How much heat required to convert 36 grams (2 moles) of liquid water at 4.0 oC to liquid water at 50 oC? The delta-H of fusion for H2O is 6.02 kJ/mol and the heat capacity of liquid water is 4.18 J/goC
A) 36 kJ
B) 12.04 kJ
C) 8.36 kJ
D) 4.18 kJ
E) 6.9 kJ
How much heat required to convert 36 grams (2 moles) of liquid water at 4.0 oC...
How much heat required to convert 36 grams (2 moles) of ice at 0 degrees C to liquid water at 50 degrees C? delta-H of fusion for H2O is 6.02 kJ/mol and the heat capacity of liquid water is 4.18J/g degrees C. A) 12 kJ B) 7524 J C) 8728 J D) 19564 J
How much heat (in kJ) is needed to convert an 18.0-g cube of ice at 0.0 oC into liquid water at 20.0 oC? For water (H2O): heat of fusion = 6.02 kJ/mol, specific heat capacity (liquid) = 4.18 J/g oC. a) 1.61 kJ b) 7.52 kJ c) 89.6 kJ d) 518 kJ
of water fusion How much heat in kJ would be required to convert 31.6 grams of ice from -10.0°C to liquid water at 0.00*C. The AH, is 6.02 kJ/mole, the specific heat of ice is 2.01 Jg*c", and the specific heat of liquid water is 4.18 jgc". Report your answer to 1 decimal place.
Assume 12,500 J of energy is added to 2.0 moles (36 grams) of H2O as an ice sample at 0 °C. The molar heat of fusion is 6.02 kJ/mol. The specific heat of liquid water is 4.18 J/(g • °C). The molar heat of vaporization is 40.6 kJ/mol. The resulting sample contains which of the following? A) only ice B) ice and water C) only water D) water and water vapor E) only water vapor
Calculate the heat required in Joules to convert 18.0 grams of water ice at a temperature of -20° C to liquid water at the normal boiling point of water. Given: -specific heat of ice = 2.09 J/g°C -specific heat of liquid water = 4.184 J/g°C -specific heat of water vapor = 2.03 J/g°C -molar heat of fusion of water = 6.02 kJ/mol -molar heat of vaporization of water = 40.7 kJ/mol
How much energy is required to heat 36.0 g H2O from a liquid at 55.0°C to a gas at 150.0°C? The following physical data may be useful. Molar Mass(H2O) = 18.0 g/mol ΔHvap = 40.7 kJ/mol Cs;liquid = 4.18 J/g oC Cs;gas = 2.01 J/goC Tb(H2O) = 100.0 oC
how much heat is released when 10.0 g of steam (water vapor ) at 105.0 C is cooled to liquid water at 25 C? S(water) = 4.18 J/g.C. ... S(steam) = 2.01 j/ g.C the heat of fusion of water is 6.02 KJ/ mol. The heat of vaporization of water is 40.7 KJ/mol
34 kJ of heat was added to 6.5 moles of solid H2O at –27 oC. What will be the final temperature of the H2O? The molar heat capacity of H2O (s) is 37.8 J mol–1 oC–1, and the molar heat capacity of H2O (l) is 75.2 J mol–1 oC–1. The heat of fusion of water is 6.03 kJ mol–1.
How much energy (heat) is required to convert 52.0 g of ice at -10.0 C to steam at 100 C?Specific heat of ice: 2.09 J/g * C DHfus = 6.02 kJ/molSpecific heat of water: 4.18 J/g * C DHvap = 40.7 kJ/molSpecific heat of steam: 1.84 J/g * C
"How much energy is evolved to convert 36g H2O(g) at 135 degreees C to H2O at -15 degrees C?" My question was previously answered but I don't understand why q5 is negative when the the enthalpy is positive ? no of moles of water = W/G.M.Wt = 36/18 = 2 moles The steam convert 135°c to 100°C q1 = mc T = 36*1.84"(100-135) = -2318.43 = -2.3184K) The heat of vaporization of water q2 = nAHvap = -2*40.7 = -81.4K)...