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DrBob222
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DrBob222
Questions (1)
You didn't substitute correctly for Keq. You should calculate
log Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+) You made log Keq = 0.27. You
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Answers (243)
You're right. It is confusing but thanks for trying to explain. I can't answer BECAUSE the atomic number of Rn is 86 or perhaps that IS the answer. The equation isn't balanced because the number 80 should be 86. If it were 86 the equation would balance.
First I want to ask where you were told that the name for ZnCO3 is zinc trioxocarbonate(iv). That is a made up name and it is not sanctioned by IUPAC. If you made it up let me know. If a teacher told you let me know. If you read it in some journal, book,
Let's talk about this. I have a book which I used in school that gives all of those lines but you know I'm just not willing to do something so mundane as to count those lines. Sorry. Is there some terrific reason why the number of lines in so many elements
delta x*delta p = h/4*pi where delta p = mv v = 121.8 km/s. Convert to m/s so delta p = mdelta v = 9.11E-31g*121.8 km/s x (1000 m/km) = ? and substitute into delta x = 6.626E-31/4*3.1416*delta p Post your work if you get stuck.
I don't understand exactly what you're asking. molar mass ethane is 30 g/mol so -1560 kJ x 111/30 = ? kJ. If you want MJ you divide kJ by 1000.
mm = molar mass grams H2O liberated = 2.70 g Ni(CN)2.4H2O x (4*mm H2O/mm Ni(CN)2.4H2O) = ? g H2O Then mols H2O = grams H2O liberated/mm H2O and substitute into the following for n and solve for V in liters. Remember to convert T to kelvin which is 273 +
Jade, I don't know your horseshoe prowess but when I play my three throws land about 20, 30, and 10 feet from the stake. Poor accuracy and poor precision.
what reaction is the delta H rxn = 33.9 kJ?
You didn't substitute correctly for Keq. You should calculate log Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+) You made log Keq = 0.27. You should follow the instructions I gave. (Pb) = 1 (Ni^2+) = 0.27 (Ni) = 1 (Pb^2+) = x so the equation is 0.020 = 0.12 -
I don't have an answer to 1 except to say that's the nature of the beast. 2. Salt causes a "salting out" effect because it is so much more soluble so the salt dissolves and the soap, which was in solution, precipitates. This is a relatively easy and cheap
1 m = 39.37 inches so 1 cubic meter = 39.37 in x 39.37 in x 39.37 in = almost 61023.34 cubic inches. So, 1 cubic meter x (61023.34 cubic inches/cubic meter) x (1 gallon/231) = ?
Rachel, Charles, whoever. See your post above. Done the same way. Post your work if you need help.
HPO4^2- + HOH ==> H2PO4^- + OH^- as a base since it adds a H^+.. HPO4^2- + H2O --> PO4^3- + H3O^+ as an acid since it donates a H^+. Ka2 for HPO4^2- = 6.2E-8 from the data so Kb = Kw/Ka = 10E-15/6.2E-8 = 1.6E-7 Which is the larger number? FYI. Solutions of
What kind of data do you have and what is it?
See my response to your repost above. Hope that helps.
Pb^2+ +2e^- ==> Pb Eo red = -0.13v (copy to below) Ni^2+ +2e^- Ni Eo red =-0.25 (reverse this and change sign of Eo red to Eo ox) Pb^2+ +2e^- ==> Pb Eo = -0.13 v Ni ==> Ni^2+ + 2e Eo = +0.25 v (Add these to get Eo cell)
Changing the concentration of reactants or products does NOT change the value of the equilibrium constant. After all, it's a constant. Changing temperature WILL change K but concentration, pressure, volume, will not. Note that changing volume, pressure or
Let's call phenol HP and it ionizes as ....................HP ==> H^+ + P^- I...........2.55E-6........0..........0 C..................-x............x..........x E..........2.55E-6-x.........x..........x I don't know how advanced this course is but if you
1. Calculate the pOH of 0.093 M LiOH. Since LiOH is a strong base it will ionize completely and (OH^-) = 0.093. Then pOH = -log (OH^-). Substitute and solve for pOH. 2. Calculate the hydroxide ion concentration, [OH-], in 1.36 M HNO3. HNO3 is a strong
You have too many error in your post. Electrons not right, charges not right.
2. On both 1 and 2 please post your work if you get stuck. For 2 you have 15.5 g NaOH in 500 mL solution which is 15.5 g/40 g/mol = 0.3875 mols/0.500 L = 0.775 M So NaOH is strong base, it ionizes completely so (OH^-) will be 0.775 M. Convert that to pOH
1.Let's call the weak acid HA, then (HA) in mols/L = 0.0017 moles/0.500L = 0.0034 M. Then ..................HA ==> H^+ + A^- I................0.0034..........0.....0 C..................-x.............x......x E............0.0034 - x.......x......x pH =
The type is gases
Two elements combining in a chemical change makes a compound.
www.google.com/search?client=firefox-b-1-d&q=what+is+the+name+of+1%2F2+of+a+chromosome
Titration step is 3Mn^2+ + 2MnO4^- + 4OH^- ==> 5MnO2 + 2H2O ----------------------------------------------------------------------------- mols MnO4^- used = M x L = ? mols Mn^2+ in the sample = mols MnO4^- x 3/2 = ? grams Mn = mols Mn x atomic mass Mn = ?
Cu wire to Cu^2+ to Cu^+. Then, assuming acid conditions, 5Cu^+ + MnO4^- + 8H^+ --> Mn^+ + 5Cu^2+ + 4H2O mols MnO4^- used = M x L = ? mols Cu+ titrated = moles CuO = 5*moles MnO4^- grams CuO = mols CuO x molar mass CuO %CuO = (grams CuO/mass sample)*100 =
What about it?
During the combustion of 5.00 g of octane, C8H18, 239.5 kcal (1002 kJ) is released. 1. So 5.00 g releases 239.5 kcal, how much is released by 17.7 g? Just convert. Two ways to do this. 239.5 kcal x (17.7 g/5.00g) = ? kcal or set up a proportion like this.
Forgot to note that the acid and base are on the reactant side while the conjugate base (of the acid) and conjugate acid (of the base) are on the product side. That is acid1 + base2 ==> conjugate base of acid1 + conjugate acid of base 2
I don't want to do all of your work for you but if you will identify a problem and tell me what you don't understand about it I'll be glad to walk you through it. And I'll do each of them with you. I want you to learn it; i.e., not copy it.
Pick out the two pairs. The acid of the pair is the one with the larger number of H atoms. The conjugate base of the pair is the one with the fewer H atoms. I'll be glad to check your work if you post it.
No they don't maintain their properties BECAUSE Fe becomes Fe2O3 and O2 becomes Fe2O3. I would answer that Fe is oxidized and O2 is reduced.
First I must tell you that calcium trioxonitrate (4) is not a correct name for CaCO3 (or anything else). Several names are acceptable by IUPAC for CaCO3 but that name is not one of them. Calcium carbonate is an acceptable name by IUPAC. Please tell me what
You should Google this question. You will have all of the information you need.
2Mg + O2 ==> 2MgO mol Mg = g/atomic mass = 3.00 g/24.3 = 0.123 mols O2 required = (1/2)*0.123 = 0.0617 grams O2 = 0.0617 x 32 g = 1.97 Law conservation of mass says you don't gain or lose mass. So Total MgO produced should be 3.00 + 1.97 = 4.97 g NOTE: You
You need to proof the problem because something ain't right. There is no such thing as Ag2NO3. You react with AgNO3 with HCl and you don't get H2CO3. I'll bet you meant Ag2CO3. So read what you posted. Correct. Repost.
Jeff, here is a web site by Chemteam that discusses almost any chem subject you wish to discuss. They are excellent and I urge you to copy this link. I can't post the entire link so you will need to add the h t t p s://www. without the spaces.
heat lost by Al + heat lost by Fe + heat gained by H2O = 0 [mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0 Substitute and solve for the only
If you are looking for a career path for "pure math, life science,CAT ,visual arts", I don't think one exists. You find career paths in pure math or in visual arts, or in life sciences or in ......, but not (usually) in all of them at once. For specific
Why not look on a map instead of asking us to look on a map and for the information to you?
pH = -log (H^+) pH = -log (3.5E-6) Punch -key then log key then 3.5E-6. The answer should be 5 something. Post here if you want to check your answer.
Sounds more like geography to me.
I see a statement but no question
ml1 x M1 = mL1 x M2 mL1 x 3.25M = 400 mL x 0.250 Solve for mL1 volume, take that volume of the stock, transfer to a 400 mL volumetric flask and make to the mark with distilled water. Stopper. mix thoroughly, label. Done.
MgCl2 + AgOH is interesting. I suspect the author of the problem is thinking MgCl2 + 2AgOH ==> Mg(OH)2(s) + 2AgCl(s) and the net ionic equation is Mg^2+ + 2Cl^- + 2Ag^+ + 2OH^- --> Mg(OH)2(s) + 2AgCl(s) However, I don't think this will happen. First, the
That 1 1/2 year vs 2 year vs some other number is a moving target. Where I live it's been 1 year and 10 months but it has been longer than that at other places. Technically, neither the starting date nor the starting place has been definitely determined.
Interesting but I don't see a question.
The following equation will work almost all or part of your normality problems mL x N x milliequivalent weight (mew) = grams 13.9 mL x N x 0.2042 mew = 0.394 g
