The value of ΔS° for reaction is - 22.2 J/K.mol
→ 
Calculation,
Given value of S°(J/K.mol) for
= 248.5
= 240.5
= 210.6
= 256.2
Formula used:
ΔS° (Reaction) = ∑S°(Product) - ∑S°(Reactant)
ΔS° = (256.2 + 210.6 ) - ( 248.5 + 240.5) = 466.8 - 489 = - 22.2 J/K.mol
The change in stander entropy of reaction is - 22.2 J/K.mol. The negative sign indicates the that entropy of reaction is decreases when reactant converted into product.
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I’m not sure it’s correct but I think it’s D) He used voltage adjustments to make charges oil drops float
Answer:
1 g
Explanation:
The half-life of Am-242 (16 h) is the time it takes for half of it to disappear.
We can make a table of the mass left after each half-life.

The mass remaining after 48 h is 1 g.