The molar solubility of C a ( O H ) 2 C a ( O H ) 2 was experimentally determined to be 0.019 M. Based on this value, what is the K s p K s p of C a ( O H ) 2 C a ( O H ) 2 ?

Ksp = 2.74 x 10⁻⁵

Explanation:

The solubility equilibrium for Ca(OH)₂ is the following:

  Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)

I                          0                0

C                       + s               + 2s

E                        s                   2s

According to the ICE table, the expression for the solubility product constant (Kps) is:

Ksp = [Ca²⁺] x ([OH⁻])² = s x (2s)² = 4s³

Then, we calculate Ksp from the solubility value (s):

s = 0.019 M

⇒ Ksp = 4s³ = 4 x (0.019)³ = 2.74 x 10⁻⁵

yes

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(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