This question asks to compare the energy emitted by a piece of iron at T = 603K with the energy emitted by the same piece at T = 298K.
Then you need to use the Stefan–Boltzmann Law
That law states that energy emitted (E) is proportional to fourth power of the to the absolute temperature (T), this is E α T^4 (the sign α is used to express proportionallity.
Then E (603) / E (298) = [603K / 298K]^4 = 16,8
Which meand that the Energy emitted at 603 K is 16,8 times the energy emitted at 298K.
Answer:
28 grams CaH₂(s) is required for production of 15L H₂(g) at 25°C and 825Torr.
Explanation:
CaH₂(s) + H₂O(l) => Ca(OH)₂(s) + H₂(g)
Using ideal gas law, PV = nRT
=> moles H₂(g) = PV/RT = [(825/760)Atm](15L)/(0.08206L·Atm·mol⁻¹·K⁻¹)(298K) = 0.6659 mol H₂(g)
From stoichiometry of given equation,
=> 0.6659 mol H₂(g) requires 0.6659 mole CaH₂(s)
Converting moles to grams, multiply by formula weight,
=> 0.6659 mole CaH₂(s) = 0.6659 mole CaH₂(s) x 42g/mole = 27.966 grams CaH₂(s) ≅ 28 grams CaH₂(s) (2 sig. figs.)