(g) + 1/2O2(g) --> H2O(l) that's the heat formation liquid H2O in kJ/mol and my text shows that as -285.8 kJ/mol Then H2O(l) ==> H2O(s) is another 6.0 kJ/mol to form ice and freezing water to ice is exothermic so that ...
May 19, 2014

chemistry
PV = nRT
May 19, 2014

Chemistry
That's the best choice.
May 19, 2014

Chemistry
delta T = i*Kf*m So you want the highest delta T. i is the van't hoff factor, Kf is constant, molality is constant in this case so it appears to depend upon i. i = 1 for A i = 1 for B i = 2 for C i = 2 for what you have but I suspect you made a typo and that should be ...
May 19, 2014

Chemistry
I don't know. I looked up the numbers in my text. Although they are slightly different I obtained essentially the same values you did; i.e., 128.2, 6.05, and -58.2 all in kJ/mol. I also looked up dGo in the same text and it was 130.4 at T = 298 and using dG = dH - TdS I ...
May 19, 2014

Chemistry
heat gained by cold water + heat lost by warm water = 0 [mass cold H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass warm H2O x specific heat H2O x (Tfinal-Tinitial)] \ 0 Substitute and solve for the only unknown, which is Tf.
May 19, 2014

chemistry
Look up the reduction potentials and add the oxidation half cell to the reductin half cell to obtain the Eocell. If it is + the rxn is spontaneous; if Eocell is - it is not spontaneous.
May 19, 2014

chemistry
What are you starting with?
May 19, 2014

Chem
mols Pb(NO3)2 = M x L = approx 0.28 mols PbCl2 formed = mols Pb(NO3)2 (see the 1:1 ratio for Pb(NO3)2/PbCl2 Then g PbCl2 = mols PbCl2 x molar mass PbCl2.
May 18, 2014

Chemistry
http://www.jiskha.com/display.cgi?id=1400465598
May 18, 2014

Chemistry
If you post your work so we know what you did perhaps we can find the error. Especially important is what are you using for Eo values.
May 18, 2014

chemistry
You should make a habit of looking to see if an earlier post has been answered. I answered this about 5-6 hours ago.
May 18, 2014

science help
For 1 read about tectonic plates. For 2 you didn't list them.
May 18, 2014

Chemistry
Mg^2+ + 2e --> Mg 2Cl^- ==> Cl2 + 2e
May 18, 2014

chemistry
If you are talking about HNO3, nitric acid, a suitable procedure is to bubble gaseous N2O5 into distilled water until saturation is achieved.
May 18, 2014

Chemistry
If you will show what you did perhaps we can find the problem. Other than guessing, however, we don't know what you've done.
May 18, 2014

Chem
You don't give enough information to know if you want the Eocell for a spontaneous or non-spontaneous cell. I assume you want the spontaneous reaction. Al(s) ==> Al^3+ + 3e Eo = +1.66 Fe^2+ + 2e ==> Fe(s) Eo = -0.440 -------------------- 2Al + 3Fe^2+ ==> 2Al^3+ + ...
May 18, 2014

Chemistry
dGorxn = (n*dGo formation products) = (n*dGo formation reactants)
May 18, 2014

Chemistry
mols AgNO3 = grams/molar mass = approx 0.06 but that's only an estimate as are all of the other calculations I do here. mols NaCl = M x L = approx 5E-4 .......NaCl + AgNO3 ==> AgCl + NaNO3 I......5E-4......0........0......0 add...........0.06.................. C.....-5E...
May 18, 2014

Chemistry
You want 15 g CaCl2. You will need 15 g CaCl2 x [(molar mass CaCl2.2H2O/molar mass CaCl2)] = ? g CaCl2.
May 18, 2014

chemistry
It may be possible depending upon how many other metals have specific heats close to the unknown metal. In any case, however, you probably can narrow the choice to one of a few and more importantly you can eliminate many metals.
May 18, 2014

Chemistry
I see a problem below that has been answered by Mathmate that isn't the same problem as this one. This one doesn't allow water to be added and both of the solutions available are more concentrated than the final desired one. I don't think this one is possible.
May 18, 2014

Chemistry
This can be done?? If you use 2L of the weaker solution of 150g/L and zero L of the more concentrated solution, then you have 300g/2L and that is too concentrated.
May 18, 2014

Chemistry
I don't believe this can be done with the data listed. I think at least one equation is missing.
May 18, 2014

chemistry
A catalyst lowers the activation energy
May 18, 2014

Chemistry
2.07L, yes.
May 17, 2014

Chemistry
(V1/T1) = (V2/T2) Remember T must be in kelvin.
May 17, 2014

Chemistry
To find pH for each equivalence point. For HNO3 + NaOH ==> NaNO3 + H2O That is a strong acid vs a strong base. Neither Na^+ nor NO3^- is hydrolyzed; therefore, the pH at the equivalence point is just that of pure H2O which is 7.0. For NaOH + HCN ==> NaCN + H2O The CN^- ...
May 17, 2014

add on--Chemistry
Don't forget to make that 107 g.
May 17, 2014

Chemistry
Print these steps. 1. Write and balance the equation. NH3 + HCl ==> NH4Cl (The problem doesn't say it reacts with HCl and you only need to know that 1 mol NH3 is in 1 mol NH4Cl but I think an equation makes it more obvious.) 2. Convert what you have, in this case NH3, ...
May 17, 2014

AB Chemistry4
Yes, that's the only way to find the number of mols in 36.0 g H2O then to convert mols MnCl2 to g MnCl2.
May 17, 2014

chemistry
This 4-step procedure will work all simple stoichiometry problems. Limiting reagent problems take an addition one or two steps. 1. Write and balance the equation. You have that. 2. Convert what you have (in this case H2O) to mols. mol = g/molar mass = 133.6/18 = approx 7.5 but...
May 17, 2014

chemistry
This is a buffer problem. Use the buffer equation. Use the Henderson-Hasselbalch equation.
May 17, 2014

Simple question about music
Here is an article you may want t read. I don't know how accurate the figures are. I googled "number copies Thriller sold" and came up with many web sites to read. This is just one of them. http://www.newyorker.com/online/blogs/culture/2013/01/did-michael-...
May 16, 2014

Chemistry
For a. eqn 1. mols NaOH = M NaOH x L NaOH eqn 2. mols acid = mols NaOH eqn 3. M acid = mols acid/L acid You're right, the NaOH is diluted which means you will add MORE NaOH when you titrate the acid. Look at eqn 1, If you add more volume you THINK you have more mols NaOH, ...
May 16, 2014

Science help PWEASEEEE
I would pick c.
May 16, 2014

Choosing Courses
I would think it would be ok, especially if you did well in algebra I.
May 16, 2014

chemistry
Look up the reduction potentials and calculate Ecell. If Ecell is positive the reaction is spontaneous. The element oxidized is the element that lost electrons. Reduction is the gain of electrons.
May 16, 2014

chemistry
There are 33+18+2 atoms in 1 mol of C33H18O2. How many moles do you have? That's 33g/molar mass C33H18O2
May 16, 2014

chemistry
http://wiki.answers.com/Q/What_is_the_difference_between_biochemical_pharmaceutical_and_diagnostic_chemical_reactions_in_health_care?#slide=1
May 16, 2014

chemistry
You are absolutely correct and I goofed. I didn't do what I preach so much about making sure the units come out right. I should have written it as 4.7 L/s x ? sec = L to be pumped. Solve or ? sec.
May 15, 2014

chemistry
This is a math problem, not a chemistry problem. I would convert everything to cm. 50 m = 5000 cm 25 m = 2500 cm 2 m = 200 cm Volume of pool is width x length x depth = ? cc. Convert to L. 1000 cc = 1L = initial volume of pool in L. You want to lower by 5 cm so now the ...
May 15, 2014

analytical
Google them. You can get their formula and their properties.
May 15, 2014

chemistry
Nope. You didn't substitute. And you know the answer can't be right. You start with 0.343 mol, it decomposes and you have more than you started with after 43 mn. I don't think so.
May 15, 2014

chemistry
The units of min^-1 tell you that the reaction is first order. So ln(No/N) = kt. No is starting. N is end k from problem t = 42 min
May 15, 2014

chem
looks ok to me.
May 15, 2014

chem
7.42E-5 kg x (1000g/1 kg) x (1000 mg/g) = ?
May 15, 2014

chemistry
yes. You reversed the sign because the reaction is reversed. So that makes it positive. Then double that for e.
May 15, 2014

chemistry
You have the reaction as shown. You want the reverse reaction and double that. So take dH that you have, change the sign (because of the reversal) and double it.
May 15, 2014

Math 2
I would ask 1/4 what of fabric?
May 15, 2014

chemistry
Read and correct your post. Clarify. It makes no sense as is. delta T in kJ?
May 14, 2014

chemistry
You can see from your previous ost how this is done. I'll be glad to check your numbers.
May 14, 2014

chemistry
Calculate mg morphine in each of the choices. a. 20% w/v means 20 g morphine/100 mL. I wouldn't recommend this choice (if you want the patient to live). b. 2M solution. How many mols is that. mols = M x L = 2M x 0.050 L = 0.1. g = mols x molar mass = 0.1 x 283 = 28.3 g or ...
May 14, 2014

Chemistry
The two half reactions are 2Al ==> 2Al^3+ + 6e Eo = 1.66 3Fe^2+ + 6e ==> 3Fe Eo = -0.44 -------------------- Add half cells to obtain rxn. Add Eo to obtain Eocell.
May 14, 2014

Chem
dGorxn = (n*dGo products) - (n*dGo reactants)
May 14, 2014

Chemistry
n stands for number of mols. It is 1 for CaO, 1 for CaCO3, and 1 for CO2. Look up dHo formation CO2, CaO, and CaCO3. Calculate dHo reaction. Look up dSo for CO2, CaO, and CaCO3. Calculate dSo reaction. Then dGorxn = dHorxn - TdSorxn
May 14, 2014

Chemistry
dHorxn = (n*dHoformation producs) - (n*dHoreactants) dSo rxn = (n*dSo formation products( - (n*dSo formation reactants) dGrxn = dHrxn - TdSrxn dG<0 spontaneous dG=0 equilibrium dG>0 not spontaneous.
May 14, 2014

NO SCHOOL
I presume this is a tax, probably Goods and Services Tax in Canada. Your post isn't clear. If you have 100 to spend on an item it will cost you 100 + 10% of 100. Or you may mean that the total cost of an item, including GST, is 100.00. And you want to know the cost without...
May 14, 2014

chemistry
Surely it's in your text. You're right. Bases are proton acceptors according to the Bronsted-Lowry theory of acids/bases.
May 14, 2014

chemistry
This isn't in your text/notes. The first half is According to this theory, acids are proton donors. So bases must be proton ........?
May 14, 2014

chemistry
Metals with a small heat capacity heat faster and cool quicker than metals with a higher heat capacity.
May 14, 2014

chemistry
That's right. % = (mass solute/total mass)*100 = ?
May 14, 2014

chemistry
You have half of it right. The other half you have ignored. q1 = heat needed to raise T of water from 15C to 100 C is what you have calculated. q2 = heat needed to change T of water at 100 C to steam at 100 C. That is q2 = mass H2O x heat vaporization.
May 14, 2014

math
Change meters to cm. 1 meter x (100 cm/1m) = 100 cm 100 cm + 5 cm = ?
May 14, 2014

Chemistry
I would say an oxidizing agent, a reducing agent, and a salt bridge/porous barrier. If you want to get technical I suppose you might add a cell (something to hold the oxidizing agent/reducing agent and the salt bridge).
May 14, 2014

chemistry
H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
May 14, 2014

Chemistry
What does the equation tell you? That 89.3 kJ are absorbed when 1 mol (76.1g CS2) are formed. You have 6.33 g; therefore, 89.3 kJ x (6.33/76.1) = ? kJ
May 14, 2014

science
60 meters in 34 seconds. That is avg velocity of 60m/34s = ? m/s.
May 14, 2014

chemistry
20L propane = 20L x (1 mol/22.4L) = about 0.892 mols propane. You get 2220 kJ for 1 mol; therefore, you will get 2220 x (0.892 mol/1 mol) = ? kJ released.
May 14, 2014

Chemistry
1. Write the equation. 2. Balance it. 3. dHrxn = (n*dHo formation products) - (dHo formation reactants)
May 14, 2014

Chemistry
You're cluttering the board now. Spend your time solving problems instead of typing.
May 13, 2014

Chemistry
Same 'ol same 'ol. See your other posts.
May 13, 2014

Chemistry
I don't see anything new here. See your other posts.
May 13, 2014

Chemistry
See your other posts.
May 13, 2014

Chemistry
Zn^2+ + 2e ==> Zn from the table is -0.76v. Your half cell reaction is Zn ==> Zn^2+ + 2e E = +0.76v Ag^+ + e ==> Ag E = 0.80 v. --------------------------- Add rxnx; add half cell oxidation half to reduction half. Cell rxn is Zn + 2Ag^+ ==> Zn^2+ + 2Ag Ecell = 0....
May 13, 2014

Chemistry
n = 2 and you know that from the equation I wrote earlier of Zn + 2Ag^+ ==> Zn^2+ + 2Ag The term you are talking about is (RT/nF)ln[(Zn^2+)/(Ag^+)] and if you want to substitute R, T, n, and F in there you may. However, if you do all of that AND multiply by 2.303 to convert...
May 13, 2014

Chemistry
It really helps us help you if you use the same screen name. Tom, Tyler, help ASAP etc is not helpful at all. You know from your previous post how to calculate the standard Zn/Ag^+ cell potential. That is Eocell and that is for standard concentrations of 1M Zn^2+ and Ag^+. ...
May 13, 2014

Chemistry ASAP
yes
May 13, 2014

Chemistry ASAP
Look up the reduction potential for Zn^2+ + 2e ==> Zn(s) in your text. Reverse the sign. Add that to the reduction potential for Ag^+ + e ==> Ag(s) That potential will be for the reaction of Zn(s) + 2Ag^+(aq) ==> 2Ag(s) + Zn^2+(aq)
May 13, 2014

Chem 130
E= 2.180E-19 J x (1/n^2)
May 13, 2014

Chem 130
[Kr]4d25s2 http://www.webelements.com/zirconium/atoms.html
May 13, 2014

Math 6th Grade
Do you honestly think -15 = -21? How can that be. 2 = 2 and 6 = 6 and -2 = -2 but -15 = -21?
May 13, 2014

chemistry
You have two typos in your equation. The correct equation, balanced, is CaCO3 + 2HCl ==> CaCl2 + H2O + CO2 mols CaCO3 = grams/molar mass mols CO2 = mols CaCO3 from the coefficients in the balanced equation. Now g CO2 = mols CO2 x molar mass CO2.
May 13, 2014

Chemistry
I would first convert 40% HCl into molarity. 40% (I guess that's w/w) HCl means 40g HCl/100 g solution. 40g/36.5 = approx 1.10 mols HCl mass = volume x density so volume = mass/density. That 100 g soln = 100/1.2 = approx 80 mL or 0.08 L; therefore, the molarity is 1.1/0....
May 13, 2014

science
You can add the phases. You have it partly balanced but I am removing the 2 and 3 and start over. First you write the products (and that isn't considered part of the balancing). H3PO4(aq) + CaCl2(aq) ==> Ca3(PO4)2 + HCl You do these by trial and error; basically that ...
May 13, 2014

SIG FIGS
1.75E2
May 13, 2014

college chemistry
To melt ice requires q1 = mass ice x heat fusion. To raise T of melted ice at zero C to Tfinal requires q2 = mass melted ice x specific heat liquid H2O x (Tfinal-Tinitial) To cool the water from 75C to Tfinal requires q3 = mass liquid water x specific heat x (Tfinal-Tinitial...
May 13, 2014

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