Question

1. A 20.0-8 piece of aluminum at 0.0°C is dropped into a beaker containing 100.0 g of water. The temperature of the water drops from 90.0°C to 85.7°C. What quantity of energy did the aluminum absorb? What is the specific heat of the aluminum? 2. If 55.0 ml of water at 22.5°C is mixed with 100.0 ml of water at 95.2 'C, what would you expect for a final temperature? 3. Cold packs (the kind that get cold when you break an inner compartment) are often made with ammonium nitrate in one package and water in the other. If the mass of water in the pack is 100.0g, the mass of ammonium nitrate is 25.0 g, and the temperature change when the two components are mixed is -20.0°C, what is the enthalpy for dissolution of ammonium nitrate? 4. Calculate the enthalpy change for the formation of 1.00 mol strontium carbonate from its elements. Sr(s) + C(graphite) + 3/2 02(g) → SrCO3(s) The experimental information available is: Sr(s) + O2(g) → Sro(s) Sro(s) + CO2(g) → SrCO3(s) C(graphite) + O2(g) → CO2 (g) AH = -592 kJ AH XD =-234 kJ AHX0 = -394 kJ 5. Using enthalpies of formation (AH) determine the reaction enthalpy (AHx) for the following: a. 2 H2S (g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) b. CzHc(g) + 3.5 O2(g) → 2 CO2(g) + 3 H2O(g)

Answer 1

1. MAL=20.0 g, Tip = 0.0°c, TJAL = 85.7°C, CAL = ? MH₂O = 100.09, TH=90.00, To H₂O = 85.7°C, CHO = 4. 185/9°C quantity of energy aluminium absorb, = - MHzoX CH₂0 x (T& H20 - Tian) =-100.0 x 4.18 (85.7 - 9010) =-(-1797.4J - 1797.4 J aluminium - 1797.4 specific heat of 2010x (85.7-en 1800 2019-11-201 = 1.04865811 Tlgoc S 1.05 3/8°C