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.
Ice floats after it crystallizes because ITS DENSITY IS LESS THAN THAT OF WATER.
When a quantity of water is cools down by reducing its temperature, the molecules of the water lose kinetic energy and slow down in their movement. As the water is cooling down, it is volume is expanding. When the temperature reaches zero degree Celsius, the water becomes ice. At this point, the ice can float on water because its density is less than that of water; this is as a result of the spaces that now exist in the ice structure.
Carbon dioxide (CO2)
in the process of respiration, oxygen and glucose react to form carbon dioxide and water.
Answer:
The answer to your question is: SiCl₄
Explanation:
Data
amount of Si 1.71 g
amount of Cl 8.63 g
MW Si = 28 g
MW Cl = 35.5
Process (rule of three)
For Si For Cl
28 g of Si ------------------ 1 mol 35.5 g of Cl --------------- 1 mol
1.71g of Si --------------- x 8.63 g of Cl -------------- x
x = 1.71 x 1 / 28 = 0.06 mol x = 8.63 x 1 / 35.5 = 0.24 mol
Now, divide both results by the lowest of them.
Si = 0.06 mol / 0.06 = 1 molecule of Si Cl = 0.24 / 0.06 = 4 molecules of Cl
Finally
Si₁ Cl₄ or SiCl₄
Answer: The standard entropy of vaporization of ethanol is 0.275 J/K
Explanation:
Using Gibbs Helmholtz equation:
For a phase change, the reaction remains in equilibrium, thus 
Given: Temperature = 285.0 K
Putting the values in the equation:
Thus the standard entropy of vaporization of ethanol is 0.275 J/K
