Calculate the work, w, gained or lost by the system when a gas expands from 15 L

Question

Calculate the work, w, gained or lost by the system when a gas expands from 15 L to 45 L against a constant external pressure of 1.5 atm. 1 L atm = 101 J 0-6.8 kJ +4.5 kJ +6.8 kJ -4.5 kJ A 4.50-g sample of liquid water at 25.0 C is heated by the addition of 112 J of energy. The final temperature of the water is degree C. The specific heat capacity of liquid water is 4.18 J/gK. 25.2 129 -19.0 5.95 31.0 A 50.0-g sample of liquid water at 25.0 degree C is mixed with 35.0 g of water at 79.0 degree C. The final temperature of the water is degree C. 27.4 161 0269 52.0 47.2 What is the enthalpy change (in kJ) of a chemical reaction that raises the temperature of 250.0 ml of solution having a density of 1.25 g/ml by 7.80 degree C? (The specific heat of the solution is 3.74 joules/gramK.) 6.51 kJ 8.20 kJ -7.43 kJ -12.51 kJ -9.12 kJ For a particular process that is carried out at constant pressure, q = 135 kJ and w = - 25 kJ. Delta E = 135 kJ and Delta H = 110 kJ Delta E = 110 kJ and Delta H = 135 kJ. Delta E = 160 kJ and Delta H = 135 kJ. Delta E = 135 kJ and Delta H = 160 kJ. Therefore, How much heat is absorbed/released when 35.00 g of NH3(g) reacts in the presence of excess O2(g) to produce NO(g) and H2O(1) according to the following chemical equation? 4 NH3(g) + 5 O2(g) righatarrow 4 NO(g) + 6 H2O(l) Delta H degree = 1168 kJ 600.1 kJ of heat are released. 600.1 kJ of heat are absorbed. 2400 kJ of heat are released. 2400 kJ of heat are absorbed.

Explanation / Answer

1.1.5*3*101=4.5 kJ 2. 112/(4.18*4.5)+25 =31 3. 50*(t-25)=35*(79-t) t=(35*79+50*25)/(50+35) =47.2 4. 3.74*1.25*7.8*250=-9.12 5.delta E=160 delta H=135 6.1168*35/(17*4) =600.1kJ released

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