Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Answer:
Ne has the larger ionization energy
<h2>Please mark me as brainliest</h2>
False, pushing a sofa on a hardwood floor would be easier than on carpet.
Answer:
the atomic packing factor of Sn is 0.24
Explanation:
a = b = 5.83A and c = 3.18A.
Volume of unit cell = a²c
= (5.83)² * 3.18 * 10⁻²⁴ cm³
= 1.08 * 10⁻²²cm³
Volume of atoms =
(∴ BCC, effective number of atom is 2)
Volume of atoms =
= 2.55*10⁻²³cm³
<h3>therefore, the atomic packing factor of Sn is 0.24</h3>
