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.
Explanation:
Elements in the same group have same number of valence electrons. And we know, the elements which have same number of valence electrons, have similar physical and chemical properties. Hence, the elements in the same group have similar physical and chemical properties.
Answer:
5.64×10²³ atoms C
Explanation:
Convert moles of H to moles of C:
2.81 mol H × (2 mol C / 6 mol H) = 0.937 mol C
Convert moles of C to atoms of C:
0.937 mol C × (6.02×10²³ atoms C / mol C) = 5.64×10²³ atoms C
Answer:
<h2>A. Mercury's orbit is shorter than Earth</h2>
