If a sample is packed poorly, the sample will not heat evenly and will take longer to melt.
Answer:
ΔS = +541.3Jmol⁻¹K⁻¹
Explanation:
Given parameters:
Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹
Standard Entropy of C = 5.7Jmol⁻¹K⁻¹
Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹
Standard Entropy of CO = 198Jmol⁻¹K⁻¹
To find the entropy change of the reaction, we first write a balanced reaction equation:
Fe₂O₃ + 3C → 2Fe + 3CO
To calculate the entropy change of the reaction we simply use the equation below:
ΔS = ∑S
- ∑S
Therefore:
ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1
ΔS = +541.3Jmol⁻¹K⁻¹
1. C
2. C
3. In elastic deformation, the deformed body returns to its original shape and size after the stresses are gone. In ductile deformation, there is a permanent change in the shape and size but no fracturing occurs. In brittle deformation, the body fractures after the strength is above the limit.
4. Normal faults are faults where the hanging wall moves in a downward force based on the footwall; they are formed from tensional stresses and the stretching of the crust. Reverse faults are the opposite and the hanging wall moves in an upward force based on the footwall; they are formed by compressional stresses and the contraction of the crust. Thrust faults are low-angle reverse faults where the hanging wall moves in an upward force based on the footwall; they are formed in the same way as reverse faults. Last, Strike-slip faults are faults where the movement is parallel to the crust of the fault; they are caused by an immense shear stress.
I hope this helped :D
Troposphere has most of our clouds temperature etc
