In order to completely neutralize 20 mL of a solution of HCl 0.1 M, 40 mL of a solution of NaOH must be added. What is the M of the NaOH solution?
No, this working is incorrect. You tripped up on two points.20 milliliters is what decimal fraction of a liter?milli- means one thousandth, so 20 mL = 20/1000 liters20/1000 = 2/100 = 1/50 literDivide one by fifty:50 ) 1 wont go , result 0.? so try50) 10 won’t go result 0.0? so try50)100 goes twice result 0.02So 0.02 L = 20mL Result for VaYou made the same mistake with Vb = 40 mL SHOULD BE = 0.04 LThe second mistake was more important because in this equation the first two scaling mistakes canceled out!You wrote: 1 mol x 0.1 M x 0.20 L / 1 mol x 0.40 L = 8x10-3M1 x 0.1 x 0.2 / 1 x 0.4 = 0.05M NOT 8 millimol !You multiplied (x) instead of dividing ( / )
Calculate the solute potential for a 0.1 M NaCl solution at 25 degrees C. If the concentration of the NaCl?
The solute potential = – iCRT, where i = the ionization constant, C = the molar concentration (a.k.a. osmolarity), R = the pressure constant (R = 0.0831 liter * bars/mole * K), and T = the temperature in K (273 + °C). At atmospheric pressure and 25°C has an osmotic potential - A 0.15 M NaCl solution contains 2 ions, Na+ and Cl- (where sucrose stays as one particle); therefore i = 2, and the water potential = -7.4 bars. Water will move into the cell by osmosis from 0.1 to 0.15.
How would one calculate the amount of (NH4) 2SO4 which must be added to 500ml of 0.2M NH3 to yield a solution of pH=9.35, given pKb (NH4OH) =4.74?
5.37 or near to it !Hope this helps !!!Best of luck
Is 1 M (molar concentration) more concentrated than 1 m (molal concentration)?
Yes, the concentration of a 1 molar solution would be slightly higher than that of a 1 molal solution, even when the solvent is water.To prepare one litre of a 1 molar solution, 1 mole of the solute is first completely dissolved in a limited quantity of water, and the solution is then made up to 1 litre in a marked standard flask by adding required amount of water. Thus, 1 litre of a 1 molar solution contains exactly 1 mole of the solute.To make a 1 molal solution, 1 mole of the solute is simply dissolved in 1 kg (1 litre) of water. In this case, the actual volume of the final solution would be slightly more than 1 litre, as the dissolution of the solute is usually accompanied by a definite increase in volume. So, the actual amount of the solute present in 1 litre of the solution would be slightly less than 1 mole.
Which of the following aqueous solution has the highest osmotic pressure (please help)?
break all the sub. into ions CaCl2 = Ca2+ + 2Cl- = 3 x 0.030 = 0.090M NaF=Na+ + F- = 2 x 0.060 = 0.120M HCl = H+ + Cl- = 2 x 0.075 = 0.150 C12H22O11=no change(neither dissociation nor association) = 1 x 0.10 = 0.10M in case of CH3COOH association takes place as it forms a dimer so its conc. is 1/2 x 0.050 = 0.025 now O.P=CRT = osmotic pressure is directly proportional to concentration so , HCl has highiest osmotic pressure and CH3COOH has lowest osmotic pressure.
What will the molarity of CI ions be when 0.2 mol NaCl and 0.2 mole BaCl2 are dissolved in a 500ml solution?
First calculate total number of cl- ions.One mole Nacl will dissociate to give one mole Na+ and one mole cl-. So .2 mole will give .2 mole cl-One mole bacl2 will dissociate to give one mole ba2+ and 2 mole cl-. So .2 mole will give .4 mole cl-So the total moles of cl- are .4+.2=.6Molarity is total moles of solute divided by volume of solution in L. So the molarity is .6/.5= 1.2M
Which solution has a higher osmotic pressure, 10% urea, 10% sugar, or 10% glucose?
The osmotic pressure is proportional to the molar concentration of solutes. In the given example, a 10% solution of urea contains more moles urea than does 10% glucose than does 10% sugar (sucrose). Thus, osmotic pressure of urea > than that of glucose > than that of sucrose.
How much PCL5 must be added to one litre vessel at 250C in order to obtain a concentration of 0.1 mole of Cl2 at equilibrium.KC is 0.414?
PCl5 dissociates according toPCl 5→ PCl3+Cl2Initial conc. : a(let) 0 0Final conc : a(1-x) ax axwhere x is the degree of dissociation[Cl2] = 0.1 = ax ….......(i)kc= ([PCl3][Cl2])/ [PCl5]=> (ax. ax)/a(1-x) = 0.0414=> ax2/(1-x) = 0.0414=> ax2=0.0414 { 1-x=1 , since x=> 0.1 x = 0.0414 { From (i) }=> x=0.414From (i) ,a = 0.1/0.414 = 0.2415So, moles of PCl5added =0.2415 moles.
Does anyone have information about hypertrophic scars?
I have several small hypertrophic (raised) scars on the back of my shoulder & just paid a plastic surgeon $100 for 10 min of his time for him to tell me it would cost me $3000 to have them removed. does ANYONE know anything about how else I might take care of the problem w/out the insane expense?