A 5.0 g sample of aluminum with a specific heat of 0.90 J/(g °C) was heated from 22.1°C to 32.1°C. How much heat, to the nearest
1 answer:
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When the enthalpy value is given, we can calculate how much heat is use or produces in a given equation.
67.6 kCal ---> 67.6 kCal= 1 mol of reaction
1 mol of reaction= 1 mol of CO (based on the coefficient)
so 1 mole of CO gives us 67.6 kCal of heat.
calculation:
1 mol CO
67.6 kCal ---> 67.6 kCal= 1 mol of reaction
1 mol of reaction= 1 mol of CO (based on the coefficient)
so 1 mole of CO gives us 67.6 kCal of heat.
calculation:
1 mol CO
Answer:
c. ΔH° is positive and ΔS° is positive.
Explanation:
Hello,
In this case, as the Gibbs free energy for a reaction is defined in terms of the change in the enthalpy and entropy as shown below:
Thus, as the reaction becomes spontaneous (ΔG°<0) at temperatures above 1100K (high temperatures), it necessary that c. ΔH° is positive and ΔS° is positive as the entropy will drive the spontaneousness as it becomes smaller than TΔS°.
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