Calculate the solubility at 25°c of agcl in pure water and in a 0.0080m agno3 solution. you'll find ksp data in the aleks data tab. round both of your answers to 2 significant digits. the ksp value of agcl is 1.77 times 10 to the -10.a. solubility in water: b. solubility in 0.0080 agno3 solution:

a) S = 1.33 E-5 M

b) S = 2.21 E-8 M

Explanation:

           S            S          S

a) in pure water:

⇒ Ksp = [ Ag+ ][ Cl- ] = ( S )( S ) = S²

∴ Ksp AgCl = 1.77 E-10

⇒ S² = 1.77 E-10

⇒ S = √1.77 E-10

⇒ S = 1.33 E-5 M

b) in 0.0080 M AgNO3:

         0.008     0.008        0.008

          S        S + 0.008         S

⇒ Ksp = 1.77 E-10 = (S + 0.0080)*( S )

⇒ Ksp = 1.77 E-10 = S² + 0.008S

⇒ S² + 0.008S - 1.77 E-10 = 0

⇒ S = 2.21 E-8 M

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

your parents i think this could be it