Thursday
July 30, 2015

Posts by DrBob222


Total # Posts: 49,590

Chem
Devron is right. You need to know the order of the reaction. I copied the following from Wikipedia. For order zero, the rate coefficient has units of mol·L−1·s−1 (or M·s−1) For order one, the rate coefficient has units of s−1 For ...
April 23, 2015

Chemistry
The word I was trying to think of when I wrote "temporary" was conditional. Many authors write conditional units; I suppose that's because they don't like to see numbers with no units. Technically, however, activities don't have units so neither do Kc, Kp...
April 23, 2015

Chemistry
It depends upon the k. Ksp, Ka, Kb, Keq, Kc, Kp and the like have no units because activities are supposed to be used for the species and activities have no units. Some authors use units and some argue these Ks have units but they do not. The units used (and they vary) are ...
April 23, 2015

Chemistry
If it's a strong acid it will ionize 100% so the pH will be -log(acid). If it's a weak acid the acidity will be less than that and you can gauge that with the pH.
April 23, 2015

Chemistry
Same process as your Hvap problem.
April 23, 2015

Chemistry
How many mols in 8.51E10 molecules? That's 8.51E10 moleceles x (1 mol molecules/6.02E23 molecules) = ? Then q in kJ = mols H2O x Hvap(kJ/mol)
April 23, 2015

Science
refrigerators, freezers, microwaves/ranges to boil water, etc.
April 23, 2015

chemistry
I2 + 2S2O3^2- ==> S4O6^2- + 2I^- mols S2O3^2- = M x L = ? Convert mols S2O3^- to mols I2 using the coefficients in the balanced equation. Then M S2O3^2- = mols S2O3^2-/L S2O3^2-
April 23, 2015

chemistry
C6H8O6 + I2 + H2O → C6H6O6 + 2I^- + 2H^+ mols I2 = M x L = ? mols vit C = mols I2 since the ratio is 1:1 in the balanced equation. g vit C in the titrated portion = mols x molar mass. Convert to mg vit C the multiply by 500/25 to give mg vitamin C in the tablet.
April 23, 2015

Science
I do these limiting reagent problems the long way. HCl + NaOH ==> NaCl + H2O mols HCl = M x L = about 0.056 but you need a better answer than that estimate. That will produce 0.056 mols NaCl mols NaOH = M x L = approx 0.0625. That would produce approx 0.0625 mols NaCl. In ...
April 23, 2015

chemestry
Convert 1.453 g benzoic acid to mols. mols = grams/molar mass = ? q = Ccal x (Tfinal-Tinitial) +3227000 x mol = Ccal x (delta T) You know everything but Ccal. I wonder if that is 2.265 T increase.
April 23, 2015

chemestry
Note that you need to spell chemistry correctly. The heat of solution which is the heat lost when KCl dissolves, is given by the decrease in temperature of the water as q = mass H2O x specific heat H2O x (Tfinal-Tinitial) That q is for 0.75g which is 0.75g/molar mass KCl = ...
April 23, 2015

Physical Science
In multiplication and division, which is all you have in this problem, the rule is that you may have as many significant figures in the answer as you have in any of the numbers. So where did the 4.49 come from? 5.00 x (750/760) x (273./300.) = 4.49. You're allowed 3 s.f., ...
April 23, 2015

DrBob222-re: chem question
[H^+](OH^-] = Kw = 1E-14 You have OH^- and Kw, solve for H^+
April 23, 2015

Chemistry
This is the equilibrium equation. Are you looking for the Keq expression? Keq = (Fe^2+)(H2S)(H2O)^2/H3O^+)^2 Some may not include the H2O in the constant; the general rule is it isn't included EXCEPT where it is involved in the reaction. Since it is produced in the ...
April 23, 2015

Chem - Acid & Bases
I see two statements but no question. And is tht [H-] a typo?
April 23, 2015

chemistry
What's the powder? Is there a reaction? What's the product?
April 23, 2015

chemistry
mL x (1 L/1000 mL) x M x molar mass = grams or you can do it in pieces. mols Mg(NO3)2 = grams/molar mass Then M = mols/L. You know M and mols, solve for L and convert to mL.
April 23, 2015

chemistry
Read it from the graph.
April 22, 2015

chemistry
http://www.jiskha.com/display.cgi?id=1429763582
April 22, 2015

chem
% w/v = (g solute/mL)*100 = ?
April 22, 2015

Chemistry
See your post below on NO/NO2
April 22, 2015

Chemistry
dGorxn = (n*dGo formation products) = (n*dGo formation reactants) dGorxn - is spontaneous. dGorxn + is not spontaneous. dGorxn = 0 reaction is at equilibrium.
April 22, 2015

Thank you
You're welcome.
April 22, 2015

chem
dE = hc/wavelength
April 22, 2015

chemistry
Use the Henderson-Hasselbalch equation. Get data from the equation and an ICE chart. millimoles NaOH = M x mL = apprx 38 but you need a bettr answer than that millimols CH3COOH = x .......CH3COOH + NaOH ==> CH3COONa + H2O I........0.......38.........0.........0 add......x...
April 22, 2015

Chem - Molarity
mols NaCl = grams/molar mass M NaCl = mols NaCl/L solution = approx 1.5 M but that's an estimate to use in the next part. Then mL1 x M1 = mL2 x M2 mL1 x approx 1.5 = 1300 x 0.1M mL1 = ?
April 22, 2015

Chemistry
Of course not. If you didn't convert mols glucose to mols CO2 it can't be right.
April 22, 2015

Chemistry
No. mols glucose is right. You didn't convert mols glucose to mols CO2; i.e., you left out the step before the x 22.4
April 22, 2015

Chemistry test can't fail please help (senior)
John, you work all of these stoichiometry problems with 3 or 4 steps. 1. Write and balance the equation. 2. Convert what you have to mols. There are a couple of ways to do this. a. If you have grams, then mols = grams/molar mass. b. If you have a solution, then mols = M x L...
April 22, 2015

Chemistry
See my response above.
April 22, 2015

Chemistry
How do you get to that number? Details please.
April 22, 2015

Chemistry
C'mon John, Bob Pursley told you how to work the problem, I showed you in more detail how to do it, now this is the second post after all of that. You've had all of the help you can get from us EXCEPT I shall be happy to explain anything you don't understand. The ...
April 22, 2015

Chemistry
......N2 + 3H ==> 2NH3 I.....1.....3.......0 C....-x...-3x......2x E....1-x..3-3x.....2x Ptotal = 1-x+3-3x+2x = 4-2x Then XNH3 = (2x/4-2x)= 0.21 Solve for x, then evaluate pNH3, pN2 and pH2. Finally, Ptotal = the sum of the partial pressures.
April 22, 2015

Chemistry test help please
You work this exactly like the problem with glucose and oxygen. 1. Write and balance the equation. 2. Use the coefficients to convert mols of what you have to mols of what you want. 3. Convert mols of what you want to either grams or volume. g = mols x molar mass volume = mols...
April 22, 2015

Chemistry
1. mols glucose = grams/molar mass 2. Using the coefficients in the balanced equation, convert mols glucose to mols CO2. 3. Now convert mols CO2 to L. L CO2 = mols CO2 x 22.4 L/mol = ?
April 22, 2015

science
First, how are you going to get water at 135C?
April 22, 2015

science
c = cool w = warm [mass w H2O x specific heat H2O x (Tfinal-Tinitial)] + [mass c H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Subsitute and solve for Tfinal.
April 22, 2015

science
Use the same formula for the next question up, substitute, solve for specific heat metal. Post your work if you get stuck.
April 22, 2015

AP Chemistry
Plug and chug into the HH equation. These problelms, where no concentration is given for the buffer, is a wide open problem with no restrictions. Here is what you do. Let x volume(in mL) CH3COONa and 10-x = volume in mL of CH3COOH. pKa CH3COOH is about 4.74 but you need to use...
April 22, 2015

AP CHEM
Normally buffering range is considered to be +/- 1 pH unit from the pKa value.
April 22, 2015

chemistry
pH = 6.5 = -log(H^+). Solve for (H^+). Convert to M H2SO4. Then mols acid = (H2SO4) x L = ? Using the coefficients in the balanced equation, convert mols H2SO4 to mols CaCO3, then convert to grams CaCO3.
April 21, 2015

chemistry
pH = ? or (H2SO4) = ?
April 21, 2015

Chemistry
I don't think so. 0.00215 is right and 0.022 is right. mols sodium acetate should be 0.00025 If you will post your set up I'll find the error.
April 21, 2015

Chemistry
Use the Henderson-Hasselbalch equation. You have (acid). Look to see how much of the sodium acetate was formed and calculate concn of that. Plug in HH equation and pH comes out as the answer.
April 21, 2015

Science
Move the flaps. Moving them down, although increasing the lift, also increases the drag.
April 21, 2015

Chemistry
%w/w, %w/v. From the way the problem is stated I assume you want %w/v. % w/v = (grams solute/mL volume) x 100 = % w/v (40/220)*100 = ?
April 21, 2015

Chemistry
Let me forget the steps you go through and go through your problem step by step; then you ask questions if needed. But I think all will be clear. First, F is a gas. The periodic table consists of 5 gases (+ the noble gases but let's not worry about them right now), 2 ...
April 21, 2015

Chemistry
PV = nRT
April 21, 2015

Science
http://www.nasa.gov/audience/foreducators/k-4/features/F_Four_Forces_of_Flight.html
April 21, 2015

Chemistry
If you don't care what the strength of the buffer is you can do it this way. Then when you finish you can calculate the strength of the solution you have prepared. let x = mL base; then 1000-x = mL acid. 3.5 = 3.74 + log (0.1x)/[(0.1)(1000-x)] Then solve for x and 1000-x. ...
April 21, 2015

Chemistry
It would help if you gave the values of Ka when problems like this is posted. Since Ka values differ from text to text (and from website to website) we can't get the same answer you get if we don't use the same numbers you have. Anyway, I took your 0.575 and calculated...
April 21, 2015

To Tracey
Regarding the FeO==> Fe(OH)3 problem below, take a look at the additional post. Here is the link. http://www.jiskha.com/display.cgi?id=1429650539
April 21, 2015

Chemistry
Your complete balanced equation is correct. Good work! Your complete ionic equation is correct except for omitting the (aq) on S^2-(aq) on the left. I'm sure that's just a typo. Still a good job! Now, how do you turn a complete ionic equation into a net ionic? You'...
April 21, 2015

Chemistry
The empirical formula is the simplest formula that can be written for a compound; in this case what you have written is already in the lowest form since 7 is a prime number and can't be divided by anything except 7.IF it were (and it isn't) C6H6O3 it could be divided ...
April 21, 2015

Chemistry
Generally you will be right doing it that way. If the material is soluble and you have a liquid solution (such as HCl, H2SO4, HNO3, etc are all in aq solution) so the solubility rules work very well. Yes, gases are easy. If you're mixing solutions that makes it easy, too.
April 21, 2015

Chemistry
While eating supper it occurred to me that something was amiss about that problem. How do you get an aqueous solution of FeO. FeO is a solid and very little of it will dissolve. This may be one of a series of reactions for the formation of rust. Anyway, I wonder if that should...
April 21, 2015

Chemistry
1. It won't balance because you don't have the right formula down. The left side is ok but iron(III) hydroxide is Fe(OH)3, 2. It is a pptn reaction because aq/g/l on the left goes to a solid on the right. However, it ALSO is a redox reaction. Fe on the left is +2 and ...
April 21, 2015

science
http://www.merckmanuals.com/home/hormonal-and-metabolic-disorders/biology-of-the-endocrine-system/endocrine-function
April 21, 2015

Please Read
Thank you.
April 21, 2015

AP Chemestry
You didn't list the units of k but from your work I'm guessing that is in hours and you used minutes. When I substituted 2.5 hours in for time I obtained 3.82 M. That answer makes no sense when both A and B are listed as 3.82 M but that could be a printing error. Check...
April 21, 2015

AP Chemestry
What are the units on k = 4.10E-2 what? If units are not available is this a first order reaction? You have the same two answers for A and B? Is that right? If you will answer those question AND show your work I expect I can find your error. Assuming first order reaction I ...
April 21, 2015

chemistry
Please explain because I don't understand. You have the data and the formula. What's left? Plug and chug.
April 21, 2015

Chemistry AP
I didn't look at any of these individually but the subject is right. Click on these and see if any of them help. https://www.google.com/search?q=chemical+equilibrium+practice+problems.&ie=utf-8&oe=utf-8 As an addition, I have had luck with the Google of "Calculations ...
April 21, 2015

Physical Chemistry
I showed you how to do that when you first posted. All you need to do is punch those numbers into your calculator. I'll be glad to explain anything you don't understand but re-posting the question doesn't do any good and wastes our time.
April 21, 2015

Chemistry
It is not a pure sample because the melting point range is too large.
April 21, 2015

Chemistry
looks good to me.
April 20, 2015

Chemistry
Add water, dissolve the K2SO4, filter, the K2CO3 solid will not go through the filter paper. Dry the paper to recover the K2CO3. Evaporate the water from the filtrate to recover the K2SO4.
April 20, 2015

Physical Chemistry
238 amu x (1.66E-24 g/1 amu) = ? grams.
April 20, 2015

Science
http://www.lselectric.com/differences-between-electric-motors-and-generators/
April 20, 2015

To Kristin
I responded to that pKa problem. Here is a link. http://www.jiskha.com/display.cgi?id=1429545787
April 20, 2015

CHEM
See this. http://www.jiskha.com/display.cgi?id=1429576498
April 20, 2015

Laboratory Chemistry
To make this clear, the DPM is 1000/0.01 mL
April 20, 2015

Laboratory Chemistry
1000 is right.
April 20, 2015

Science (Chemistry)
Ionic materials are polar. Water is a polar solvent. Hexane is a non-polar solvent.
April 20, 2015

Chemistry
Multiply equation 2 by 2 and add equation 1 to equation 2. Then cancel anything (mole wise) that is common to both sides. For example 2SO2 on the right and 2SO2 on the left will cancel.
April 20, 2015

Science
1 mol of anything is 6.02E23 of those anythings. 1 mol S contains 6.02E23; therefore, if we have a different number, that is mols S = 9.3E24/6.02E23 = ?
April 20, 2015

Science
mols H2O = 1,000/18 = ? One (l) mole contains 6.02E23; therefore, ? molx x 6.02E23 = # H2O molecules.
April 20, 2015

AP Chemistry
whew! I'm glad I didn't goof.
April 20, 2015

AP Chemistry
I don't know. I don't think it can be PbCl2 forming since the two beakers are connected only with a salt bridge. The cathode surely must be the Cl2 since the Pt is an inert electrode and no conditions are given for the H2O there (that is I don' think we would worry...
April 20, 2015

AP Chemistry
If you know the anode and cathode already you can get the reaction from that. At the anode you have Pb ==> Pb^+ + 2e At the cathode you have Cl2 + 2e ==> 2Cl^- Add the two half equations for the cell reaction. You can add the phases.
April 20, 2015

san jac
V proportional to n V = 3.5 when n = 5 Therefore, if n = 2.5, V must be ...?
April 20, 2015

maths lit,life sciences,business studies,history
And what qualifications do you have?
April 20, 2015

Chemistry
Yes but I would keep the 6.29 since the 536 allows for three significant figures. Don't throw numbers away unless it's necessary.
April 20, 2015

chemisty
251J x (1 L*atm/101.325 J) = 2.48 L*atm work = -p(V2-V1) -(-2.48) = 2.7(V2-0.35) Solv for V2 in L.
April 20, 2015

Chemistry
You can reason this out. pH = pKa + log (base)/(acid) So if pH = pKa, then log (base)/(acid) must be zero. Set that as an equation to get log (base)/(acid) = 0 and 10^0 = ? ? must be 1; how do you get 1 out of log(base)/(acid). The only way to obtain 1 is for (base) = (acid) ...
April 20, 2015

Chemistry
Yes, 5.75E-8 is the answer for pH = 7.24. You should do pH = -log(H3O^+), then 7.24 = -log(H3O^+) -7.24 = log(H3O^+) (H3O^+) = 10^-7.24 = 5.75E-8
April 20, 2015

Chemistry
You calculated the pOH. The question asks for pH. After you have pOH, then pOH + pH = pKw = 14. You know pKw and pOH, solve for pH.
April 20, 2015

Chemistry
You took my response correctly and the correct answer is the HCN/CN pair because if Ka = 4.9E-10 then pKa = 9.31 which is quite close to the desired pH of 9.2; therefore, it is easy to make solutions in which the ratio of base/acid is close to 1. If you use the HH equation, ...
April 20, 2015

Chemistry
Actually you don't need to use the HH equation at all. The rule is that to make a buffer you want pKa to be within about 1 unit of the desired pH. So convert each of the Ka values to pKa values and choose the one that is within about 1 unit. And if you are good at scanning...
April 20, 2015

chemistry
mols ZnO desired = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols ZnO to mols O2. Then convert mols O2 to grams. g = mols x molar mass = ?
April 20, 2015

Chemistry
Please note the correct spelling of celsius. delta T = i*Kf*m The ice T is -0.4 so delta T is 0-(-.4) = 0.4 i for NaCl is 2 (two particles) Kf for water is 1.86 solve for molality.
April 20, 2015

Chemsitry
Use the Henderson-Hasselbalch equation.
April 20, 2015

gravimetric
[mass BaCrO4 ppt x (atomic mass Ba/molar mass BaCrO4)/mass sample)]*100 = %Ba For mass BaCrO4 substitute the following: (mg BaSO4 x 100/1000) into the top equation and solve for mass sample. Post your work if you get stuck.
April 20, 2015

gravimetric prob
AgCl is converted to Ag metal in the photodecomposition so the 10 mg that decomposed weighs 10mg x (atomic mass Ag/molar mass AgCl) = ? The other 10 mg does the same = ? So the new weight of the AgCl ppt is 704.7 mg - ? - ? = x mg AgCl + Ag.
April 20, 2015

gravimetric prob
Difference between initial and final weight perchlorate = mass H2O Difference between initial and final weight ascarite = mass CO2 %H = [(mass H2O x 2*atomic mass H/molar mass H2O)/mass original sample]*100 = ? % C = [(mass CO2 x atomic mass C/molar mass CO2)/mass original ...
April 20, 2015

Chemistry
NaCl added to ice lowers the freezing point of ice to below zero C, melting the ice takes energy, the energy must come form someplace; it comes from the water/ice mixture. The result is that the temperature of the ice/salt/water mixture is below zero. It can be MUCH lower than...
April 19, 2015

Chemistry
That looks ok to me but you're not allowed that many significant figures. If it were me I would round to 40.2 times.
April 19, 2015

chemistry
What do you think and why? Hint: Which will NOT change the concn of the substance in any way?
April 19, 2015

Science
No.
April 19, 2015

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