12.825 g of C12H22O11 in 0.045 M solution  in mL. The answer that I have is 832.6mL. Is this correct.?
603764
Correct ... +2 points for me
887797
It looks like you got it. This is how it came out for me:\n\nThis is the reaction:\nC10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)\n\nand Hc=(Summation of Hf of products)-(Summation of Hf of reactants)\nSo...\n\nHc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)\n-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)\n-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)\n-5156.8kJ/mol=(-5078.6)- Hf(C10H8)\n-78.2kJ/mol=-Hf (C10H8)\n78.2kJ/mol=Hf (C10H8)\n\nI believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.
818293
im pretty sure but  not positive\nC3H8 + 5 O2  -->  4 H2O + 3 CO2\n\nso 5 mol O2 to 1mol C3H8
527028
This time i ans ur question but next time plz do ur homewrk ur self.\n\n1.\nWeight of NaOH = (4*.05) = .2 grams\nMoles of HCl =  Moles of NaOH = (.2/40) = .005\nVol. of HCl = .005*22.4 = .112 l = 112 ml\n\n2.\nWell let the grams be x\n1000x = 74.5*2.25*(75+x)\n(1000-167.25)x = 167.25*75\nx = 14.98 grams
532149
I don't know if this is right ... but check it and tell me\n\nV*C KOH = V*C HBr\n\nV KOH = 18.6ml C KOH = 0.752M\nV HBr = 22.1ml C HBr = ?\n\n18.6*0.752 = 22.1*C HBr\n\nC HBr = 0.6329M\nIts seems reasonable cause the difference between concentrations isn't significant
472386
(250 X 0.35) + (1350X0.75) + (500X0.15) / (250+1350+500) = \n87.5+1012.5+75/(2100)\n1175 m mol/2100\n1.175 MOLES / 2100ml = 115g of sulfuric acid\n\n1.175moles /2.1 L = 0.58 M SOLUTION
313620
M(conc.base) x vol.(conc.base) = M(dil. base) x vol.(dil. base)\n12 x vol. = 2.5 x 350\n vol = 72.9 ml \n\nIn a graduated cylinder, take 72.9 ml of concentrated NaOH. And in a standard flask, take 350 - 72.9 = 277.1 ml of water. To this flask, add the concentrated NaOH (72.9ml) from the graduated cylinder and shake the flask for uniform concentration. Now you've prepared 350 ml of diluted NaOH solution.
942373
From the reaction\n\nS8(g) ===> 4S2(g)\n\nKp = (pS2(g))4/pS8(g) where p is the partial pressure\n\nGiven the partial pressure of S8(g) at equilbrium 0.25 atm, clearly 0.75 atm of S8(g) has been consumed to produce (from the equation) 4 volumes of S2(g), Given the volume was fixed this mean dS2(g) = 3 atm\n\nSubstituting into Kp yields\n\nKp =3^4/0.25 = 324\n\nThus Kp at 1325 K is 324
646163
Presuming that the base is monobasic.\n\na) with 25 ml added, you are at the equivalence point.  The pH is 7.\nb) after adding 25.01 ml, you will have 0.01 ml of base that did not get reacted.  The molarity of the base in solution will then be 0.1 * (0.01/50.01) = 2.0 x 10^-5.  Your pOH is 4.70, pH would be 9.30\nc) Now, it's 0.1 ml of unreacted base.  Molarity is 0.1 * (0.1/50.1) = 2.0 x 10^-4.  Your pOH is 3.70, pH is 10.30.\nd) Now 1 ml of unreacted base.  Molarity is 0.1 * (1/51) = 2.0 x 10^-3.  pOH is 2.70, pH is 11.30.\ne) Now 25 ml of unreacted base.  Molarity is 0.1 * (25/75) = 3.33 x 10^-2.  pOH is 1.48, pH is 12.52.\n\nIn each case, you have to figure out the amount of unreacted titrant and recalculate the molarity as if you were simply diluting it, using McVc = MdVd, where the volume is now the total volume of the solution (the unreacted amount plus the 25 ml of acid and the 25 ml of base that have neutralized each other).
858181
It would be 1.8N.  \n\nThree times in the case of Phosphoric Acid.
38679
C6H8O7\nassume a 100 gram sample. meaning there's that many grams of each.  now you divide each percent by the atomic mass of the element. this gives you the number of moles of each.\nC: 37.51/12.01 = 3.12\nH: 4.20/1.008 = 4.16\nO: 58.9/16.00 = 3.68\nnow you know the number of moles, so all you have to do is get them as whole number coefficients.  so first, divide by the lowest number this way at least one of them is a 1.\nC: 3.12/3.12 = 1\nH: 4.16/3.12 = 1.333\nO: 3.68/3.12 = 1.18\nnow you see that the H has a .333 which is a third, so you multiply them all by three to make the H a whole number too...\nC: 1.0 x 3 =   3\nH: 1.333 x 3 = 4\nO: 1.78 x 3 = 3.54\nall that's left is the O, and that's close to 3.5, so we multiply them all by two getting:\nC: 3.0 x 2 = 6\nH: 4.0 x 2 = 8\nO: 3.5 x 2 = 7\nso the emperical formula is C6H8O7 (where the numbers are subscripts)
871536
Start with 2.948g K2Cr2O7. End with how many mL. K=39, Cr=52, O=16, K2Cr2O7=294. Multiply by 1molK2Cr2O7/294gK2Cr2O7. The gK2Cr2O7 cancel, leaving molK2Cr2O7. Multiply by 1000mLK2Cr2O7/0.5molK2Cr2O7. This comes from 0.5M, which means 0.5 mol in 1000mL. The molK2Cr2O7 cancel, leaving mL solution to be evaporated.\n\nStart with 45mL NaCl, end with gNaCl. Na=23, Cl=35.5, NaCl=58.5. Multiply by 0.125molNaCl/1000mLNaCl. This comes from 0.125M. The mLNaCl cancel, leaving molNaCl. Multiply by 58.5gNaCl/1molNaCl. The molNaCl cancel, leaving the gNaCl that is the answer.\n\nStart with 100g glucose, G. End with L of G. C=12, H=1, O=16, C6H12O6=180. Multiply by 1molG/180gG. The gG cancel, leaving molG. Multiply by 1L G/0.5molG. This comes from 0.5M G. molG cancel, leaving L of G. Which is the answer.
352923
Come on.\n\n2 mol KClO3 = 3 mol O2\nSo 2*4 mol KClO3 = 3*4 mol O2\nSo 8 mol KClO3 = 12 mol O2\n\nAlgebra is fun! lol...not!
773826
the answer that i got is 3.10 M. \n\n3.25 g NaOH/40.00 g NaOH = .0813 mol NaOH\n\n26.2 mL / 1000mL = .0262 L\n\nM= .0813 mol / .0262 L = 3.10 M
952215
[H3O+][OH-] = 1.0 x 10^-14 M always at 25C\nso if [H3O+]= 5.0 x 10^-10M\n[OH-]= 2.0x10^-5
934849
291.13 L is the correct volume.\n\nCheck your algebra when you are solving for V1.
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T