a solution was prepared by dissolving 177 mg of potassium sulfate (k2so4, mw = 174.24 g/mol) in 775 ml of water. calculate the following:
a) moles of k2so4
b)millimoles of k2so4
c)molarity of k2so4, k+, so4(2-)
d)ppm of k2so4
e)%(w/v) k2so4
f)pk+
g)pso4(2-)

Explanation:

a) Mass of potassium sulfate = 177 mg =0.177 g

1 g= 1000 mg

Molecular mass of potassium sulfate = 174.24 g/mol

Moles of potassium sulfate,n =

b) 1 mol = 1000 milli moles

0.001016  mol =0.001016 × 1000 milli moles = 1.016 milli moles

c)

n = moles of substance

V = Volume of the solution in L

Molarity of of potassium sulfate:

Volume of the solution = 775 mL = 0.775 L (Mass of solute is very less)

1 mol potassium sulfate gives 2 mol of potassium ions and 1 mole of sulfate ions.

Molarity of potassium ions:

Molarity of sulfate ions:

d)

Volume of solution = 0.775 L

Amount of potassium sulfate = 177 mg

e) Volume of the solution = 775 mL

Mass of potassium sulfate = 177 mg =0.177 g

f)

(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