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Calculate the amount of heat needed to melt 2.0 kg of iron at its melting point (1809 K) given that: ΔH<sub>fus</sub> = 13.80 kJ/mol. A. 27,600 kJ B. 25,000kJ C. 494 kJ D. 27.6 kJ
C. 494 kJ
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Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1 ∘C. At what temperature does benzene boil when the external pressure is 410 torr?
The external pressure of the benzene is: Pb = 410 torr = 54662.20 PbThe molar enthalpy of vaporization of benzene is: ΔHa =30.72 Kj/mol = 30720 J/molThe normal boiling...
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Calculate the standard entropy of melting of ice ,if it's melting 0oc and given the standard enthalpy of melting point 600j/mol
ΔH = 600 J/molT = 273 Kwe know, S= ΔH/TS = 600 Jmol-1/273 K = 2.19 Jmol-1k-1 Hence,the standard entropy of melting of ice is 2.19 Jmol-1k-1
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What is the effect of a small amount of impurity on the melting point of an organic compound?
Decreases melting point
Elevates melting point
Stagnates melting point
None of the above
A small amount of impurity decreases melting point an organic compound
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From an initial mix of: 2 mol Cl2, 1 mol Br2, 0.5 mol I2, 0.5 mol Cl−, 1 mol Br−, 2 mol I−; the total quantity of I2 present after all possible reactions occur is
0.5 mol I2 and 0.5 mol Cl- will not react with anything; 2 mol I- will react with one mol of Cl2 producing one mol of I2: 2I- + Cl2...
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Calculate the heat required to vaporize 7.87 g of benzene at its normal boiling point.
Heat of vaporization (benzene) = 30.7 kJ/mol
Find the amount of benzene (Mbenzene = 78 g/mol) ν = 7.87/78 = 0.1 (mole) Find the heat required: Q = 0.1 × 30.7 = 3.07 (kJ)
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Calculate the heat required to vaporize 7.87 g of benzene at its normal boiling point.
Heat of vaporization (benzene) = 30.7 kJ/mol
Heat = kJ
Solution: Molecular weight of benzene: 78.11 g/mol Amount of benzene: 7.87 / 78.11 = 0.1 mol Energy needed (heat): 0.1 ∙ 30.7 = 3.07 kJ Answer: Heat = 3.07 kJ
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calculate the entropy change of melting of ice,if its melting point is 0C,Given that the standard entalpy of melting ice is 600Jmol
ΔH = 600Jmol-1 T = 2730K S - ? S = ΔH/T S = 600Jmol-1/2730K = 2,20 Jmol-1K-1
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How much heat does it take to melt 18.1 g of phenol at its melting point? The heat of fusion of phenol is 122 J/g.
Q=m*ΔHf; Q - heat required; m - mass of phenol; ΔHf - heat of fusion;Q=18.1*122=2208.2 J;Answer: 2208.2 J
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Poly melts at approximately at 327 degrees celsius. If the heat of fusion is 5.28 kJ mol ^-1, what is the molar entropy change oc fusion?
Temperature T=327OC=600K Heat, q=600×5.28kJ Entropy change, ΔS=K 600×5.8kJ=3480KJ(3480)/(6.023)(1023) = 5.778-21
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