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50.0 mL H 2 O x 1 g H 2 O x 1 mol H 2 O x 715 kJ =

Given the enthalpy change for the reaction below as shown, how much heat is released when 50.0 mL of liquid water is produced from the combination of hydrogen and oxygen gas? 2H 2 (g ) + O 2 (g) → 2H 2 O(l) + 715 kJ. 50.0 mL H 2 O x 1 g H 2 O x 1 mol H 2 O x 715 kJ =

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50.0 mL H 2 O x 1 g H 2 O x 1 mol H 2 O x 715 kJ =

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  1. Given the enthalpy change for the reaction below as shown, how much heat is released when 50.0 mL of liquid water is produced from the combination of hydrogen and oxygen gas? 2H2(g) + O2(g)→ 2H2O(l) + 715 kJ 50.0 mL H2O x 1 g H2O x 1 mol H2Ox 715 kJ = 1 mL H2O 18.02 g H2O 2 mol H2O = 992 kJ released

  2. The reaction that occurs in heat packs used to treat sports injuries is shown below. How much heat is released when 1.00g of iron is reacted with excess O2 if the change in enthalpy for the reaction is -1652 kJ? 4Fe(s) + 3O2(g) → 2Fe2O3 1.00g Fe x 1 mol Fe x 1652 kJ = 7.39 kJ released 55.85 g Fe 4 mol Fe

  3. Use the thermochemical equations shown below or the standard heats of formations provided to determine the enthalpy for the reaction: H2CO + O2 → H2CO3 A) H2CO3 → H2O + CO2 ∆H = 77.5 kJ H2O + CO2 → H2CO + O2 ∆H = 62.5 kJ Substance Standard Heat of Formation (kJ/mol) H2CO -1011 H2CO3 -1151 BOOK METHOD (SYSTEM OF EQUATIONS) - -A + -B = -77.5kJ + -62.5 kJ = -140 kJ DIRECT METHOD (HESS's LAW EQUATION) - ∆H = (-1151kJ/mol) - (-1011 kJ/mol) = -140 kJ/mol

  4. Is this reaction always, sometimes or never spontaneous? (reply A, S or N) 4Fe(s) + 3O2(g) → 2Fe2O3 + 1652kJ ∆H is negative, ∆S is negative, which means it is sometimes spontaneous. This reaction will take place at low temperatures, but not at high temperatures.

  5. A 2.6 g sample of a metal requires 15.2 J of energy to change its temperature from 21oC to 34oC. Use the table to determine the identity of metal. Q = s m ∆T s = Q m ∆T s = 15.2 J (2.6g) (34oC - 21oC) s = 0.45 J/goC Metal is iron

  6. A sample of gold loses 3.1 J of energy when it cools from 27 oC to 19oC. What is the mass of this sample of gold? Q = s m ∆T m = Q s ∆T m = - 3.1 J (0.129 J/goC ) (19oC - 27oC) m = 3 g Au

  7. If 294 J of heat is transferred to a 10.0 g sample of silver at 25oC, what is the final temperature of the silver? (sAg = 0.24 J/goC) Q = s m ∆T Tf-Ti = Q s m Tfinal - 25oC = 294 J (0.24 J/goC ) (10.0g) Tfinal = 150oC

  8. Exothermic Reaction Reactants  Products + Energy Energy of reactants Energy of products Reactants Energy -DH Products Reaction Progress

  9. Endothermic Reaction Energy + Reactants  Products Products Energy +DH Endothermic Reactants Reaction progress

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