Achemist dissolves 274.mg of pure barium hydroxide in enough water to make up 50.ml of solution. calculate the ph of the solution. (the temperature of the solution is 25°c.)

12.81.

Explanation:

M = (no. of moles of solute)/(Volume of the solution (L))

M = (mass/molar mass) NaCl / (Volume of the solution (L))

mass of Ba(OH)₂ = 274 mg = 0.274 g, molar mass of Ba(OH)₂ = 171.34 g/mol, Volume of water = 50.0 mL = 0.05 L.

M = (mass/molar mass) Ba(OH)₂ / (Volume of the solution (L)) = (0.274 g / 171.34 g/mol) / (0.05 L) = 0.03 M.

Ba(OH)₂ → Ba²⁺ + 2OH⁻,

Every 1.0 mol of Ba(OH)₂ gives 2.0 moles of OH⁻.

∴ [OH⁻] = 2(0.032 M) = 0.064 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.064) = 1.194.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.194 = 12.81.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied