Monday
June 27, 2016

Posts by DrBob222

Total # Posts: 52,582

Chem 1410
Use PV = nRT
April 14, 2016

chemistry
In some cases these are complete questions. But this is a laundry list. We don't do complete assignments. If you will tell us what you don't understand about one or more of these perhaps we can help you through them.
April 14, 2016

Chemistry
Didn't we go over this yesterday? You need to learn to do this. I gave you a link. If you will identify your hang up I can help you through it.
April 14, 2016

Chem LAB
What you have written is already in the form of the net ionic equation.
April 14, 2016

CHEM
This is a limiting reagent problem Do it just as the other LR problem except this is with heat involved. O2 is the limiting reagent. Since the sign of dH is - that means heat is released.
April 13, 2016

Chemisty
mols Cd(NO3)2 = grams/molar mass = ? and M Cd(NO3)2 = mols/L = ? Next,you will need to calculate (OH^-) from 0.800 M NH3. Then Qsp = (Cd^2+)(OH^-)^2 Substitute M Cd^2+ and M for OH^- into Qsp. If Qsp>Ksp, there will ppt. If Qsp<Ksp, no ppt. The idea with Qsp is to see if...
April 13, 2016

chemistry
That's ok but you have too many significant figures in the answer.
April 13, 2016

chemistry
5890 J x (2*23/0.567) = ?
April 13, 2016

chemistry
AgBr in HNO3 is no effect (or the same solubility). You're right, there is no common ion and the only rxn that you need to consider is combination of H with Br. IF (and it isn't) HBr were a weak acid it would be more soluble but since HBr is a strong acid there is no ...
April 13, 2016

chemistry
Le Chatlier's Principle says that when a system at equilibrium is disturbed it will shift so as to undo what we've done it it. AgBr in 0.1 M HBr. ......AgBr ==> Ag^|Using the balanced chemical equation, calculate the equilibrium constant for this system. AgBr(s) ==&...
April 13, 2016

chemestry
This is a limiting reagent (LR) problem and you know that because amounts are given for BOTH reactants. First, make sure the equation is balanced. mols H2O2 = grams/molar mass = ? mols N2H4 = grams/molar mass = ? Next, using the coefficients in the balanced equation, convert ...
April 13, 2016

chemestry
Where is the equation? And please spell chemistry right next time.
April 13, 2016

@ color wheel--Chemistry
How many mols do you want? That's M x L = ? Then mols = grams/molar mass. You know mols and molar mass; solve for grams. What you want to do is weigh out that amount of Cu(NO3)2.3H2O, place it in a 100 mL volumetric flask, add some water, swirl to dissolve all of the salt...
April 13, 2016

Chemistry
No. The reaction takes place at 25 C so the Keq will Keq at 25C. The problem doesn't make that clear. If the problem were stated correctly it would say, "Using the balanced chemical equation, calculate the equilibrium constant for this system at 25 C."
April 13, 2016

Chemistry
Kat, this is nothing more than plugging the numbers into the Keqp expression and evaluating it. What do you not understand about it. The Keq expression you wrote is correct.
April 13, 2016

Science
See your post below.
April 13, 2016

Chemisty
See your post above.
April 13, 2016

chemistry
mols H2 gas = grams/molar mass. Then each mols has 6.02E23 molecules. Each molecule has two H atoms. Each H atom has 1 electron, 1 proton and no neutrons.
April 13, 2016

Chemistry
I don't know enough of what you have to be absolutely sure. If that is 43.5 mL H2 gas collected OVER WATER the answer is different that if that's 43.5 mL dry H2 gas. I will assume it is 43.5 mL H2 gas collected over water so here is what you do. PV = nRT. Ptotal = pH2...
April 13, 2016

chem
millimoles Br2 used = mL x M = 4*0.23 = approx 0.9 1 mol Br2 adds to 1 mol alkene; therefore, mmols Br2 alkene = approx 0.9. mols in 50 cc at stp = 50/22,400 and millimols = 50*1000/22,400 = approx 2. % alkene = (mmols alk*100/mmols gas) = ? % alk = mols alkene/
April 13, 2016

Chemistry
What is that 4.2 H2C2O4?).
April 13, 2016

Chemistry
I don't have any idea of what your experiment was. My crystal ball is hazy. I think you should repost this as a physics question and please provide a little info so those physics guys/gals will know what you did.
April 13, 2016

Chemistry
Oxidation is the loss of electrons. Do you know know to determine the oxidation state? H on the left is zero and on the right is +1. N on the left is zero and on the right it is -3. Here is a link that will show you how to calculate the oxidation state. http://www.chemteam....
April 13, 2016

Chemisty
This looks so straight forward. What is your problem with this. Exactly what do you not understand.
April 13, 2016

chemisty
or grams?
April 13, 2016

Chemistry
First I would relabel the bottle as potassium nitrate. Then place some KNO3 in an open dish, place in an oven at 105 degrees C, leave for 2-3 hours (or constant weight if you prefer), place in a desiccator and allow to cool.
April 13, 2016

chemistry
Reverse equation 1 and add to equation 2. Divide the final equation by 2. Add the dH values; when an equation is reversed, change the sign of dH.
April 12, 2016

chemistry
I wonder why phosphoric acid is not as good as hydrophosphoric acid. Also, note that Zn is Zn(II) phosphate etc. 3Zn + 2H3PO4 ==> Zn3(PO4)2 + 3H2 mols Zn = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols Zn to mols Zn3(PO4)2. Then g Zn3(...
April 12, 2016

Chemistry
6 L CO2. Knowing that 1 mol of any gas occupies 22.4 L at STP, then 6/22.4 = ? mols CO2. Then using the coefficients in the balanced equation, that many mols CO2 x 2 = mols HCl used Then M HCl = mols HCl/L HCl = ?
April 12, 2016

Chemistry
Is that an aqueous solution? 1337.6 kJ/0.0125 = ? kJ = heat released by 1 mol MgO I didn't check the mol calculation for MgO.
April 12, 2016

Chemistry
Do you mean 0.052 molar? pH = -log(HCl) = -log(0.052M) = ?
April 12, 2016

Chemistry
http://www.jiskha.com/display.cgi?id=1460431080
April 12, 2016

Chemistry
This is the H2A problem and I worked that for you re the link for the post just above.
April 12, 2016

Chemistry
You have no concns listed so you can't get Keq that way. Use dGo = -nEoF You will need to look up the value of Eo for the reaction. Obviously the one with Eo a positive number will give you a negative dGo and that is the one that is spontaneous. If you want Keq, you can ...
April 12, 2016

Chemistry
Is D density? Zn + 2HCl ==> ZnCl2 + H2 Calculate mols Zn and convert to mols H2. Then use PV = nRT P will be 1.234 atm - vp H2O (in atm)[or convert 1.234 atm to mm Hg and use 21.0 mm for vp H2O) Everything else is as usual and V will be ihn liters.
April 12, 2016

Chemistry
(NaOH) = 1 M NaOH x (100 mL/1,500) = ? The N is the same as the M (in this case).
April 12, 2016

a note for olereb48---chemistry
Using the quadratic (H^+) from ka1 is 0.023 and not 0.027 which makes H3PO4 then 0.10-0.023 = 0.077 M Yes, (H^+) from ka2 is = ka2 and that is insignificant. Yes and No for part ka3. (H^+) is that from part 1; i.e., 0.023 and not = k3. Likewise, (HPO4^2-) is known from ka2 ...
April 12, 2016

chemistry
H3PO4 ==> H^+ + H2PO4^- H2PO4^- ==> H^+ + HPO4^2- HPO4^2- ==> H^+ + PO4^3- Write the expressions for ka1, ka2, and ka3. You solve k1 as if it were a monoprotic acid; ie. ......H3PO4 ==> H^+ + H2PO4^- I......0.1......0......0 C......-x.......x......x E.....0.1-x...
April 12, 2016

Chemistry-Dr.Bob222
H2A + 2NaOH ==> Na2A + 2H2O mols NaOH = M x L = ? mols H2A = 1/2 x mols NaOH (look at the coefficients in the balanced equation). Then mols H2A = grams H2A/molar mass H2A. You know mols and grams, solve for molar mass.
April 11, 2016

Chemistry-Dr.Bob222
Refer to the H2A problem above BUT you only titrate the first H on H2SO4. Post your work if you get stuck.
April 11, 2016

Chemistry
PV = nRT
April 11, 2016

Chemistry
yes. Good work.
April 11, 2016

Chemistry- Dr.Bob222
I would not throw away the places after 5 that are eligible to be retained. I came up with 5.04 to three significant figures.
April 11, 2016

Chemistry
(p1v1/t1) = (p2v2/t2)
April 11, 2016

Chemistry
Robin,you've done a terrific job but you need a couple of changes. Is F^- a weak base; if so it would not attract H^+ to make HF. Then HCl is a strong acid that IONIZES COMPLETELY which makes Cl^- a weak conjugate base of HCl. Since HF is a weak base the added H^+ makes HF...
April 11, 2016

Chemistry
1250.0 kJ/mol x # mol = 352.40 kJ Solve for # mols. Then mol = grams/molar mass. You know mols and molar mass, solve for grams.
April 11, 2016

Chemistry
SO3 + H2O --> H2SO4 mols SO3 = 4.88E24/6.02E23 = approx 8 but you need more than an estimate. Convert mols SO3 to mols H2SO4 using the coefficients in the balanced equation. Since 1 mol SO3 produces 1 mol H2SO4, then mols SO3 = mols H2SO4. Convert mols H2SO4 to grams. g ...
April 10, 2016

chemistry redox reactions
I don't understand the post. For example is R(e-) and R^- the same?
April 10, 2016

Chemistry
Before anyone can answer this for you we need to know the procedure you used in detail.
April 10, 2016

Chemistry
How many mols CH3COOH (HAc) do you have? That's M x L = 1.2 x 1.5 = 1.8 .........HAc + OH^- ==> Ac^- + H2O I........1.8...0........0.......0 add............x.................. C.........-x..-x........+x...... E.......1.8-x...0........x Now substitute the E line into the...
April 10, 2016

Chemistry
Since it's only 92% UO2 then you have only 10 kg x 0.92 = ?kg UO2. Convert that many kg UO2 to ? kg U. That's ?kg UO2 x (atomic mass U/molar mass UO2) = ? kg of U.
April 9, 2016

Chemistry
2SO2 + O2 ==> 2SO3 mols SO3 needed = grams/molar mass = ? Using the coefficients in the balanced equation, convert mols SO3 to mols O2. Now convert mols O2 to grams. g O2 = mols O2 x molar mass O2.
April 9, 2016

Analytical chemistry
Fe2O3 ==> 2Fe^3+ and reduce to Fe^2+. 6Fe^2+ + Cr2O7^2- ==> 6Fe^3+ + 2Cr^3+ I balanced only the redox portion of the equation which is sufficient for what we want to do. mols Cr2O7^2- = M x L = ? mols Fe = 6 x that (look at the coefficients in the equation above). g Fe...
April 9, 2016

Chemistry
You were right on the (H^+) = 10^-1.57 = 0.027 mols/L.But wrong on the rest of it. You want 0.027 mols/L so you will need 1/4 that for mols in 250 mL or 0.027 x (250 mL/1000 mL) = 0.0069 Then mols = grams/molar mass. You know molar mass and mols, solve for grams. I estimated ...
April 9, 2016

Science Please help ASAP
No, cold ocean water want do it. If that were true we would have hurricanes in the winter.
April 8, 2016

Chemistry
p1v1 = p2v2
April 8, 2016

Chemistry
Keq = Kf*Ksp = [Ag(NH3)2]*(Cl^-)/(NH3)^2 From my viewpoint, the 2 for NH3 does play a role.
April 8, 2016

Chemistry
(V1/T1) = (V2/T2) Remember to convert T1 and T2 to kelvin.
April 8, 2016

Chemistry
Would you believe 200.0 mL? Isn't 400 k and 400 K the same? Technically it should be written as 400 K.
April 8, 2016

Chemistry
A fairly long solution. Start with two equations. The first is to find the ratio of base to acid. pH = pK3 + log (base)/(acid) Substitute and solve for b/a. Equation 2 is b + a = 0.01. Solve those two equations simultaneously for a (acid) and b (base). Since those are M concns...
April 7, 2016

Chemistry please help!
I thought Bob Pursley showed you how to do this. Add equation to the reverse of equation 2. That gives you kJ/mol graphite to diamond. If you want 1 kg, then kJ/mol from the above x (1000 g/12.01) = ? heat absorbed to make 1 kg. For part b. ?kg from above = mass H2O x specific...
April 7, 2016

Chemistry
The first number is the atomic number(it's a subscript); the second is the mass number (it's a superscript). 84Po210 = 2He4 + 82X206 Note the subscripts add up on each side; i.e., 94 = 82 + 2 and the superscripts add up. That is 210 = 26 + 4. So X must be Pb.
April 7, 2016

Chemistry
Jessie, Mark, Chris or whomever. Just follow the solution for Mark.
April 7, 2016

chemistry
See your posts above.
April 7, 2016

Chemisty
I wonder how "common lab equipment" is defined. I wonder how accurately this is to be done. Does that mean we can't use any other chemical? I would prepare a 0.1M baking soda solution (NaHCO3) and fill a buret. Add phenolphthalein indicator, then add NaHCO3 from ...
April 7, 2016

Chemistry
Refer to your other posts. Almost all of these can be worked with two thoughts. #1. k = 0.693/t1/2 #2. ln(No/N) = kt where No = initial and N = final, k is from the #1 calculation and t1/2 is the half life.
April 7, 2016

Chemistry
Cool 212 water to what. zero? 10? 99. 211? Pick a number and you'll be right.
April 7, 2016

Chemistry
(g solute/g solution)*100 = % mass/mass
April 7, 2016

Chemistry
% w/w of 20.1% means 20.1 g NaBr in 100 g solution. So g NaBr = 20.1 g solution = 100 g = g NaBr + g H2O or g H2O = 100-20.1 = 79.9 n NaBr = grams/molar mass = ? n H2O = grams/molar mass = ? Total mols = n NaBr + n H2O XNaBr = nNaBr/total mols.
April 7, 2016

chemistry
124 kcal x (11.2 g HF/molar mass H2) = ?
April 7, 2016

Chemistry
2Na + 2H2O -- 2NaOH + H2 mols Na = grams Na/atomic mass Na. You know atomic mass Na and mols Na, solve for grams Na. The 50 cc NaOH has nothing to do with the problem assuming that the water used was in excess.
April 7, 2016

Chemistry
http://www.jiskha.com/display.cgi?id=1459868645
April 6, 2016

Chemistry
Diagram of the solution? We can't do that on this forum.
April 6, 2016

Chemistry
Will you please explain, in detail, exactly what you don't understand about each step?
April 6, 2016

Chemistry
K2O + HOH ==> 2KOH mols KOH needed = grams/molar mass = ? mols K2O = (1/2)*mols KOH from the coefficients in the balanced equation. grams K2O needed = mols K2O x molar mass K2O = ?
April 5, 2016

Chemistry
See your post on Si.
April 5, 2016

Chemistry
657 kJ x (130/atomic mass Si) = kJ released.
April 5, 2016

chemistry 107
You must mean precipitate and not participate. BaSO4 is the ppt that forms.
April 5, 2016

Chemistry
The molecular formula is not 44*0.5. If 0.5 mol weighs 44 g then a whole mole must weigh 88 g. An acid has a -COOH group which is 45 so the rest of the molecule must be 88-45 = 43. A -CH3 group is 15 so 43-15 = 28 and CH2 groups are 14 each so the molecular formula must be ...
April 5, 2016

Chemistry
And the 1.3? What about it?
April 5, 2016

Chemistry
Your addendum makes no sense. That's k = 1.3 for what. Is is Mg(OH)2? If so 1.3 isn't right. You've made a typo surely.
April 5, 2016

Chem
That's the specific heat for H2O. q = m*c*delta T.
April 5, 2016

chemistry
mL x density x (%/100) = 13.37 Solve for mL of the solution.
April 5, 2016

Chemistry
This forum can't handle drawings.
April 5, 2016

Chemistry
Why don't you look these up on Google. Better for you to do and for me to do it.
April 5, 2016

Chemistry
See your other post above.
April 5, 2016

Chemistry
What reaction?
April 5, 2016

CHEM
What's wrong with what I gave you last night? The directions are there; I made 1 L of solution so adjust to 100 mL for this and you have it. I'll be glad to help you through it but I don't want to do the work for you.
April 5, 2016

Chemistry
2HCl + Ca(OH)2 ==> 2H2O + CaCl2 mols HCl = M x L = ? mols Ca(ON)2 = 2 x mols HCl (look at the coefficients in the balanced equation). Then M Ca(OH)2 = mols Ca(OH)2/L Ca(OH)2. You know M and mols, solve for L.Convert to mL if needed.
April 5, 2016

CHEM
Usually you know the pH but you don't show that. You know pKa for NH3 to be about 9.26 but use the number in your text/notes. If you want the buffer to be between 0.01 and 0.5, why not choose say 0.3. If base + acid = 0.3, then base = 0.1 and acid must be 0.2 so pH = pKa...
April 5, 2016

Chem
(p1v1/t1) = (p2v2/t2) Remember t must be in kelvin.
April 5, 2016

chemistry
Calculate X for A(which I'm calling a) and X for B (which I'm calling b). Poa and Pob are Poa and Pob respectively. pa + pb = 40 mm pa = Xa*Poa Knowing pa you calculate pb Then pb = Xb*Pob You know pb and Xb, solve for pob. Post your work if you get stuck and I can ...
April 5, 2016

Chemistry
This is a little hard to show in writing but here goes. ......M^2+ + 4CN^- ==> [M(CN)4]^2- I..0.150....0.870.......0 C.-0.150...-0.600....+0.150 E.....0.....0.270.....0.150 The first equilibrium shows the formation of the complex on the right. With such a large Kf of over ...
April 5, 2016

Chemistry
Pb(OH)2 ==> Pb^2+ + 2OH^- Al(OH)3 ==> Al^3+ + 3OH^- Calculate OH^- needed for a 0.21M Pb^2+ solution to ppt Pb(OH)2, Calculate OH^- needed for a 0.41M Al^3+ solution to ppt Al(OH)3. You will want the (OH^-) to be less than you calculate to prevent pptn of Pb(OH)2. ...
April 5, 2016

Chemistry
........PbSO4 ==> Pb^2+ + SO4^2- ........SrSO4 ==> Sr^2+ + SO4^2- You know Ksp for each and you know metal is 0.039. I like to solve for OH^- needed to ppt each. Using my values for Ksp (which won't be the same as yours probably since texts differ), SO4^2- for Pb = 4...
April 5, 2016

Chemistry
http://chemistry.elmhurst.edu/vchembook/568denaturation.html
April 5, 2016

Chem
100.0 g CO2 x (molar mass O2/molar mass CO2) = ? g O2 in 100.0 g CO2.
April 5, 2016

Chemistry
Add eqn 1 to eqn 3 to the reverse of eqn 2. Add the dH values; when you reverse the rxn change the sign of dH for that rxn. Post your work if you get stuck.
April 5, 2016

Chemistry
Probably several but you can't beat this one for simplicity. Any other method that will test the acidity/basicity will work.
April 5, 2016

Chemistry
Add water, test with litmus paper.
April 5, 2016

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