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
Answer:
Correct option is B)
According to de-Broglie,
λ=mvh=6.023×10232×5×104cm/sec6.62×10−27ergsec=4×10−8cm=4Ao
Calcium carbonate has the formula: CaCO3
From the periodic table:
mass of calcium = 40 grams
mass of carbon = 12 grams
mass of oxygen = 16 grams
Therefore,
molar mass of CaCO3 = 40 + 12 + 3(16) = 100 grams
molar mass of carbonate = 12 + 3(16) = 60 grams
One mole of calcium carbonate contains one mole of carbonate. Therefore, 100 grams of CaCO3 contains 60 grams of CO3.
If the 0.5376 grams of the unknown substance is CaCO3, then the amount of carbonate will be:
amount of carbonate = (0.5376*60) / 100 = 0.32256 grams
Based on the above calculations, the sample is not CaCO3
Number 9 is A and 10 is A
