what is normality & molarity of 10% pure hydrochloric acid?
109301
Normality generally refers to molecular weight of asa substance in grams dissolved in 1 litre of solvent.\nIn the present case 3.65 grams of HCl dissolved in 1 litre of water.\n\nMolarity refers to molecular weight of a substance dissolved in gram molecular weight of solvent.\nIn the present case 3.65 gram of HCl dissolved in 18 gram of water.
313383
Molarity= moles/liters
970448
N = Normality...in this case the concentration of Sulfuric Acid in a solution.\n\nThe concentration of Sulfuric Acid in the 1.00 N Solution is almost 4 times as concentrated as the 0.255 N Solution.\n\nIn practice Normality is "similar" to Molarity, except Normality takes into account the actual number of protons exchanged in a reaction.  Normality = Molarity times the number of protons exchanged.  In Sulfuric Acid, two protons are available for exchange...therefore, a one molar solution would be a two normal solution.  It is useful when doing titrations to determine concentrations.
851693
The concentration in the final solution will be:\n\n1x10^-3 g/10 ml = 1x10-4 g/ml\n\nThere were 3 10 fold dilutions, so you will need to multiply by 10^3\n\n1 x 10^-4 x 10^3 = 0.1 g/ml\n\nThe concentration of BSA in the original stock solution is 0.1 g/ml
969572
M (molarity) is defined as "mole per L (mole/L)" of solution\n\nglucose = C6H12O6, its molecular weight is 180.16g/mole (say about 180g/mole)\n\n1000mM = 1000mmole / L\n--> (1000mmole / L ) * (180g / mole)\n--> cancel out the "mole"\n--> 180000 mg/L\n\nsame logic for 970mM = 970mmole / L\n--> (970 mmole / L) * (180g / mole) = 174600 mg / L\n\nnote: you can leave the "m (milli)" there in your calculations, because, remember, "m" simply means 10^(-3), which is 0.001
245305
First of all, NaX is not an acid, but it could be a salt. Second of all, it doesn't matter what is their standard molarity, you need the volume of the HX and NaX. If you assume that each is 1L, then HX will be 0.490M and NaX would be 0.500M.
285303
If you mean mg/100mL, then it should be the same thing, since 1 dL = 100 mL\n\nIf you mean mg/mL, then:\nMultiply through by ratios to cancel the units:\n\n[ ] mg/dL * [10 dL / 1 L] * [1 L / 1000 mL]=
959929
A.\nmole of HCl availible = 1.0 * 100 = 100 mmole\nmolarity of diluted HCl = 100 / 1000 = 0.10 M\n\nB.\nmole of HCl availible = 1.0 * 10 = 10 mmole\nmolarity of diluted HCl = 10 / 500 = 0.02 M\n\nC.\nmole of HCl availible = 1.0 * 50 = 50 mmole\nmolarity of diluted HCl = 50 / 100 = 0.50 M\n\npeace\nvixklen
808549
M=molesof solute/Liter of solution\n\nM= molarity
965686
10% w/v could be 10g per 100ml of NaCl >> 10g/100ml\n1 mole of NaCl has 58.1g >> 1mol/58.1g\n\n\n[(10g NaCl)  (1 mole)] / [(100ml NaCl) (58.1 g NaCl)]\n\n1.72 M
795448
normality = molarity x ionizable H\n\n18 M x 2 H = 36 N
807507
(2.5 x 10-3)(100 mL) = M(10 mL)\ngiving you M = 2.5 x 10-2 M as the final molarity\nNumber of g of alanine is (2.5 x 10-2mol/L)(10 x 10-3 L) ( 89.1 g/mol)
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
744434
Commercial vinegar is (5.0 ± 0.5)% acetic acid by weight i.e. 50g per litre. If the molecular weight of acetic acid is 60.05 g/mole, then the molarity would be 50/60.05 = 0.83 molar.
916567
STP is 1 atm, 273 K.\n\nmoles of HCl = PV/RT = 1 atm * 1.1 L/ (0.08206 * 273K) = 0.049 moles\n\nSince HCl and NaOH is a 1:1 reaction, 1 mole HCl gives 1 mole NaCl, so you get 0.049 moles NaCl (MW 58.5), or 2.87g.
968497
The initial concentration of the acid is C=mole/V=1/2=0.5 mole/L\nLet's assume that the acid is HA\n\n.. .. .. .. .. .. .. HA <=> H(+) + A(-)\nInitial .. .. .. .. 0.5\nDissociate .. .x\nProduce .. .. .. .. .. .. .. x .. .. .. x\nAt Equil. .. . 0.5-x .. .. . x .. .. .. x\n\nKa=[H+][A-]/[HA] =x^2/(0.5-x)\n\nbut x=[H+]=3.4*10^-2 so\n\nKa= ((3.4*10^-2)^2) / (0.5-3.4*10^-2) =2.48*10^-3
1179708
10 milliliters (mL) = 1 centiliter (cL) \n10 centiliters = 1 deciliter (dL) = 100 milliliters \n10 deciliters  = 1 liter1 = 1000 milliliters \n10 liters = 1 dekaliter (daL) \n10 dekaliters  = 1 hectoliter (hL) = 100 liters \n10 hectoliters = 1 kiloliter (kL) = 1000 liters \n\nSo Answers are\n-------------------------------------------------------------------------------\n\n750 mililiters (mL)\n=7.5 centiliter (cL)\n=0.75 deciliter (dL)\n=0.075 liter\n=0.0075 dekaliter (daL) \n=0.00075 hectoliter (hL)\n=0.000075 kiloliter (kL)\n-------------------------------------------------------------------------------