Sunday
August 31, 2014

Posts by DrBob222


Total # Posts: 43,419

Chemistry
NOPE. You changed SO4^2- on the right to S2O4^2- and you can't do that. Also you have the e on the wrong side. Other than that it is ok. 2OH^- + SO3^2- ==> SO4^2- + 2e + H2O
May 4, 2014

Chemistry
I shall be glad to help you through this but you need to learn to do these yourself. I do these the looong way. The first step is to determine the oxidation step of each element. I will help you here. S on the left is +4 and on the right is +6. Add electrons to the appropriate...
May 4, 2014

chemistry
It will take mass H2O x heat vaporization to turn the 30g into steam at 100 C.
May 4, 2014

Chemistry
This one is right. If you haven't already do so go back and look at the sulfite ==> sulfate because the one is not right.
May 4, 2014

Chemistry
I think the trouble you are having is caused by not putting the charges in. That H on the right should be 4H^+. The MnO4 on the right should be MnO4^-. When you do that the e need on the right is not 8 but 3 so the charge on the left is zero and on the right is (-1+4-3) = zero...
May 4, 2014

Chemistry
Again you need to learn to do these. Mn on the left is +4 and on the right is +7. Cu on the left is 2+ and on the right is zero. Balance the electron change first. Mn on the left goes to Mn on the right with ? change of e. Cu on the left goes to Cu on the right with ? change ...
May 4, 2014

CHEMISTRY
I worked this for you yesterday.
May 4, 2014

Chemistry
mols malic acid = grams/molar mass M = mols/L solution.
May 4, 2014

Chemistry
I worked it using the numbers you provided and came up with the same answer you did of 0.963 so there is nothing wrong with your math. The problem is that the reduction potential of Ni is not 0.26. It is -0.26.
May 4, 2014

chemistry
I don't believe you didn't know how to do any of this. 1. delta Tb = impure b.p.- pure b.p. 2. delta T = Kb*m Substitute and solve for m. 3. Isn't the mass given in the problem. Convert to kg. 4. m = mols/kg solvent. You know m and kg solvent, solve for mols
May 4, 2014

chemistry
See your post above.
May 4, 2014

chemistry
I suggest you read these things before you post them. You provided absolutely no information other than it is in the introductory information. But I don't see any introductory information.
May 4, 2014

Chemistry
You say you've looked up these values. My values won't be the same as yours because I'm using an old book but they will be close. Here is how you do the first one. Cr==> Cr^3+ + 3e You look in your table and find this potential. It will be listed "backwards...
May 4, 2014

chemistry
(H3O^+) = (A^-)
May 4, 2014

chemistry
pH = -log(H3O^+) 4.95 = -log(H3O^+) -4.95 = log(H3O^+) Another way of writing this is (H3O^+) = 10^-4.95 Solve for (H3O^+)
May 4, 2014

chemistry
An interesting problem. Three equation and three unknowns but we can work it around with just two since these are related. Let X = mass BaO and Y = mass CaO and mm stand for molar mass. ---------------- eqn 1 is X + Y = mass whatever that may be. Convert to sulfates and you ...
May 4, 2014

chemistry
Mg + 2HCl ==> MgCl2 + H2 mols Mg = X mols H2 produced = X (look at the coefficients; 1 mol H2 produced for 1 mol Mg used) Volume = mols*22.4L/mol volume 22.4X liters.
May 4, 2014

to Bob P--Chemistry
I've gone cross eyed with these large numbers but I don't believe H and O balance in your equation.
May 4, 2014

Chemistry
2C60H87O23N12P +173O2 ==>120CO2 + 72H2O + 24HNO3 + 2H3PO4 Check this carefully. mols biomass = grams/molar mass Using the coefficients in the balanced equation, convert mols biomass to mols CO2. Now convert mols CO2 to grams. g = mols CO2 x molar mass CO2
May 4, 2014

chemistry
pH = -log(H^+)
May 4, 2014

Chemistry
The easiest way to do these problems is in parts and you need three formulas to do it. q at a phase change; i.e., solid to liquid or liquid to gas. For solid to liquid (melting point) it is q = mass solid x heat fusion. For liquid to gas (boiling point) it is q = mass liquid x...
May 4, 2014

CHEMISTRY
I worked this for you last night. Did you look at that?
May 4, 2014

chemistry
I believe you forgot the coefficients (e.g. 8). 8CH4 ==> C8H8+ 12H2 dGfrxn = (n*dGfC8H8 + n*dGfH2)-(n*dGfCH4) 620.735 = [(1*214.417) + (12*0)] - (8*dGfCH4) 620.735 = 214.417 + 0 - 8x Solve for x.
May 3, 2014

Chemistry
dH = mass H2O x specific heat H2O x (Tfinal-Tinitial) This will give you the enthalpy change for the reaction as stated in the problem and that is for 0.05g P. Generally these problems don't want that; they want the enthalpy change per mol but this problem doesn't ask ...
May 3, 2014

chemistry
I suppose you are to use M and m as the same. 0.121g/100 gH2O = solubility. mol N2O = 0.121/60 = approx o.002 and 0.002/0.1kg solvent = 0.02m p = KcC 1 atm = Kc*0.02 Kc = 1 atm/0.02 = 50 atm/m p = KcC 1.600 = Kc from above *C Solve for C in molality and convert to g/100 if ...
May 3, 2014

Science
In warmer climates the winters are warmer. The moisture from a hot shower is in the form of vapor (gas); however, it could be a liquid if it hasn't had time to completely vaporize and the water droplets are still visible (and you can feel those droplets on the skin).
May 3, 2014

Chemistry
mols H2SO4 = M x L = approx 0.1 mols NaHCO3 = M x L = approx 0.05 2NaHCO3(aq) + H2SO4(aq) ==>Na2SO4(aq) + 2H2O(l) + 2CO2(g) This should get you started. Post your work if you get stuck and clearly explain what you don't understand.
May 3, 2014

Chemistry
m = mols/kg solvent 2.0 = mols/3.5 Solve for mols, then mol = grams/molar mass. You know m and molar mass, solve for grams.
May 3, 2014

Chem Check Please!!!
I don't agree with 4. I agree with 5 and 6.
May 3, 2014

Chem Check Please!!!
I don't agree with 2. I agree with 3.
May 3, 2014

Chem Check Please!!!
I don't agree with #1.
May 3, 2014

Chemistry
I think Ni is initial N. and Nf is final N so the electron is making the transition from N = 1 to N = 2 so this is an absorption but the formula I list is for emission. That won't change frequency or wavelength but it will change the sign of E. delta E = R(1/1 - 1/4) where...
May 3, 2014

Chemistry
Take 100 g sample for 58.5g C 4.1g H 11.4 g N 26.0 g O Convert to mols. I'll use approximations. 58.5/12 = 4.875 4.1/1 = 4.1 11.4/14 = 0.814 26/16 = 1.625 Now find the ratio in small whole numbers with the smallest number being 1.00. The easy way to do that is to divide ...
May 3, 2014

General Chem 2
2Al^3+ + 6I^- ==> 2Al(s) + 3I2(s) dGo = -nFEo n = 6
May 3, 2014

Chemistry - Science (Dr. Bob222)
x = 2 y = 1 How did I do that? V on the left has a oxidation state of +5 and on the right it is +4. S on the left is +4 and on the right is +6. V has changed by 1e S has changed by 2e So V couple must be multiplied by 2 and S couple by 1 to make each lose/gain 2 electrons.
May 3, 2014

Chemistry
mols = grams/molar mass
May 3, 2014

Chemistry - Science (Dr. Bob222)
Thanks. I just didn't think the 50% could be right. The concept behind disproportionation is that an element in an "intermediate" oxidation state (in this case Cu^+ is halfway between Cu below it and Cu^2+ above it) can "react with itself" if the ...
May 3, 2014

chemistry - Science
I would do this. Calculate g Zn that goes into solution and g Cu that are deposited from solution. Coulombs = amperes x seconds = ?. 95,485 coulombs will use (65.4/2)g Zn and deposit (63.5/2)g Cu. Calculate g Zn used, convert to mols, and add to the 0.1 mol already there. ...
May 3, 2014

chemistry - Science
See your other post.
May 3, 2014

Chem Check Please!!!
2 is ok 3 is ok I don't agree with 4.
May 2, 2014

Chem Check Please!!!
1. Cris, I don't agree with C as an answer. Here is the way you keep these straight. Just remember the definition of oxidation and reduction. Oxidation is the loss of electrons. Reduction is the gain of electrons. The substance oxidized is the reducing agent. The substance...
May 2, 2014

chemistry
By the way, it's past my bed time but I'll be around tomorrow.
May 2, 2014

chemistry
87 is right. 37 and 50 are right Check your numbers for PEN. You can't possibly get 74 something. 50 x 1+ is about 50. 37 x 1+ = about 37 and add in a pinch for electron mass and you have AT LEAST 50+37+pinch = 87. And it can't be less than 86.909187 because it LOSES ...
May 2, 2014

chemistry
I can't follow what you've done. I have mass p + mass N + mass e = 87.723173 and that - 86.909187 = 0.813986 amu and that converted to kg = 1.3516563E-27 kg. Let me know if we are on the same page to this point and please show your work for as far as you are and we can...
May 2, 2014

chemistry
How would you answer these and why?
May 2, 2014

Chemistry
d2sp3 but obviously those atoms without d orbitals are not eligible for this.
May 2, 2014

chemistry
You can deposit 196.97/3 = approx 66 g Au with 96,485 coulombs of electricity. So how much will be required to deposit 19.7g 96,485 x (19.7/66) = approx 29,000 coulombs. coulombs = amperes x seconds. Substitute coulombs and seconds and solve for amperes. Note those numbers I ...
May 2, 2014

Chemistry
TiO2 + 2C ==> 2CO + Ti We don't need to make any corrections for molar mass and kg if we keep everything in kg. kg mols C = 14.0/12 = approx 1.17 kg mols TiO2 = 44.0/79.88 = approx 0.5 Using the coefficients in the balanced equation, convert kg mols of each to kg mols ...
May 2, 2014

Science
I'm not positive how to answer this. Certainly b and d are not driven by humans and can be eliminated. Global warming, c, is attributed to human impact so that's a definite yes. A is the one I'm unsure about. The greenhouse effect is the name we have given to gases...
May 2, 2014

CHEM 101
We can't draw pictures, structures, diagrams, etc on this forum.
May 2, 2014

chemistry
CaCO3 ==> CaO + CO2 mols CaCO3 = grams/molar mass. Using the coefficients in the balanced equation, convert mols CaCO3 to mols CO2. Now convert mols CO2 to grams. g = molls x molar mass.
May 2, 2014

chemistry chem science
time in seconds is 35 min x (60 seconds/min) = ? amperes = 160 mA = 0.160 A. Calculate Q = # coulombs = amperes x seconds. Then you know that 96,485 coulombs will deposit 112.4/2 grams Cd (that's molar mass Cd/valence). (112.4/2)g Cd x ?coulombs from above/96,485 = ?
May 2, 2014

chemistry
Instead of someone just giving you the answers, I would like to know what it is you don't understand.
May 2, 2014

Chemistry 1
The highest freezing point will be the one with the smallest delta T and that will occur for the material with the smallest number of particles; i.e., highest van't Hoff factor x molality or i*m A. 0.1M KCl im = 2*0.1 B. 0.05M NaCl im = 2*0.05 C. 1M AlPO4 im = 1*3 D. 0....
May 2, 2014

chemistry
(P1V1/T1) = (P2V2/T2) You don't need to use moles of gas if you use this formula. Notice that you have no initial volume and no actual final volume (just twice the initial). I would pick a convenient number (any number) for V1 and multiply that by 2 for V2.
May 2, 2014

Chemistry
2NaOH + H2SO4 ==> Na2SO4 + 2H2O mols H2SO4 = M x L = ? Using the coefficients in the balanced e4quatioin, convert mols H2SO4 to mols NaOH. Now M NaOH = mols NaOH/L NaOH. You know mols and M, solve for L NaOH and conert to mL if you wish.
May 2, 2014

Chemistry
mols C2H5OH = grams/molar mass m = mols/kg solvent
May 2, 2014

Chemistry
C2H8 + 5O2 ==> 2CO2 + 4H2O When working with gases we can use a shortcut in which we use L as if L = mols. Then we make the conversion form mols propane to mols O2 as usual. ?L O2 = 3.8 L C3H8 x (5 mols O2/1 mol C3H8) = 3.8 x 5 = ?
May 2, 2014

Chemistry
delta T = i*Kf*m 2 = 1*Kf*0.5 Solve for Kf.
May 2, 2014

chemistry
You know we can't draw pictures on this forum but the half reactions are MnO4^- + 3e + 2H2O ==>MnO2 + 4OH^- NO + 4OH^- ==> NO3^- + 3e + 2H2O I have written the MnO4^- as a reduction an the NO as an oxidation so as to keep the electrons on opposite sides. I think you ...
May 2, 2014

questions
When the car burns fuel (either gasoline or diesel) it combines with oxygen and under the best of conditions would produce CO2 and H2O. However, with the heat of the engine (due to the combustion that is necessary to power the car) some carbon monoxide, NOx which is NO, NO2, ...
May 1, 2014

Chemistry- is this correct?
To two significant figures, yes. If that 0.30 was really 0.300 (and you just omitted a zero) then you could have 3 s.f. and that would be 0.125M. As posted, however, 0.13 is right.
May 1, 2014

To JJ
Your problem is somehow intertwined with another and I can't post an answer there. Solve the problem by applying the dilution formula of mL1 x M1 = mL2 x M2
May 1, 2014

chemistry
Don't you have a graph or a table that gives you solubility at various temperatures?
May 1, 2014

Chemistry
You want 5L of O2. And if you could get 100% yield from the reaction you could calculate how much NaNO3 would be required to produce 5L O2. However, the problem says the reaction is only 78.4% yield so we must start with more NaNO2 so that at the end with just 78.4% yield we ...
May 1, 2014

Chemistry
2NaNO3 ==> 2NaNO2 + O2 5L O2 at 78.4% yield would need to be 0.784 = 5/x and x = about 6.4 6.4L at STP = ? mols mol = 6.4L/22.4L = approx 0.29 Convert mols O2 to mols NaNO3. That's 0.29mol O2 x 2 mol NaNO2/1 mol O2) = approx 0.57 mol NaNO3. Then g NaNO3 = mols x molar ...
May 1, 2014

chemistry
Use PV = nRT. T must be in kelvin. I don't know what 0.01m is. If you mean for that to be 0.01 mols, that will be n
May 1, 2014

Chemistry
M = mols/L. 0.142M NaI = 0.220 mols NaI/ L solution. Solve for L.
May 1, 2014

Chemistry
mL1 x M1 = mL2 x M2
May 1, 2014

Chemistry
Let's call this BNH2 to save some typing. ..........BNH2 + HOH --> BNH3^+ + OH^- I........0.223.............0.......0 C.........-x...............x.......x E.......0.223-x............x.......x But you know x = 9.7E-6. Evaluate 0.223-x and substitute into the Kb ...
May 1, 2014

chemistry
Please clarify the question. What is 5.75gmolecules?
May 1, 2014

chemistry
.........NH4HS(s) ==> NH3(g) + H2S(g) I........solid.........0........0 C........solid.........x........x E........solid.........x........x Kc = (NH3)(H2S) 1.8E-4 = x^2 x = 0.013M. Does this exhaust the 5.00 g NH4HS? 0.13 mols/L x 3L = 0.040 mols H2S and 0.040 mols NH3. How...
May 1, 2014

typo--Chemistry
Specific heat is 4.184 and not 4.194. I typed it right in the equation but wrong above that.
May 1, 2014

Chemistry
It would have helped if you had re-posted the question; however, I found it and here is the response I made. heat lost by pebble + heat gained by water = 0 heat lost by pebble = 532.1 J heat gained by water = mass H2O x specific heat H2O x (Tfinal-Tinitial) Set q = heat gained...
May 1, 2014

Science
I think this is nonsense but the answer probably is B. That's the only one of the four that doesn't have the same thing on the right that's on the left.
May 1, 2014

Chemistry
I expect the difference in #1 is that I didn't read anything on the calculator but kept everything in until the final answer.
May 1, 2014

Chemistry
I agree with #2. For #1 I obtained 1.1E-11
May 1, 2014

Chemistry
If you have used the screen name Sha in the last day or so I worked this problem earlier. I may be able to find it if you didn't.
May 1, 2014

Chemistry
You made a typo in th equation. It should be CH3NH2 + HOH ==> CH3NH3^+ + OH^- I obtained 2.9E-2 for OH^- but we may not have used the same Kb. I used 5E-4.
May 1, 2014

Chemistry
NH4^+ is the only one hydrolyzed. You know NO3^-, Cl^- and K^+ are not hydrolyzed BECAUSE those hydrolysis products would be STRONG acid/bases. For example, Cl^- + HOH ==> HCl + OH- or K^+ + HOH ==> KOH + H^+ and that doesn't happen. Why NH4^+? BECAUSE it forms a ...
May 1, 2014

chemistry
Let's just eliminate some answers. You know A can't be right because there are a number of "essential" inorganic compounds. You know B can't be right--why do we need H2 and O2? What about C? D is ludicrous. I know people who live to be 100 and never at ...
May 1, 2014

Chemistry
My tables aren't quite the same as your (but I have an old old text). But dSrxn = (n*dSproducts) - (n*dSreactants) dSrxn = (2*NaOH + H2) - (2*Na + 2H2O) What stands out is the -S(H2) you have. Shouldn't that be a +. If this doesn't fix the problem please repost ...
May 1, 2014

Chemistry
You have two sets of P, T, V. (P1V1/T1) = (P2V2/T2)
May 1, 2014

college chemistry
When you can do the one before this you can do this one.
May 1, 2014

College chemistry
Where in the world are you getting your numbers. I don't see 2.02 or 12.01 in the problem. Wait, I suspect you are using molar mass H2 and molar mass C and molar mass. That isn't what the problem is all about. dGrxn = (n*dGproducts) - (n*dGreactants) They give you ...
May 1, 2014

College chemistry
Instead of me flying in the dark why don't you post your work on the products-reactants you've tried and let me find the error in your work.
May 1, 2014

Chemistry
For #2. The half rxn are Zn ==> Zn^2+ + 2e and Cu^2+ + 2e == Cu To find Ehalf cell for Zn use E = Eocell - (0.05916/n)*log[1/(Zn^2+)]. Plug in the (Zn^2+), calculate Ehalf cell (using the reduction potential you find in your standard potential table), then reverse the sign ...
May 1, 2014

Chemistry
You haven't asked a question here but more like a book of questions. Do you just want to "compare" answers or do you have a real problem with understanding a particular part? Clarify what you don't know and explain in detail and we can help you through it. I ...
May 1, 2014

science
You apparently have no concept of how the freezing point of a pure material such as water is affected by the solute. Out of four choices you guessed wrong on three so th answer obviously must be A. Also it is obvious you didn't look up anything about colligative properties...
May 1, 2014

science
B is not true. C is not true. Look up colligative properties.
May 1, 2014

Buying A Car
I agree with Ms. Sue.
May 1, 2014

8th grade science
Check out horizon.
May 1, 2014

chemistry
You can reason through it this way PV = nRT. Since P, R and T are constant, then V = n*constant which I'll call k and V = nk. Evaluate k by 31.4 = 1.5k and k = 31.4/1.4 = 20.93. Then V = nk 41.4 = n*20.93 and n = 41.4/20.93 - 2.03 mols. OR you can reason that a higher ...
April 30, 2014

Chemistry
heat lost by pebble + heat gained by water = 0 heat lost by pebble = 532.1 J heat gained by water = mass H2O x specific heat H2O x (Tfinal-Tinitial) Set q = heat gaine by water and solve for mass H2O.
April 30, 2014

Chemistry
heat gained by cold water + heat lost by hot water = 0 [mass cold H2O x specfic heat H2O x (Tfinal-Tinitial)] + [mass hot H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute the numbers and solve for Tf.
April 30, 2014

Chemistry - check answers
These look good to me. Good work.
April 30, 2014

Chemistry
My guess is that you didn't take into account the total volume. millimols HCl=mL x M = 25 x 0.250 = 6.25 mmols NaOH = 15 x 0.35 = 5.25 NaOH + HCl ==> H2O + NaCl mmols HCl = 6.25 mmols NaoH = 5.25 difference is excess HCl by 1.00 mmol. So you have 1.00 mmols HCl in a ...
April 30, 2014

chemistry
You need to write the entire problem. There isn't enough information here to answer anything.
April 30, 2014

Chemistry
millimoles NH3 = mL x M = 100 x 0.1 = 10 mmols NH4^+ = 100 x 0.1 = 10 mmols HCl added = 5 x 0.1 = 0.5 .......NH3 + H^+ ==> NH4^+ I.....10.....0........10 added.......0.5............... C....-0.5..-0.5......+0.5 E.....9.5.....0......+10.5 pH = pKa + log(base)/(acid) pH = pKa...
April 30, 2014

CHEMISTRY HELP !!!!
Your post said "E measured" so I assume you have that given to you. If it isn't given, add the oxidation half cell to the reduction half cell to obtain Ecell.
April 30, 2014

Pages: <<Prev | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | Next>>

Search
Members