Question

4) Calculate the molar solubility of a barium fluoride solution that contains 6.5 x 10 M barium nitrate. 5) Calcium nitrate is slowly added to a solution that contains 0.10 M PO ions and 0.10 M OH ions. a) Which ion will precipitate first? Kp of Ca (PO4)2 1.2 x 10 25, Kp of Ca(OH), 4.68 x 10 b) When the second precipitate starts to form, what is the concentration of the first ion left in solution? 6) What treatment would precipitate Ba ion out of a solution? a) Nacl b) Li,S c) К,Cо, 7) By inspection, determine whether the following processes would have a positive or negative entropy change. a) sublimation of dry ice b) sugar dissolving in water c) condensation of water vapor 8) From the provided tabulated data, determine if the process is spontaneous at 25.0°C by calculating a) ASuniverse and b) AGo system CaCO3 (s) -1206.9 CaO (s) + CO2 (g) ДНР, К/mol So, J/mol K -635.6 -393.5 213.6 92.9 39.8 476 J/mol K. At what temperature will it become 9) A system has a AH spontaneous? 199.5 kifmoi and a ASye 10) The molar heat of fusion of benzene is 10.9 kl/mol. Calculate the entropy change when solid benzene phase changes at its melting point, 5.5°C. 11) For the reaction, N;Oa (g)2NO2 (g), K = 0.113 at 298 K and AG 5.4 kJ/mol. At some non- equilibrium condition, [N;O] = 0.453 M and [N2] = 0.122 M. Find AG under these conditions. 12) What is the purpose of the Zeroth Law of Thermodynamics? a sys- are n The c mal e qulhbrium wiTh Tu 4 3ystem, Then Thoyr two are in Thernal eqilileriuas Te each OTher. There is no ut heat flous e Heat is fundamental and weasuredhy, ProPerty 3 ma ter 13) How much heat is required to heat 1.50 kg of water from 25.0°C to 100.0°c? C 4.18 J/g °C 0.235 J/g °C) at 58.5°C is submerged in 100.0 g of water at 24.8°C. The final temperature 14) A block of silver (C of the mixture is 26.2°C. What is the mass of the silver block?

can you please show me the detailed steps for solving each of the problems
thanks

Answer 1

Question 4.

The solubility product of BaF2 (Ksp) = 10^-6

Here, Ksp = [Ba²⁺][F^-]²

i.e. 10^-6 = [Ba(NO3)2]*[F^-]²

i.e. 10^-6 = 6.5*10^-3 *[F^-]²

i.e. [F^-]² = 0.153846*10^-3

i.e. [F^-] = 0.0124 M

Therefore, the molar sollubility of BaF2 in 6.5*10^-3 M Ba(NO3)2 = 0.0124/2 = 6.2*10^-3 M

Question 5.

a) Ca3(PO4)2 will precipitate first because the Ksp of Ca3(PO4)2 is less than that of Ca(OH)2.

b) Ksp of Ca3(PO4)2 = 1.2*10^-25 = [Ca²⁺]³*[PO₄^3-]² = (3s)³*(2s)² = 81 s⁵

i.e. s = 4.307*10^-6 M

i.e. [Ca²⁺] = 3s = 1.292*10^-5 M

Now, Ksp of Ca(OH)2 = [Ca²⁺][OH^-]²

i.e. 4.68*10^-6 = 1.292*10^-5 * [OH^-]²

i.e. [OH^-]² = 0.362

i.e. [OH^-]= 0.602 M