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:
12.0108408
Explanation:
Denote the element with a letter like say X. Since it has a subscript of 5, then, X5.
Molecular mass=102.133g/mol.
% of X in compound =58.8/100
=0.588
Mass of X in the compound = 0.588*102.133 ( the % of X in compound * molar mass of compound)
= 60.054204
X5=60.054204
Then element X has a mass of 60.054204/5=12.0108408
Answer:
98.3 gradius Celsius
Explanation:
This problem is solved using the Ideal Gas Equation
pV = nRT
...
Please see the step-by-step solution in the picture attached below.
Hope this answer can help you. Have a nice day!
Answer:
a
Explanation:
the answer is a i'm pretty sure it might be wrong tho i'm sorry