What volume (mL) of   0.157 M   KOH   contains   23.0 g   of   KOH ?
954213
KOH  Formula weight is 56.11  following that,\n\n56.11 gm of KOH in 1000 mL gives 1 Molar solution\n\nThen 23 gm in how many mL (?) gives 0.157 Molar solution\n\n\n? = (23 gm x 1000 mL x 1 M) / (56.11 gm x 0.157 M) = 2611 mL
709481
Wrong. 2.0623x10^23 should be divided by 6.02x10^23\n\nthe answer comes out to be 0.342 moles.
321421
0.215 kL = 215L\n(1kL = 1000L)
835275
KOH is a strong base and HBr is a strong acid.\ngreq acid= N*V=a*M*V. HBr is monoprotic so a=1 and \ngreq acid=1* 0.1*30*10^-3=3*10^-3\n\ngreq base= a*M*V. KOH is monovalent base so a=1 and \ngreq base= 1*0.1*30*10^-3=3*10^-3\n\nYou have greq acid= greq base which means you reached the equivalence point. Since both the acid and base are strong the final pH is 7 (there is no hydolysis of the salt) -remember the H+ and OH- coming from the base and acid have completely reacted but [H+] is never 0 since you have the equilibrium for the self dissociation of water.\n\nYou could have come to the same conclusion with moles (1:1, stoichiometry, same number of moles thus neither the acid nor the base is in excess, and there is no salt hydrolysis since both are strong electrolytes) but equivalents is the best way to solve titration problems.
165248
(.002mol /20ml) (1000ml / 1L ) = .10 M NaOH\n\n\n pOH =  -log 0.10 =  1\n\n pH = 14.00 - pOH   = 14 - 1  = 13 = D
137179
you need the balanced rxn which has a 2:1 ratio of Mg: O2\n 314 gms Mg = 12.9 moles divided by 2 = 6.46 moles O2. At STP that = 144.7 L volume.
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
697635
1) Calculate the number of moles of NaOH in 300 mL\n3.81 g /( 22.99+16+1.008) = 0.0955 mol \n\n2) Calculate the number of moles of NaOH in the 25 mL transfered\n0.0955 mol * 25/300 = 7.96 x 10^-3 mol\n\n3) Calculate the molarity \n7.96 x 10^-3 mol / 100 mL = 0.0796 mol/liter
929350
Do you mean M instead of N?\nIf you mean M, then\n1,00M/L*0,500 L= 0,500M\n0,500 moles NaOH*(40g NaOH/1 mol NaOH)= 20,0g of NaOH\nIt is c).
1167519
there would be approximately 2.648*10^24 acetone molecules.  This was done by first finding the number of gram in the volume by multiplying volume by the density.  This would give the number of grams which is then converted to moles.  Then you can use avagadro's number to multiply the number of moles to find the number of molecules.  The long equation is below\n325 ml * 0.7857 g/ml * 1/58.0798 mol/g * 6.0221 X 10^23 moleucles/mol\n\n0.7857 is the density of acetone\nacetone has a molecular weight of 58.0798 g/mol \nand avagadros number is approx. 6.0221 * 10^23\n*Note you must divide by the molecular weight to get the correct units at the end
837288
Use the formula M1V1=M2V2\n\nThat is the first Molarity times the first volume are equal to the second Molarity times the second Voume.\n\n.2098M x70mL = .662M x V2\n\nSolve for V2.\n\n(.2098 X 70)/.622 = V2\n\nV2 = 23.61mL NaOH to titrate the solution to neutrality.
227577
Molar mass is 61.97894 g/mol.
50903
The molar mass of NaCl is 23 g Na + 35.5 g Cl= 58.5 g NaCl/mol. \n8 g NaCl*(1 mol/58.5 g)=.137 mol NaCl.
704981
V = LWH\nV = (200cm)(200cm)(1cm)\nV = 40000 ml\nV = 40 Liters
239849
.7M x 20mL = .25M x (Volume of new IA solution)\nVol = (.7 x 20) / .25\nVol= 56mL
773889
Suspected trick question.  The volumes included are irrelevant.  You know the molecular weight (66000 Da) and you know how many micromoles you need total in each solution.  What does it matter if the one micromole is in 1 ml or 10K gallons?  You know it only contains one micromole.  Divide the molecular weight by 10^6 and you're done.  (or you have not stated the problem correctly)
835897
Assuming KOH is a strong base,\npOH = - log [-OH] = - log (1.00 M) = 0\npH + pOH = 14\npH = 14 - pOH\npH = 14\npH = - log [H+]\n[H+] = antilog(-pH)\n[H+] = 1 x 10^-14 M