What is the density of NaOH?
969947
The density of solid sodium hydroxide is 2.130 g per cubic centimetre.
374994
The density of diesel fuel varies from one refined lot to the next. The density of diesel fuel ranges from 82%-95% that of water.\n\nSince the weight of water is 8 lbs/gallon, it would stand to reason that a gallon of diesel fuel would be between 6.5 lbs and 7.6 lbs.
329838
Density = mass / volume\nSo 514 g/500mL = ??? g/mL\nAnd that's all of your homework I'll do for you. ;)
110989
Density of hexane - .6548 g/l\nDensity of methanol - .791 g/l\n\nMethanol has a higher density.\n\nUse chemfinder.com to find info on chemicals.
58191
Density - 1.151 \nNot easy to find
685262
the answer depends on how many H+ need neutralising by the NaOH.\nassuming it is only 1 per molecule of the acid then you can use the following equations:\n\nno.of moles = (vol/1000)*concn\nno. of moles = (40.11/1000)*0.240 = 0.0096264\n\nno. of moles = mass(g) / RMM\ntherefore\n0.0096264 = 0.55 / RMM\ntherefore RMM = 0.55 / 0.0096264\nRMM = 57.134\n\nif you have 2H+ that need neutralizing then the molecular weight = 2*57.134 = 114.268 etc etc
199332
Can't give you a definite answer but as the primary ingredient will be water it will have a density very similar to that of water which is 1.000.
927975
15%(m/v) means 15grams of NaOH in 100ml of solution.\nHence 120*100/15= 800ml will contain 120 g of NaOH
657714
NaOH + HCl --> NaCl + H20\n\nn(NaOH) = 1.00 x 32.6cm3\n        = 1.00 x 32.6 x 10^-6\n        = 3.26 x 10^-5  mol\n\nn(HCl) = 1.00 x 25cm3\n       = 1.00 x 25 x 10^-6\n       = 2.5 x 10^-5  mol\n\ntherefore, NaOH is in excess.. hence the final solution is alkaline, which means the pH is more than 7.\n\nDue to the limitation of my knowledge, i can't give you the exact pH value. Just hope that it will help you to continue solving...
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).
820519
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...
879131
To determine density you also have to know the temperature.  The colder a substance gets, the more dense it becomes (ie. the particles come closer together).  Water is the one odd ball, it becomes most dense at 4 degree C instead of below freezing like most substances.
868152
1 mol NaOH and 40 g NaOH are equivalent.
582005
pH=13.36\n\nNaOH is in excess (.0397 mol-.025 mol = .0147 mol)\nDivide by total volume (.0147/(.0397L + .025L) = 2.27x10^-1)\nCalculate pOH = -log (2.27x10^-1) = .6435\nSubtract 14 to find pH 14-.6435 = 13.36
732197
The density can be assumed to be 1.0 g/ml.  \n\nSo you have 25 ml x 1.0 g/ml = 25 g H2O\n\nThe molar mass of water is 18 g/mol, so you have:\n\n25 g H2O/18 g H2O/mol H2O = 1.4 mol H2O
272450
The density must be 1.09 g/mL, not g/mol\nM = mol solute/ L sol'n\n\n8.50% sol'n NaC2H3O2 = 8.50 g NaC2H3O2/100 g sol'n\n\nTherefore, \nM = (8.50 g NaC2H3O2)/(100g sol'n) x (1 mol/82.0 g)\nx (1.09 g/1 mL) x (1000 mL/1 L)\nM = 1.1298780 M = 1.13 M ^_^
572696
density of pure water is 62.4 lbs/sq ft.\n\nSo, load = dAgh\n\nor A = 40000/(62.4*5*32) = 4.01 sq ft.