help on chemistry/ math type problem?
919487
I think the general idea is to report as much information as possible from each measuring device. Otherwise you are just throwing away information. So for the ones that read out some digital display, you would simply report all the digits that it reads out. The other ones, where you have some choice in the matter, you would read to the nearest marking, plus estimate one more digit. \n\nAs an example, take the ruler marked with 1 mm markings:\n\ncase A. Widget length lines up exactly 10 mm \ncase B. Widget length is halfway between the 10 mm mark and the 11 mm mark. \n\nYou could actually measure a difference between A and B, so the number you report should reflect that difference. So case A would be reported as 10.0 mm, where as B would be reported as 10.5 mm. If you report both as just 10 mm (because that is how the ruler is marked) you would be missing the fact that you can measure a difference between the two. But generally, you can't estimate anymore that one place past how the ruler is marked, so you wouldn't try to report 10.55 mm, for example.
612107
a. 2k+ = .20moles , SO42-=.10moles\n\nb.Fe3+=.10moles,3NO3-= .30moles\n\nc.2Al3+=.20moles,3SO42-=.30moles\n\nd. Ni2+=.10moles,SO42-=.10moles
740485
Look up in Hand book of chemistry and mathematics, you school library will have it. Your professor should also have it.
799202
for those questions, all you have to do is translate them into the equation\ncheck chemtutor.com for hellp, it explains it much clearer than i can
304632
Okay, so the first thing that happens it that the NaF completely dissociates:\nNaF ---> Na+ + F-\n\nCool.  Now, the Na+ is just floating around, so we can ignore it.  Let's deal with the F-.  The F- is in the water, so it's going to do this:\nF- + H2O <---> HF + OH-\nNote that the arrow is going in both directions; it's an equilibrium problem, so I need to set up an equilibrium equation.  \n\nBut first, because I'm dealing with OH- ions, I'll need a Kb value.  Since Kw, the product constant of water, equals Kb * Ka, and Kw= 1E-14:\n(1E-14)/(6.8E-4)= Kb = 1.47E-11\n\nNow,\nKb=[OH-][HF]/[F-]\nor\n1.47E-11 = x^2/(.00339-x)\nsolve for x (I just set it equal to zero, graph it, and find the x-intercpets on my calculator):\nx=2.23E-7 M \nSince x=[OH-], [OH-]=2.23E-7 M\n\nWe can calculate pOH by taking the negative log of [OH-]:\n-log 2.23E-7 M = 6.65\nSince pH + pOH = 14.00:\npH + 6.65 = 14.00\npH = 14.00-6.65\npH = 7.35, the answer within one one-hundredth, which is probably some minor rounding discrepancy.  Yay!
429906
Never trust materials science majors ...\n\n1. NaCl (s) --> Na+ (aq) + Cl- (aq)\n2. Ca (s) + 2H2O (l) --> Ca(OH)2 (s) + H2 (g)\n3. Ca(OH)2 (s) --> H2O (g) + CaO (s)
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
877763
OK, I think I have this one pegged:\nThe original mixture is 10% acid and 90% water and the final mixture is 60% acid and 40% water after pouring some amount of the original mixture into 5 liters of pure 100% acid. Set up two equations:\nfor the acid: x (10) + 5 (100) = y (60) where x = original mixture volume and y = volume of resulting mixture.\nfor the water: x (90) + 0 (100) = y (40)\nthen these can be reduced to:\n10x + 500 = 60 y     and  90x = 40y\nsolve the simultaneous equations to find x and y\nThe volume of water in the final mix is equal to y times .40\nGood luck
268703
Use the overall rate equation: rate = k[A]^x[B]^y...\n1) define for rate1 and rate 2 with the appropriate concentrations.\n2) divide both sides (e.g. rate1/rate2 = ...\nNote that this cancels k\n3) Solve for x, y, ...  These are your respective orders.\n4) Add x, y, ...  This is your overall order.\n\nYou can get the value of k by:\n5) Plug in values of x, y, ... to either rate1 or rate2 equation.\n5) Solve for k.
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
24810
Cu + 2H2SO4 --> 2H2O + SO2 + CuSO4\nDouble the H2SO4 and the H2O.
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
367815
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
245982
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)