Thursday
December 18, 2014

Posts by DrBob222


Total # Posts: 46,131

Chemistry
Of course they do.
December 6, 2014

chemistry
Both explain how NH3 acts as a base.
December 6, 2014

caution---chemistry
I agree with 0.2278 mols; however, you are limited to 3 significant figures by the 74.1 number; therefore it should be rounded to 0.228 mols.
December 6, 2014

chemistry
mols = M x L = ? But without M it can't be done unless there are other numbers.
December 6, 2014

chemistry
The Volhard method can be used for almost any ion that forms a suitable ppt with Ag ion since it is an indirect method. In practice, you add an excess of AgNO3 to form the AgX ppt, then titrate the excess Ag added with CNS^-. Br^-, I^-, Cl^-, and a host of others.
December 6, 2014

AP Chemistry
I'm at a loss. I thought, and think, the same as you and from what I can read on the web that is still true. Also I read that solutions of sugar could sit for years without the glycosidic bond being broken. Nothing was mentioned about temperature; it could be that boiling ...
December 6, 2014

chemistry
2C2H2 + 5O2 ==> 4CO2 + 2H2O b. mols C2H2 burned = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols C2H2 to mols CO2. Now convert mols CO2 to g CO2. g CO2 = mols CO2 x molar mass CO2 = ? c is done the same way. a is done the same way. For ...
December 6, 2014

Chem
p = 1 n = 1 e = 0
December 5, 2014

Chemistry
See your other posts.
December 5, 2014

Chemistry
See your last post.
December 5, 2014

Chemistry
mols NaCl = grams/molar mass osmols NaCl = 2xmols NaCl mols KCl = grams/molar mass osmols KCl = 2xmols KCl mols CaCl2 = grams/molar mass osmols CaCl2 = 3xmols CaCl2 Take sum osmols to obtain total. That/1L = ? osmolarity
December 5, 2014

Chem
I have always called it sulfur dioxide.
December 5, 2014

Chemistry
Sodium forms the sodium ion (Na^+) and bromine forms the bromide ion (Br^-). Since bromine (Br2, oxidation state of zero) goes to bromide ion (Br^-, oxidation state of -1), it is a gain of an electron for each bromide ion formed. Gain of electron is reduction. Obviously Na ...
December 5, 2014

chemistry
B is 3 N is 3 O is 2
December 5, 2014

Chemistry (Electrochemistry)
What I do is write the two half cells. Be ==> Be^2+ + 2e....E = x anode Pb^2+ + 2e ==> Pb....E = -0.13 cathode --------------------------------- Be + Pb^2+ ==> Be^2+ + Pb E = 1.72 The cell consists of an oxidation half cell (Be/Be^2+) and a reduction half cell (Pb^2+/...
December 5, 2014

Chemistry
no unpaired electrons
December 5, 2014

chemistry
1. 15 g EDTA/100 = 15%. Scale up to 250. 15 g EDTA x (250/100) = ? 2. mL1 x %1 = mL2 x %2 Substitute and solve.
December 5, 2014

Chemistry
130 x (0.128g/washer) = ?
December 5, 2014

chemistry
I don't understand your question.
December 5, 2014

Chemistry
When using gases one may use liters directly as if they were mols. 250 L O2 x (12 mols CO2/12 mols O2) = 250 x 12/12 = 250 L CO2 produced.
December 5, 2014

chemistry
With no diagram it is impossible to know what flask C, flask D, beaker E etc are.
December 5, 2014

Chemistry
Take a 100 g sample which gives you 37.01g C] 2.22 g H 18.50 g N 42.27 g O Convert grams to mols mols = grams/atomic mass. 37.01/12 = approx 3.08 2.22/1 = approx 2.22 18.50/14 = approx 1.32 42.27/16 = approx 2.64 Now find the ratio of these elements to each other with the ...
December 5, 2014

chemistry
2NaOH + H2SO4 ==> Na2SO4 + 2H2O mols H2SO4 = M x L = ? mols NaOH = twice that M NaOH = mols NaOH/L NaOH. You know M and mols, solve for L and convert to mL.
December 4, 2014

chemistry
Use (P1V1/T1) = (P2V2/T2) Remember T must be in kelvin.
December 4, 2014

science
I agree with san but the speed of light is 3E8 m/s
December 4, 2014

chem
This really is just a substitute the numbers and solve. Tell me what your problem is. If we can clear up some confusion I'll bet you can do this on your own.
December 4, 2014

Test
Yes, write it down. I found that writing it on cards, read the stack of cards a time or two, then I would throw the cards away and write everything on new cards. Writing it multiple times helped me. For those facts that were so easy to confuse, I made mnemonics as memory aids.
December 4, 2014

Algebra B PLEASE HELP
Isn't that 0.036 or 3.6%
December 4, 2014

Chemistry honors
So post what you've done and we can direct you where you've gone wrong.
December 4, 2014

Chemistry
This is done just like the Al/H2SO4 question. 10 mols H2 x (2 mols NH3/3 mol H2) = 10 x 2/3 = ?
December 4, 2014

Chemistry
The "coefficient factor" will convert any thing to anything. 5.0 mol Al x (3 mols H2SO4/2 mols Al) = 5.0 x 3/2 = ?
December 4, 2014

Bcc
If you posted your question under statistics instead of a cryptic Bcc you might get help a lot faster
December 4, 2014

chemistry
mols = grams/molar mass. You can calculate molar mass and you're given moles, solve for grams.
December 4, 2014

chemistry
Ni^2+ + 2EDTA ==> Ni(EDTA)2 mols EDTA - 0.0075 x 0.15 = ? mols Ni = 1/2 that g Ni = mols Ni x atomic mass Ni = ? %yield = (g Ni/mass sample)*100 = ?
December 4, 2014

Chemistry
No, not the T. It could be EXCEPT Ksp is given at the T in the problem so that is ok. The problem here, I think, is that you failed to write the equation and because of that you slipped up on a tiny bit of the problem. Here is the dope. .......Ca(OH)2 --> Ca^2+ + 2OH^- I...
December 4, 2014

Science
Use (P1V1/T1) - (P2V2/T2) Remember T must be in kelvin.
December 4, 2014

chemistry
I suppose you assume the density of H2O is 1 g/mL, then 300 mL has a mass of 300 g. q = [mass H2O x specific heat H2O x delta T] That q, in Joules, is caused by 9.0 g KCl So heat solution is + (it gets colder) is q/9.0 g. Change that to kJ/g and that x molar mass KCl changes ...
December 4, 2014

chemistry
If you assume the density of H2O is 1 g/mL, then 18 mL has a mass of 18 grams. mols H2O = grams/molar mass = ? There are 6.022E23 molecules in 1 mol. There are 10 electrons in 1 molecule of H2O. BTW, Avogadro's number is 6.022E23 and not 6.023E23.
December 4, 2014

Chemistry
Solve 1b before 1a. Solve for molarity of the HCl solution. That's M = mols/L. 1.2 g/mL x 1000 mL x 0.37 x (1 mol/molar mass HCl) = M concentrated solution. 1a. Then use the dilution formula of mL1 x M1 = mL2 x M2 200 mL x 2 M = mL2 x Mfrom 1b. Solve for mL to take of the ...
December 4, 2014

typo?--Chemiry 11
Let me point out that you probably meant Cr2(SO4)3
December 4, 2014

Chemistry
The formula for ppm is (grams solute/grams solvent)*1E6 = ? ppm. (g ddt/g seal fat)*1E6 = ppm (0.00450 g/10,000 g)*1E6 = ?
December 3, 2014

Chemistry
http://www.thebigger.com/chemistry/thermodynamics/first-law-of-thermodynamics/calculate-q-w-%E2%88%86u-and-%E2%88%86h-for-the-reversible-isothermal-expansion-of-one-mole-of-an-ideal-gas-at-37oc-from-a-volume-of-20dm3-to-a-volume-of-30-dm3/
December 3, 2014

chemistry
So the equation transformed to the letters is A(aq) + B(l) ==> C(aq) + D(aq) The whole point of this problem is that when writing equilibrium constants one does NOT include pure liquids or solids so Keq for this one is (C)(D)/(A)
December 3, 2014

physical chemistry
mols manitol = grams/molar mass = ? mols H2O = grams/molar mass = ? Find total mols by adding. Then find mol fraction H2O = mols H2O/total mols. Then Psoln = XH2O*Popure H2O
December 3, 2014

CHEMISTRY
Neither. pH + pOH = pKw = 14. You know pH and pKw, solve for pOH. Then pOH = -log(OH^-). Solve for OH^- You can do this another way and get the same answer. pH = -log(H^+). Substitute pH and solve for (H^+), then (H^+)(OH^-) = Kw = 1E-14. You know H^+) and Kw, solve for OH^-
December 3, 2014

CHEMISTRY
Just remember the definition. An acid is a proton donor; a base is a proton acceptor. What does all of that mean? Look for the one on the left with more H than the one on the right with fewer H. More H means acid; fewer H means conjugate base. Crazy talk. The acid is HCO3^- ...
December 3, 2014

oops==chemistry
Your original answer of 1E-7M = (OH^-) is correct.
December 3, 2014

Chemistry
coulomgs = amps x sec c = 46.0hrs x (60 min/hr) x (60sec/min) = ? Then there are 96,485 coulombs in 1 mol of electrons. ?coulomgs x (1 mol/96.485) = ?
December 3, 2014

Chemistry
work in L*atm = -p*(Vfinal-Vinitial) Then L*atm x 101.325 = ? joules.
December 3, 2014

chemistry
heat lost by condensing steam + heat lost by condensed steam @ 100 cooling to Tf + heat gained by cool(6C) water = 0 -(2259 J/g x 0.524) + [(0.524 x 4.18 x (Tfinal-Tinitial]) + [(mass cool water x 4.18 x (Tfinal - Tinitial)] = 0 Solve for Tf. My answer is about 75 C or so.
December 3, 2014

note--chemistry
My solution assumes the the ice is at zero C.
December 3, 2014

chemistry
heat lost by ice + heat gained by melted ice + heat lost by water @ 20 cooling to 6 = 0 (x g ice x heat fusion) + [(x g melted ice x specific heat H2O x (Tfinal-Tinitial)] + [mass 20C H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Solve for x grams ice. The answer I obtained...
December 3, 2014

chemistry
With the two other heat/ice/water problems I've worked, try this one on yur own and see if you can't do it. Same principles. If you have trouble repost at the top of the page and tell us exactly what is troubling you.
December 3, 2014

Chemistry
It's difficult to write subscripts and superscripts here so here is what I do. The first number is the subscript(the atomic number) and the next number (on the right) is the superscript (the mass number). The sum of subscripts on the left must equal the sum of subscripts ...
December 3, 2014

Science
I must confess that I, too, am confused but mostly because I don't know what gamma is in your equation. I THINK, but I'm not positive, that you look in the table, find the activity coefficient for H2PO4^- and HPO4^2-. Then I would be inclined to use 2 for H2PO4^- and 3...
December 3, 2014

chemisty
Sorry. I didn't intend to post this twice.
December 3, 2014

chemisty
I believe you missed the point of my response. My response was that the first sentence waxes supreme about diiodine pentafluoride reacting spectacularly with BrF3 BUT YOUR EQUATION USES I2O5 AND IF5 is a PRODUCT and not a reactant. I assume you know this is a limiting reagent ...
December 3, 2014

chemisty
I believe you missed the point of my response. My response was that the first sentence waxes supreme about diiodine pentafluoride reacting spectacularly with BrF3 BUT YOUR EQUATION USES I2O5 AND IF5 is a PRODUCT and not a reactant. I assume you know this is a limiting reagent ...
December 3, 2014

Chemistry
The lead/acid cell is PbO2 + H2SO4 ==> PbSO4 + 2H2O I assume you mean by conditioning that you are charging the cell from scratch. So the water is electrolyzed and you would expect H2 gas to be formed at the - electrode and O2 gas to be formed at the + electrode.
December 3, 2014

chemistry
We can't draw on this forum. See if you can Google it.
December 3, 2014

chem
can't draw on this forum. I suggest you Google the name or formula and see if you can find it on the web.
December 3, 2014

chemistry
Alkanes have free rotation around the single bond; the pi bond keeps the alkenes from rotating
December 3, 2014

chemistry
I worked this problem for Tina below. http://www.jiskha.com/display.cgi?id=1417573947
December 3, 2014

chemistry
No, the solutions will not remain at 1 M. At the anode the metal ion will increase and it will decrease at the cathode (assuming of course that it is a M/M^+||M^+/M cell.
December 3, 2014

Chemistry
I showed you how to do this. It's two steps. mols HCl = grams HCl/molar mass HCl Take that number of mols you calculated and put it into the equation below. Then M = mols HCl/L H2O. What's your trouble? All you have to do is to substitute the numbers into the equations...
December 2, 2014

Chemistry
See my answer above.
December 2, 2014

Chemistry
No. 45 g is not 45 mols. mols = grams HCl/molar mass HCl as I posted above. THEN, M = mols/1.5 = ?
December 2, 2014

Chemistry
mols HCl = grams/molar mass M HCl = mols HCl/L HCl = ?
December 2, 2014

Chemistry
There may be people, including chemists, who know and/or remember these dates for these people but I'm not one of them. Frankly being able to spit out these dates, in my opinion, has little to do with knowing chemistry. The easy way to get this information is to go to ...
December 2, 2014

chem(REALLY NEED HELP )
And I don't either. Your calculation obtains the grams octane IF the entire 1000 mL is octane or grams heptane iF the entire 1000 mL is heptane but it doesn't tell you the grams of each in gasoline if gasoline were a mixture of heptane and octane.
December 2, 2014

chem(REALLY NEED HELP )
1000 mL gasoline contains how much octane and how much heptane?
December 2, 2014

chemistry-URGENT
7.654E32 atoms x (1 mol/6.022E23 atoms) = ?
December 2, 2014

Chemistry
Ecell = Eocell - (0.0592/n)*log(Q) where Q is Keq expression. You will need to calculate the pH of the solution which can be done with the Henderson-Hasselbalch equation. That's pH = pKa + log (base)/(acid). Use 0.10M for acid and 0.12M for base. Solve for pH and convert ...
December 2, 2014

Chemistry
The solubility of CaF2 is increased due to the presence of H^+ because H^+ confines with the F^- to form the weakly ionized HF. The easy to do this is as follows. First, determine the value of H^+) from the buffer part of the problem. pH = pKa + log (base)/(acid). You know pKa...
December 2, 2014

Chemistry
So write the Keq expression, substitute the concentrations given and solve for the concn of H2O. Post your work if you get stuck.
December 2, 2014

Chemistry
This is a problem that almost has to be worked "backwards" to make it work. .....Fe^3+ + 3C2O4^2- ==> [Fe(C2O4)3]^3- I....0.02....1.00...........0 C....-0.02..-0.06..........0.02 E......0.....0.94..........0.02 What I have done above is to look at the size of Kc (...
December 2, 2014

Chemistry
.....cis-butene==> trans-butene I......0.250........0.145 C.......-x...........+x E......0.250-x.....0.145+x Substitute the E line into Kp expression and solve for x, 0.250-x and 0.145+x. How did I know to make the C line -x......+x an not +x......-x; i.e., how do I know it...
December 2, 2014

Chemistry
If you care to post your work I shall be happy to go over it to find the error. But first, did you use Ea in Joules/mol and not kJ/mol (a common mistake) and did you cnvert T1 and T2 to kelvin (not so common error). The usual other problem is to reverse k1 and k2. Just make ...
December 2, 2014

Chemistry
Can't you use the Arrhenius equation for this. You have Ea, k1 along with T1 and T2; the only unknown is k2.
December 2, 2014

chem please help drbob222
Since we have so many people dancing here let me dance too. Another reason, and just as important, is to utilize the common ion effect of the excess ion. In doing so you shift the equilibrium, just as Le Chatelier said will happen, and you DECREASE the solubility of the stuff ...
December 2, 2014

chem please help drbob222
I think your answer is the one that the "prof" wants but I might clarify a little. Yes, NaNO3 is not easily determined both because most Na salts are soluble and most nitrate salts are soluble so there is no easy way to convert them to insoluble forms. Of course you ...
December 2, 2014

Honors Chem
First you must determine how much heat must be removed to make that temperature transition. q = [mass H2O x specific heat H2O x (Tfinal0Tinitial)]. q should be in J. Then calculate mols NH4Cl needed to get that q. 14,800 J/mol x # mols = q(in J) Then convert mols NH4Cl to grams.
December 2, 2014

chemistry
Potassium alum is KAl(SO4)2.12H2O 0.422g Al x (molar mass KAl(SO4)2.12H2O/atomic mass Al) = ? theoretical yield.
December 2, 2014

chemistry
Use PV = nRT
December 2, 2014

Chemistry
10 min/2 min half life = 5 So it has gone through 5 half lives. 2^5 = 32 10g initially/32 = ? what's left.
December 2, 2014

chemistry
This is done the same way as the Ba-122 problem above.
December 2, 2014

Chemistry
I think your problem is taken care of with the units. Since k has units of mol/L*atm that is M/atm and C comes out in M.
December 2, 2014

Chemistry
Do you recognize that this is a limiting reagent problem? That may be the problem. If you had posted what you did I could have pointed out the exact place you went wrong. Determine the limiting reagent. Using that substance, convert mols of the LR to mols of the product Then ...
December 2, 2014

Chemistry
0.5/0.70 = approx 0.714 g of the 70% stuff you need. Then convert grams to mL using density. Remember the %yield problems you worked early in chemistry? 70% stuff x what number = 0.5 g real stuff what number = 0.5/0.70 = about 0.7 g. Doesn't that make sense. You must take ...
December 2, 2014

Chemistry Problem
If you still want to see this done please post again at the top of the page and when you do note that the first sentence talks about diiodine pentafluoride but the reaction you give with the problem uses I2O5 (which I always called iodine pentoxide but I guess the proper name ...
December 2, 2014

Chemistry
How many mols S are in 5.60 g S? If call sulfur as S (and not S8), the mols S in 5.60 g S will be 5.60/atomic mass S = 5.60/32 = ? So you have that many mols of atoms. If you are to use S8 as sulfur, please post again at the top of the page and bring that to our attention.
December 2, 2014

Chemistry
Cd ==> Cd^2+ + 2e BrO3^- + 6e + 6H^+ ==> Br^- + 3H2O Br goes from +5 to -1 which is 6 so 6 e must be transferred. For Eo cell add the Cd part to the BrO3^- part (oxidation half cell to reduction half cell). For Ecell = Eocell -(0.0592/n)*log(Q) What's Q? That's (...
December 2, 2014

chemistry
Since L(lower case) can be any integer from 0 to n-1, then if L is 4 n must be 5. mL can be -L to + L which means -4, -3, -2, -1, 0, +1.......+4. Stick two electrons in each mL and count or get sophisticated and think it through. For s electrons, L(lower case) = 0, for p ...
December 1, 2014

Chemistry
I think the easy way to solve this problem is to recognize that the HCl titrates the Na2CO3 and leaves the NaCl as is. Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2 Total HCl added = M x L = approx 13.365 millimols but you need to confirm all of these numbers since I'm estimated ...
December 1, 2014

CHEMISTRY
yes
December 1, 2014

Chem 130
2CO + O2 ==> 2CO2 15.0 L CO at STP = 15.0/22.4 = ?mols Convert mols CO to mol O2, then use PV = nRT at the conditions listed to solve for volume (in liters) of O2.
December 1, 2014

Chemistry
You need to re-read the chapter because all I can do is repeat much of what your text has. But here is a summary. You have 1 proton in the nucleus. The single electron usually occupies the first shell which we will call n = 1. If energy is added, and there are various ways to ...
December 1, 2014

chemistry
Laila is El wrongo! Used the factor upside down to get 15.3. 2KClO3 ==> 2KCl + 3O2 mols KClO3 = mols O2 x (2 mols KClO3/3 mols O2) = 10.2 x (2/3) = ?
December 1, 2014

CHEMISTRY
(OH^-) = 10^-9 is correct.
December 1, 2014

Chemistry
The one LOSING H^+ is the acid that's the one DONATING the proton); the one GAINING H^+ is the base (that's the one ACCEPTING the proton.)
December 1, 2014

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