Data: 1 nm = 1.10^3 m; Speed of light = 3.00-10^8 m-s^1; Planck\'s constant = 6.
ID: 953969 • Letter: D
Question
Data: 1 nm = 1.10^3 m; Speed of light = 3.00-10^8 m-s^1; Planck's constant = 6.626 10^-34 J-s; Avogadros number = 6.02-10^23 mol^-1; Mass of olectron = 9.11 10^-31 kg 1. A system does 125 J of work on its surroundings. Then the system absorbs 450 J of heat from the surroundings. Next, the surroundings absorb 365 J of heat from the system. Finally. 260 J of work is done by the surroundings on the system. The total change in the internal energy of the system is J. 2. Gasoline is burned in a cylinder equipped with a piston. The initial volume of the cylinder is 5.690-10^-4 m^3, and the final volume is 6.259-10^3 m^3. The piston expands against a pressure of 1.207-10^3 Pa. The amount of work done (as applies to the system) is J. 3. The decomposition of liquid hydrogen peroxide into liquid water and oxygen gas can be represented bj the following equation: 2 H_2O_2(I) rightarrow 2 H_2O(I) + O_2(g) DeltaHdegree = -196.1 kJ The amount of heat absorbed or released when a mass of 652 g of liquid hydrogen peroxide decomposes at a constant pressure according to this equation is: a. -3759 kJ b. -3240 kJ c. -1879 kJ d.-1620 kJ e. -940 kJ f.-8l0 kJ g. +3240 kJ h. + 3759 kJ i. None of the above; the correct amount of heat is kJ 4. The molar heat capacity of aluminum is 24.4 J-mol^-1 K^-1. The amount of heat needed to heat 2.63 g aluminum from 25.0 degreeC to its melting point, which is 660.4 degreeC, is: a. 1.511 kJ b. 1.571 kJ c. 1.626 kJ d. 1.690 kJ e. 2.161 kJ f. 2.325 kJ g. 40.78 kJ h. 43.88 kJ i. None of the above; the correct amount of heat is kJExplanation / Answer
1. E = - 125 J + 450 J -365 J + 260 J = 220 J
2. w = p (Vf – Vi) = ….
3. 652 g / 34gH2O2/mol = 19.2 mol H2O2
2 mol H2O2………………..-196.1kJ
19.2 mol……………………..x
X= - 1879 kJ
4.
Qreceived = m.h.t = [2.63g x 24.4 J/(mol.K) / 27.0 gAl/mol] x (600.4-25.0)K = …..J
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