Assuming enthalpy of combustion of hydrogen at 273 K, –286 kJ and enthalpy of fusion of ice at the same temperature to be + 6.0 kJ, calculate enthalpy change during formation of 100 g of ice

n(C2H5OH)=m/M=10/46 = 0.2174 molH= (-1366.8)*0.2174 = -297.14 kJAnswer: -297.14 kJ

C. entropy

The standard enthalpy of formation is the change in enthalpy that accompanies the formation of one mole of the compound from its elements. The standard enthalpy of reaction occurs in a...

ΔHform(C2H6)=ΔHcomb(C2H6)-(2ΔHcomb(C)+3ΔHcomb(H2))=-1559.7-(2(-393)+3(-286))=84,3 kJ/mol

Answer: Enthalpy of combustion is the heat released during combustion of one mol amount of any substance by oxygen.You can calculate the combustion enthalpy of methanol using the enthalpy...

n = m / MM (C2H6) = 30.07 g/moln (C2H6) = 105 / 30.07 = 3.5 moldelta H (C2H6) = - 1251 x 3.5 = -4367 kJ

a. The balanced equation: 2 H2 + O2 = 2 H2O, therefore: the Enthalpy is: (-286) ∙ 2 = -572 kJ b. Energy released in the reaction = -Enthalpy, so:...

Find the total heat used by ice in changing state from solid to liquid(water) Q1 = 20g*80cal/g = 1600 cal 1600 cal = 6694.4 J = 6.69 kJ M'C(T'-x) =...

Temperature  T=327OC=600K Heat, q=600×5.28kJ Entropy change, ΔS=K 600×5.8kJ​=3480KJ(3480)/(6.023)(1023) = 5.778-21

Solutions: Meters (m) and centimeters (cm) can be related by the prefix modifier equality 100 cm = 1 m.So therefore, this equality given above can be represented as two conversion...