Wednesday
April 1, 2015

Posts by DrBob222


Total # Posts: 48,035

chemistry
http://www.enotes.com/homework-help/what-causes-differences-melting-boiling-points-451657
February 25, 2015

chemistry
sure. Since it is an endothermic reaction, increasing T will shift it to the right and that will produce more NO
February 25, 2015

chemistry
N2 + O2 + heat ==> 2NO You go through the same reasoning as the last problem we did. You didn't have pressure in that one but the rule on pressure is an increase in P shifts the equilibrium to the side with fewer mols gas. Since you have 2 mols on the left and 2 mols on...
February 25, 2015

Science
I believe the answer is D.
February 25, 2015

chemistry
You're welcome. I'm sorry I misunderstood.
February 25, 2015

chemistry
You should have told me you can get to that and we could have cut out half of the work. If you use atm for P, then R is 0.08206 L*atm/mol*K.
February 25, 2015

chemistry
mols KClO3 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols KClO3 to mols O2. Now use PV = nRT at the conditions listed to calculate volume in L O2.
February 25, 2015

chemistry
Le Chatelier's Principle tells us that if we do something to a system at equilibrium that the system will react so as to undo what we did to it. I will rewrite the equation like this, N2(g) + 3H2(g) ==> 2NH3(g) + heat. So if we decrease T, it will try to increase T. How...
February 25, 2015

science
I think the same conclusions would have been drawn.
February 25, 2015

chemistry
Add eqn 1 to the REVERSE of eqn 2. k1 = Keq for eqn 1. k2 = Keq for eqn 2. k' = keq for the REVERSE of eqn 2 = 1/k2 When you add equilibrium equations you multiply the k values so k new rxn = k1*k' = k1*(1/k2) = ?
February 25, 2015

Chemistry Pre-AP
Sara, I worked this for you under your othr screen name of Anon above.
February 25, 2015

Chemistry
Abby, Anon, Sarah, et al. You certainly make things difficult by not sticking to the same screen name. I've answered this above.
February 25, 2015

chemistry
mols Na = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols Na to mols H2. Then mols H2 x (22.4 L/mol) = ?L H2 gas @ STP.
February 25, 2015

Chemistry
I think it's D. I flirted with A and B as answers until I realized the problem doesn't give any solubility data and it doesn't define small and large.
February 25, 2015

Chemistry
It CAN'T be C. A SMALL amount of salt and a LARGE amount of water would be dilute.
February 25, 2015

Chemistry
You should be able to know this just by looking. What exactly do you not understand about the process?
February 25, 2015

chemistry
BTW, just a reminder that THERE ARE NO UNITS in Kp, Kc, Ksp, Keq, etc. I know some profs use units (some because they don't know better) and if they call them provisional units that makes it ok, but technically, activities go in those constants and activities don't ...
February 25, 2015

chemistry
I assume you mean you START with pN2 = 1.50 atm and START with pO2 = 0.5 atm? ..........N2 + O2 ==> 2NO I........1.50..0.5.....0 C.........-x....-x.....2x E.......1.50-x..0.5-x...2x Substitute the E line into K expression and solve for x, and the othr values.
February 25, 2015

Chemistry
You are absolutely right and for that very reason.
February 25, 2015

CHEMISTRY
Partly on the right track and partly not. . Calculate the theoretical yield of Ti(s). Ok, The atomic weight of Ti=47.88 Cl=35.45 (*4) Mg=24.31 (*2) Total Molar Mass of TiCl4=189.68g/mol ok to here to calculate TY of Ti = 47.88/189.68= 0.252 moles of Ti??? No. mols = grams/...
February 25, 2015

Chemisty
See my response above.
February 25, 2015

Chemistry
See my response above.
February 25, 2015

chemistry
http://www.enotes.com/homework-help/what-causes-differences-melting-boiling-points-451657
February 25, 2015

Chemistry
I'm not quite sure of how much "estimation" the author wants but this is what I would do. .......HSO4 ==> H^+ + SO4^2- I......0.15.....0......0 C.......-x......x......x E....0.15-x.....x......x K2 for H2SO4 = (x)(x)/(0.15-x) I would solve that exactly which ...
February 25, 2015

Chemistry
q1 = heat needed to raise T of solid ice from -15 to zero C. q1 = mass ice x specific heat x (Tfinal-Tinitial) q2 = heat to melt ice at zero C to liquid H2O at zero C. q2 = mass ice x heat fusion q3 = heat needed to raise T of liquid water from zero C to 100 C. q3 = mass water...
February 25, 2015

chemistry
You are right.
February 25, 2015

chemistry
Convert 8.59g CO2 to grams C. 8.59 x (atomic mass C/molar mass CO2) =? Convert 3.52 g H2O to g H (atoms). 3.52 g H2O x (2*atomic mass H/molar mass H2O) =? Then g O = 4.30-g C - g H = ? Now convert grams each to mols. mols C = grams C/atomic mass C mols H = grams H/atomic mass ...
February 25, 2015

Chemistry
I would classify HCl as a molecular compound. I also call HCl gas a polar covalent compound. It conducts electricity when placed in solution with H2O because it reacts with H2O to produce ions. HCl(g) + H2O(l) ==> H3O^+(aq) + Cl^-(aq)
February 25, 2015

chemistry
They are essentially covalent bonds but all of these acids ionize slightly in H2O solution.
February 25, 2015

chemistry
#4 is not true; the others are true. I why #3 has that "according to the book". There are many posts I see and even a few texts that claim to show units for Keq, Ksp, Kc, Kp. etc. but in fact according to the book or not there are no units for these constants. The ...
February 25, 2015

Chemistry
Neither sucrose nor paraffin in molten form will conduct electricity. Palmitic acid has a pKa of 4.78 which is almost the same as acetic acid so I would expect it to conduct but be a poor conductor typical of a weak acid.
February 25, 2015

Physical sciences(Chemistry)
I have never heard an explanation with more malarkey than this one. Pure baloney. Nonsense.
February 25, 2015

Physical sciences(Chemistry)
Hydrogen bonding in ethanol which isn't possible in hydrocarbons or bromoethane.
February 25, 2015

Chemistry
(P1/T1) = (P2/T2) Substitute and solve for P2, then compare with p1.
February 25, 2015

Chemistry
You slove it this way. mols FeS2 = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols FeS2 to mols SO2. Now convert mols SO2 to volume. mols SO2 x 24 dm3/mol = ? dm3.
February 25, 2015

Science
Grammar aside, the answer is D. However, I think the question confuses accuracy with precision. The D answer is for increasing precision; it may or may not improve accuracy. None of the other answers will improve accuracy either.
February 25, 2015

chemistry
Using the coefficients in the balanced equation lets you convert mols of mols of anything to anything in the equation. mols KCl = grams/molar mass = approx 5 but that's an estimate. 5 mols KCl x (3 mols O2/2 mol KCl) = 5 x 3/2 = ? mols O2. Then grams = mols o2 x molar mass...
February 25, 2015

Chemistry
pCO = (1348/760) = approx 1.8 atm but you need to do it more accurately. pH2O = 1780/760 = approx 2.3 atm. .........CO(g)+H2O(g)⇌CO2(g)+H2(g) I.......1.8....2.3.....0......0 C.......-x......-x.....x......x E......1.8-x...2.3-x...x......x Substitute the E line into Kp ...
February 25, 2015

AP Chemistry
a. rate = k(A)^2(B)^3 c. If A is doubled 2^2 = 4 so rate is 4x. d. If B is doubled 2^3 = 8 so rate is 8x. e. You get the idea
February 24, 2015

Chemistry
Remember how to work this type stoichiometry problem. Step 1. Convert what you have (in this case TiCl4) to mols. mols = grams/molar mass Step 2. Using the coefficients in the balanced equation, convert mols of what you have to mols of what you want(in this case Ti). With the ...
February 24, 2015

Chemistry
Wouldn't that be 12 cm - 4.92 cm = ? The one component with RF value of 1.0 means it is at the solvent front of 12 cm as you calculated and that is 12 cm. The other one is at 4.92. . zero cm | |x 4.92 cm | | | | +x 12 cm solvent front and one component.
February 24, 2015

Chemistry
Yes, the RF value is the ratio of the distance the solute moved to the distance the solvent front moved; so 0.41 = x distance/12 x = 12 x 0.41 = ? cm
February 24, 2015

chemistry
mols KHP = grams/molar mass mols OH = mols KHP since the equation is 1:1
February 24, 2015

chemistry
I don't understand the question. Perhaps another tutor will.
February 24, 2015

honors chemistry
Partly yes and partly no. All of the materials will give you the same number of particles except NaCl will be twice the others. # molecules will be 10 x 6.0E23. For NaCl # ions will be 2 x 10 x 6.02E23. For NaCl I'm assuming it will be in whatever form that allows us to ...
February 24, 2015

honors chemistry
No. mols = grams/molar mass. Since the molar mass of each of those substances is different then the number of mols must be different. And if the number of mols is different then mols x 6.02E23 will be different for the number of particles. The number of particles (ions) for ...
February 24, 2015

chem
And you know how to do what parts? Show your work and explain what you don't understand about those parts you need help with.
February 24, 2015

chemistry
I think something is amiss in this problem. Let's go the other way. mols C = 1/12 = 0.083333 mols H = 0.033597/1.00794 = 0.03333 ratio is C2.48 to H1.00 which rounds to C5H2 and this is an unlikely hydrocarbon. So I don't think all of the looking in the world will help.
February 24, 2015

chem
Fe2(SO4)3 + 3Ba(OH)2 => 2Fe(OH)3 + 3BaSO4
February 24, 2015

Science
ccw. counterclockwise
February 24, 2015

chemistry
What an interesting problem. I don't think I've seen anything like this in my entire career. I'll get you started. wt% = 2.87 which means 2.87 g solute/(100 g solution)]*100 = 2.87 That is [(2.87g solute)/(2.87 solute + 97.13 g C6H6)]* 100 So we multiply by 10 to ...
February 24, 2015

Chemistry
You didn't explain anything. My assumption is that for #1 you are taking 2 mL of the stock solution and adding 8 mL distilled H2O and you want to know the concentration of the final solution when you start with that 2 mL of 7.03 umol/L. Again, I assume you want it in umol/...
February 24, 2015

Chemistry
You need to explain this a little better; I don't know what you're asking. I understand the 7.03 umol/L is what you have but I don't know what you're to do with it. What's the 2,8,4,6,6,4 mean?
February 24, 2015

Chemistry 101
Go back and look at the NaCl/Cl2/H2 problem you posted earlier. Those steps are the same for solving this problem. If you need help show where your trouble is and we an help you through.
February 24, 2015

Science
I disagree with A. It's true that methane is a gas but I think that's a secondary issue. Its flammability is what you are talking about.
February 24, 2015

Chemistry
C = 31.57% H = 5.30% Therefore, O = 100%-31.57-5.30 = about 63%. Take a 100g sample which will give you 31.57g C atoms. 5.30g H atoms. 63g O atoms. Convert to mols and I'll estimate. 31.57/12 = about 2.63 mols C 5.30/1 = 5.30 mols H 63/16 = about 3.94 mols O Now find the ...
February 24, 2015

Chemistry
heat gained by ice + heat gained by melted ice + heat lost by 75 mL liquid H2O = 0 (mass ice x heat fusion) + (mass melted ice x specific heat liquid H2O x (Tfinal-Tintial) + (mass 75 mL H2O x specific heat liquid H2O x (Tfinal-Tinitial) = 0 Substitute all of the numbers and ...
February 24, 2015

Chemistry
I think for part 1 they want grams. Your 1.39 mols Cl2 is right; just multiply by molar mass Cl2. part 2. H2 done the same way you did #1. mols NaCl = 163/58.44 = ? mols H2 = 1/2 that from the balanced equation. grams H2 = mols H2 x molar mass H2 which is 2 g/mol. #2. P4(s) + ...
February 24, 2015

chemistry
See below
February 24, 2015

chemistry
It is wrong but only in places and I don't understand why you made those errors. molecular equation is H2SO4(aq) + 2NH4OH(aq) ==> (NH4)2SO4(aq) + 2H2O(l) CIE: 2H^+(aq) + SO4^2-(aq) + 2NH4^+(aq) + 2OH^-(aq) ==> 2NH4^+(aq) + SO4^2-(aq) + 2H2O(l) NIE: 2H^+(aq) + 2OH^-(...
February 24, 2015

Just a question
From a purely practical standpoint, my son received a B. S degree in mass communication, signed on with a newspaper as a part time reporter, became a full time, and finally an editor of a small newspaper. With the right connections and a good grasp of the English language and ...
February 24, 2015

Chemist
a. rate = k(NO)^2(Cl2) b. The order of the equation is 2+1 = 3
February 24, 2015

science
Yes, I suppose one could build a 1 square mile enclosed area and stick a vacuum pump on it. Or for that matter, there are many square miles of space above the atmosphere without any oxygen and no enclosure is required. But as for moving that "particular" part of ...
February 24, 2015

To B.
Stop trying to be funny. You're almost spamming the board. You are't offering any solutions and you're taking up time and space. Cut it out. You COULD be banned from this site and you may want/need to use this site one of these days.
February 24, 2015

Math
Don't be ridiculous. Or something like 2+2 = 10-6?
February 24, 2015

Chem 111
React with what?
February 23, 2015

chemistry
1. millimols = mL x M = 5 mL x 0.002M = 0.01 mmols or 1E-5 mols. mols SCN = same 2. ...........Fe^3+ + 2SCN^- ==> Fe(SCN)2^+ I..........1E-5....1E-5........0 C...........-x......-2x........x E.........1E-5-x...1E-5-2x......x x from the problem is 3E-4M and that x 0.01L (10 ...
February 23, 2015

To B.
Stop trying to be funny. You're almost spamming the board. You aren't offering any solutions but you're wasting time and using up space so cut it out. You COULD be banned from this site but think about it. You may want/need this site some day, especially if you ...
February 23, 2015

Chemistry
q = mass Cu x heat fusion Cu
February 23, 2015

chem
2AgNO3 + Na2CO3 ==> 2NaNO3 + Ag2CO3(s) Before we can do anything you must identify 3.10 what and 4.43 what. This is a limiting reagent problem (LR) and you know that because amounts are given for BOTH reactants. I work these the long way. I will assume those are grams. mols...
February 23, 2015

chemistry
heat lost by object + heat gained by water = 0 [mass object x specific heat object x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute the numbers from the problem and solve for specific heat object.
February 23, 2015

To B.
I don't know but Dawn said use only addition and subtraction and that's what I did.
February 23, 2015

3rd grade math
B= ride bikes W = walk D = driven to school ================== W = B+6 B = D W + B + D = 24 For D you can substitute B For W you can substitute B + 6 Do that and solve for B.
February 23, 2015

Math
If you add units to it and watch the number of significant figures, yes. The proper unit is cc (cubic centimeters)
February 23, 2015

Math
What's confusing about l*w*h/3 ?
February 23, 2015

5th grade math
You want one side to be s and the other side 1.6s s*1.6s = 360 Solve for s and 1.6s 15 and 24?
February 23, 2015

Chemistry
mols O2 = 8.58E21/6.02E23 = ? Using the coefficients in the balanced equation, convert mols O2 to mols FeCl2. That's ? mols O2 x (4 mols FeCl2/3 mols O2) = ? Then M FeCl2 = mols FeCl2/L FeCl2. You know M and mols, solve for L
February 23, 2015

chemistry
mmols FeCl2 = mL x M = ? mmols KOH = 2x mmols FeCl2 M KOH = mmols KOH/mL KOH. You know mmols kOH and M KOH, solve for mL KOH.
February 23, 2015

chemistry
dE = q + w w = -346J from the problem. dE increased by 6365 J. Solve for q Then q = mass x specific heat x (Tfinal-Tinitial) Substitute and solve for sp. h.
February 23, 2015

chemistry
q = mass H2O x specific heat H2O x (Tfinal-Tinitial) You had 0.310 x 0.06 = about 0.0186 mols Ba(OH)2 and that produces 2 mols H2O. dHrxn = q/0.0186 and divide that by 2 to find per mol H2O.
February 23, 2015

Chemistry
You change the sign of qcombustion rxn so that qcal = -qrxn. qcal = Ccal x delta T q = 3949*1.331/141.44 = ? Then ? = Ccal*delta T. Solve for Ccal.
February 23, 2015

Chemistry
q = heat released = heat cap x delta T q = 38.5 kJ/K x 2.93 = approx 113 kJ but you need a better answer than my estimate. That's approx 113 kJ/5.80 g candy bar = approx 19 kJ/gram. I would then convert 19 kJ/g to J/g, then to calories (that's with a small c) using 4....
February 23, 2015

chemistry
You're only choice is to work the problem and see. This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. I work these the long way. 2Na(s) + Cl2(g) = 2NaCl(s) mols Na = grams/molar mass mols Cl2 = grams/molar mass Using the...
February 23, 2015

science
http://education.nationalgeographic.com/education/encyclopedia/weathering/?ar_a=1
February 23, 2015

Solution Making
That looks ok to me.
February 23, 2015

life orientation
http://brainly.com/question/305722
February 23, 2015

Chemistry
I think the m.p. of group I and group II elements are lower than the transition metals. As for the second part, I don't know unless metals with m.p. close together tend to solidify at one time while those of widely different m.p. might segregate during cooling. The Fe, Co...
February 23, 2015

Chemistry
sp g. = density HAc/density H2O 1.040 = (density HAc)/(0.9956 g/ml) solve for density HAc
February 23, 2015

CHEMISTRY
Change in color of the indicator. Formation of ppt In some cases dissolution of ppt. Formation of gas (often CO2 with carbonate samples)
February 23, 2015

CHEMISTRY
Looks ok to me.
February 23, 2015

chemistry
Dalton's law of partial pressures. The total is the sum of each. ptotal = pH2 + pAr + pHe
February 23, 2015

AP Chem
I believe Bob P counted the H atoms wrong. It's 10 from the (N2H5)2 and 4 more from the other part of the formula for a total of 14 and not 22.
February 22, 2015

chemistry
I don't see anything wrong with this. I would have used the HH equation but what you used is perfectly ok. In fact, if you substitute those values into the HH equation you get a pH of 6.999 which should be ok. pH = 7.20 + log(96.6/153.4) = 6.999 which rounds to 7.0 and ...
February 22, 2015

Chemistry
What's the problem with taking 15 mL of the regular household bleach and adding salt water to make 30 mL of solution. That will dilute the initial bleach by a factor of 2 and it will be just 1/2 what you started with.
February 22, 2015

Chemistry
I've answered this before. Use the Henderson-Hasselbalch equation. You don't get anywhere by just recopying the problem; however, if you show what you're tried and/or tell us what you don't understand about the solution we should be able to help you thorough it.
February 22, 2015

Chemistry
heat lost by Fe + heat gained by water = 0 [mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute and solve for mass H2O
February 22, 2015

Chemistry
Kp = Kc(RT)^delta n. Solve for Kc. Convert 1 g SO2 and Cl2 to mols/L Calculate Qc and compare that with Kc. Post your work if you get stuck.
February 22, 2015

Chemistry
(NOBr) = 0.782 mols/100L = approx 0.008 but you need to do it more accurately as well as all of the othr values I've estimated. .......NOBr(g) ↔ NO(g) + 0.5Br2(g) I......0.008......0..........0 C......-x.........x.......0.5x E.....0.008-x.....x.......0.5x Substitute ...
February 22, 2015

Chemistry
You change the T of 100 g water by 10 C. So how much will T change for 200 g water? Could that be 5 C?
February 22, 2015

chemistry
heat added to melt ice + heat gained by water from ice + heat lost by 125 mL H2O = 0 (22.7g ice x heat fusion) + (22.7 g H2O x 4.184 x (Tfinal - Tinitial) + (125 x 4.184 x (Tfinal-Tinitial) = 0 I get approx 11 degrees C. Solve for Tfinal.
February 22, 2015

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