The Solubility Product Constant for calcium bromide is > 1. If calcium bromide is dissolved in water you can say that the equilibrium concentrations of calcium and bromide ions are: ... ...A. High ...B. Moderate ...C. Low The solubility of calcium bromide in water is: ... ...A. High ...B. Moderate ...C. Low

(1). If calcium bromide is dissolved in water you can say that the equilibrium concentrations of calcium and bromide ions are high.

(2). The solubility of calcium bromide in water is high.

Explanation:

We know that for a reaction its solubility product is equal to the product of concentration of ions present.

For example,

As it is given that for this reaction > 1. Hence, the product of concentration of both these ions will be greater than 1 which means that products are favored.

Therefore, equilibrium concentrations of calcium and bromide ions are high.

Also, more is the number of these ions formed more will be the solubility of the compound ().

As a result, the solubility of calcium bromide in water is high.

The work done on neon = -323 J

The internal energy change= -392.84 J

The heat absorbed by neon = -69.84 J

Explanation:

Step 1: Data given

Number of moles  = 0.500 moles

Pressure  = 1 atm

Temperature  = 273 Kelvin

The pressure will change from 1.00 atm to 0.200 atm. The temperature changes from 273 to 210 Kelvin.

a) calculate the work done on neon

W = -P(V2-V1)    

⇒ with P = the pressure = 0.1 atm

⇒ with V1 = the initial volume = nRTi /Pi

⇒ with V2 = the final volume = nRTf /Pf

W = -PnR((T2/P2) -(T1/P1))

⇒ with T2 = the final temperature = 210 K

 ⇒ with T1 = the initial temperature = 273 K

 ⇒ with P2 = the final pressure = 0.200 atm

 ⇒ with P1 = the initial pressure = 1.00 atm

W = -nR (210*(0.1/0.2) - 273*(0.1/1.00))

W = -nR*(105 - 27.3)

W= -(0.500)*(8.314)*(77.7)

W = -323 J

b) calculate the internal energy change

E = (3/2)*nRT

ΔE = Ef - Ei

ΔE =(3/2)*nR(T2-T1)

⇒ with n= number of moles = 0.500 moles

⇒ with T2 =the final temperature = 210 K

⇒ with T1 = the initial temperature = 273 K

ΔE = (3/2)*(0.5)*(8.314)(210-273)

ΔE = -392.84 J

c) Calculate the heat absorbed by neon

ΔE = q + W

q = ΔE -W

⇒ with ΔE = -392.84 J

⇒ with W = -323 J

q = -392.84 J -( -323 J)

q =-392.84 J + 323 J

q = -69.84 J