Based on the following data, what is the standard Gibbs free energy of formation of the sulfate ion at 298 K? ( R = 8.31 J/(K ∙ mol)) PbSO4(s) Pb2+(aq) + SO42–(aq); Ksp = 1.7 × 10–8

Substance ΔG°f (kJ/mol) at 298 K
Pb2+(aq)
–24.4

PbSO4(s)
–813.0
a. –788.6 kJ/mol

b. –793.1 kJ/mol

c. –837.4 kJ/mol

d. –744.3 kJ/mol

e. 832.9 kJ/mol

Ineed with this also i hope you understand it

Radon-222 or rn-222 has a half-life of 3.82 days. to determine the fraction of an original sample of rn-222 that remains unchanged after 7.646 days, we calculate the amount of rn-222 remaining after 7.646 days and divide it with the original amount. using 1 gram as a basis and using half-life calculation methods, we can determine that  0.249 or 0.25  grams of rn-222 will remain, and thus the fraction remaining is 0.25 or 1/4.

This is easy a.) transparency

The rules are different for multiplication and addition, but the least precise value always restricts the answer. first do the multiplication (as per standard order of operations): 24.6681 × 2.38 ≈ 58.7 since 2.38 has the fewest significant figures, it is considered the lease precise value then do the addition: 58.7 + 332.58 ≈ 391.3 since 58.7 has fewest figures right of decimal it is considered least precise. so final answer: 391.3 with four significant figures.