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.
What amount of heat absorbs 50 g of steel (ce = 0.115 cal / g. ° C) that
does its temperature vary by 25 ° C?
Answer:
143.75cal
Explanation:
Given parameters:
Mass of steel = 50g
Specific heat capacity of the steel = 0.115cal/g°C
Temperature = 25°C
Unknown:
Amount of heat = ?
Solution:
The amount of heat to cause this temperature change is dependent on mass and specific heat capacity of the substance.
Amount of heat = m C (ΔT)
m is the mass
c is the specific heat capacity
ΔT is the temperature change
Now insert the parameters and solve;
Amount of heat = 50 x 0.115 x 25
Amount of heat = 143.75cal
Answer:
12.50g
Explanation:
T½ = 2.5years
No = 100g
N = ?
Time (T) = 7.5 years
To solve this question, we'll have to find the disintegration constant λ first
T½ = In2 / λ
T½ = 0.693 / λ
λ = 0.693 / 2.5
λ = 0.2772
In(N/No) = -λt
N = No* e^-λt
N = 100 * e^-(0.2772*7.5)
N = 100*e^-2.079
N = 100 * 0.125
N = 12.50g
The sample remaining after 7.5 years is 12.50g
An air mass is a large volume of air in the atmosphere thats mostly uniform in temperature and moisture. Air masses can extend thousands of kilometers across the surface of the Earth, and can reach from ground level to the stratosphere/10 miles into the atmosphere. an air mass over northern Canada is a continental polar air mass and is cold and dry. One that forms over the Indian Ocean is called a maritime tropical air mass and is warm and humid.
I believe C is the balanced equation.
