It would be A. proton and electron
Missing question:
A. [3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s)] / 2
<span>B. 3.40 mol Fe2O3 (s) × 26.3 kJ/1 mol Fe2O3 (s) </span>
<span>C. 26.3 kJ/1 mol Fe2O3 (s) / 3.40 mol Fe2O3 (s) </span>
<span>D. 26.3 kJ/1 mol Fe2O3 (s) – 3.40 mol Fe2O3 (s).
</span>Answer is: B.
Chemical reaction: F<span>e</span>₂O₃<span>(s) + 3CO(g) → 2Fe(s) + 3CO</span>₂<span>(g);</span>ΔH = <span>+ 26.3 kJ.
When one mole of iron(III) oxide reacts 26,3 kJ of energy is required and for 3,2 moles of iron(III) oxide 3,2 times more energy is required.</span>
Answer:
d is the answer of this question
464 g radioisotope was present when the sample was put in storage
<h3>Further explanation</h3>
Given
Sample waste of Co-60 = 14.5 g
26.5 years in storage
Required
Initial sample
Solution
General formulas used in decay:

t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Half-life of Co-60 = 5.3 years
Input the value :
