Use the thermodynamic data at 298 k below to determine the ksp for barium carbonate, baco3 at this temperature. substance: ba2+(aq) co32–(aq) baco3(s) δh°f (kj/mol): –538.36 –676.26 –1219 δg°f (kj/mol): –560.7 –528.1 –1139 s°(j/k·mol): 13 –53.1 112

Answer : the correct answer for ksp = 1.59 * 10⁻⁹

Following are the steps to calculate the ksp of reaction

BaCO₃ →Ba ²⁺ + CO₃²⁻ :

Step 1 : To find ΔG° of reaction :

ΔG° of reaction can be calculates by taking difference between ΔG° of products and reactants as :

ΔG° reaction =Sum of ΔG° ( products ) - Sum of Δ G° ( reactants ) .

Given : ΔG° for Ba²⁺ ( product )= -560.7

ΔG° for CO₃²⁻ (product ) =- 528.1

ΔG° BaCO₃ ( reactant) = –1139

Plugging value in formula :

ΔG° for reaction = ( ΔG° of Ba ²⁺ + ΔG° of CO₃²⁻ ) - (ΔG° of BaCO₃ )

⁻ = ( -560.7 + 528.1 ) - ( -1139 )

= ( -1088.8 ) - (-1139 )

= - 1088.8 + 1139

ΔG° of reaction = 50.2

Step 2: To calculate ksp from ΔG° of reaction .

The relation between Ksp and ΔG° is given as :

ΔG° = -RT ln ksp

Where ΔG° = Gibb's Free energy R = gas constant T = Temperature

Ksp = Solubility constant product .

Given : ΔG° of reaction = 50.2

T = 298 K R = 8.314

Plugging values in formula

((Converting 2477

Since , 1 KJ = 1000 J So , ))

Dividing both side by

ln ksp =

Removing ln :

ksp = 1. 59 * 10⁻⁹

the answer is b & c i believe : )

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72