Tuesday
May 26, 2015

Posts by DrBob222


Total # Posts: 49,027

Chemistry
AgCl + 2 NH3 --> Ag[(NH3)2]^+ + Cl^- K for the rxn shown is Kf*Ksp Kf*Ksp for AgCl is about 2.9E-3 Kf*Ksp for AgBr is about 8E-6 Kr*Ksp for AgI is about 2.4E-9 As we proceed from AgCl to AgI the Ksp and Krxn shifts smaller and smaller to the solubility of the AgBr is about ...
April 26, 2015

Chem
The problem asks for kJ; convert your answer in Joules to kJ.
April 26, 2015

Chem
q = mass x heat fusion = ?
April 26, 2015

chemistry
mols K2SO4 in the 40 mL sample = g/molar mass = 5.00/molar mass K2SO4 = ? Then M of the 40 mL sample is mols K2SO4/0.04 = ?
April 26, 2015

pre ap chemistry
OK, so you can't read and follow directions. You want me to do all of the work and you just copy it. Here goes BUT you must wade through the explanation as I go. How many grams do you want? That's L x M x molar mass = grams. 100 mL is 0.1 L from the problem. M is 0.500...
April 26, 2015

pre ap chemistry
Also, I'm trying to figures out just what pre-AP chemistry is.
April 26, 2015

pre ap chemistry
How many grams do you want? That's M x L x molar mass = grams. You know L, M, and molar mass. Solve for grams. What do you not understand about making this up? You always add grams to a volumetric flask, add some water, swirl until all solid is dissolved, make to the mark...
April 26, 2015

chemistry
If this is just a dilution problem, then 0.12 mM/dm-3 x (5.0 mL/25.0 mL) = ?
April 26, 2015

chemistry
What is MB?
April 26, 2015

Chemistry
I don't understand the problem either.
April 26, 2015

Chemistry
I understand the question, vaguely, but can you clue me in on the concentrations at the end of the post. You have two mM for both. Which is right for each?
April 26, 2015

Chemistry 2
dSo system = (n*dSo formation products) - (n*dSo formation reactants) dHo system = (n*dHo formation products) - (n*dHo reactants) dSo surroundings = -dHosystem/T Plug dSo surroundings and dSo system into the following and solve for dSo universe. dSouniverse = dSosystem + ...
April 26, 2015

gravimertic prob
The problem tells you %Cl = 8.43%. Is this gravimetric analysis to check that number? Convert 0.0831g AgCl to Cl. That is easiest done by 0.0831 x (atomic mass Cl/molar mass AgCl) = g Cl Then %Cl = [(mass Cl/mass sample)]*100 =
April 26, 2015

chemistry
You prepare standard solutions of anthracene and quinine in the range you expect unknowns to be, go through your procedure, measure the fluorescence of each standard and plot that [I guess you don't call it A if it is emitting :-)] against the concn to get a standard curve...
April 26, 2015

chemistry
Without any data all I can do is get you started generally. Use the excitation spectra of each to determine the wavelength to use for excitation (either at the same time or in separate runs), then use the fluorescent data (emission data) to determine the wavelength to use to ...
April 26, 2015

CHemistry
Refer to your other problem that Jai worked for you.
April 26, 2015

chemistry
jae means ==> i followed by upper case _ is ' That Ything is delta H = ?
April 26, 2015

Chemistry 2
After reading Devron's reply I realize I read your initial question wrong. His response and mine agree; the only problem with mine is I thought you were quoting dG as -220 kJ/mol where you were quoting dH. Therefore, that part I wrote about "not agreeing with" ...
April 25, 2015

Chemistry 2
First, this is at 426K and not 298 K; therefore, any number you obtain by dGo = dHo - TdSo will not agree with 220 kJ/mol. So you substitute the values you have for dHo and dSo along with 426K for T and calculate dGo. Remember dH is given in kJ/mol and dS in J/mol.
April 25, 2015

Masinde university
https://www.google.com/search?q=lattice+energies+table&tbm=isch&imgil=WmpfcEfIGpFZQM%253A%253BahYcvs4Erq8XAM%253Bhttps%25253A%25252F%25252Fusflearn.instructure.com%25252Fcourses%25252F986898%25252Ffiles%25252F31116966&source=iu&pf=m&fir=WmpfcEfIGpFZQM%253A%252CahYcvs4Erq8XAM%...
April 25, 2015

chemistry
You don't give a reaction; I assume you just didn't type it into the post. I'll write one. CaI2 + 2AgNO3 ==> 2AgI + Ca(NO3)2 4 mols CaI2. Using the coefficients in the balanced equation, convert mols CaI2 to mols AgI.
April 24, 2015

chemistry
Have you tried to read what you post with no periods? Where the hell does the sentence end and the next one start. Fortunately I know enough chemistry that I may be able to guess. You can hope that I guess right. ........2NH3 ==> N2 + 3H2 I........0......1.3....1.65 C...
April 24, 2015

Chemistry
How much solution do you have? What is the molarity of the HCl you have?
April 24, 2015

Chemistry
First, the name depends upon where the groups are located. If that is a straight chain hexane, then the ethyl group is named first followed by the number and dimethyl as in 3-ethyl-2,2-dimethyl hexane. Here is a well illustrated site. http://www.kentchemistry.com/links/organic...
April 24, 2015

Chemistry
You can't have that without having a 5-valent carbon and that isn't allowed.
April 24, 2015

chemitry
I would do this. pH = pKa + log (base)/(acid) 4.40 = 4.76 + log (base)/(acid) base = 0.3/2 = ? acid = x/2 Solve for x = mols acetic acid.
April 24, 2015

chemistry
mols NaOH = M x L = 0.01 mols HCl = M x L = 0.01 NaOH + HCl ==> NaCl + H2O 0.01...0.01 So you will form 0.01 mols NaCl and have just NaCl and H2O in the solution at the end; therefore, the pH will be that of pure H2O.
April 24, 2015

chemistry
Is this the equation? You should have given it. .........H2S ==> H2 + S I......0.102......0...0 C........-p.......p...p E......0.102-p....p...p Substitut the E line into the Kp expression and solve for p, then evaluate 0.102-p. Finally, add the partial pressures and the sum...
April 24, 2015

chemistry
Ka = ? KaKb = Kw. Substitute Ka and solve for Kb. Compare Ka with Kb.
April 24, 2015

chem
Correct products but not balanced. You should know it isn't balanced. Let's check it. I see 2 H atoms on left and right. ok I see 2 o on left aide and 3 on the right. not ok. 2H2O2 ==> 2H2O + O2
April 23, 2015

chem 2
millimols HAc = mL x M = ? = acid mmols NaAc = mL x M = ? = base. Plug these into the Henderson-Hasselbalch equation and solve for pH. part 2. mmols HCl added = 20 I will make up some numbers for mmols HAc and NaAc to make this easier to do but these aren't real numbers. ...
April 23, 2015

chemistry
The ppt is Cu(OH)2. So Cu(OH)2 ==> black? + H2O Since the problem tells you there is no oxidation/reduction, just take out the water and see what's left. And you should know CuO is black. Cu(OH)2 ==> CuO)(black) + H2O
April 23, 2015

Chem
If that is rate = k[NO]^2[Br] It us second order with respect to NO and first order with respect to Br and 3rd overall.
April 23, 2015

chemistry
k = 0.693/t1/2 Then ln(No/N) = kt No is 9.00 N = ? k from above t = 1 y or 10 y (two separate problems.)
April 23, 2015

science
This may help. http://www.livescience.com/49594-electric-fuel-cell-vehicles-explainer.html
April 23, 2015

Chemistry
You need to identify DCPIP and write the equation. It would help if you wrote the procedure you followed.
April 23, 2015

Chemistry
First I object to saying you titrated Ca(IO3)2 with S2O3^-. You didn't do that; you added I^- to IO3^- and titrated the liberated I2 with S2O3^- Here are the equations. Ca(IO3)2 ==> Ca^2+ + 2IO3^- IO3^- + 5I^- ==> 3I2 (which isn't complete but the redox part is ...
April 23, 2015

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

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