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Chapter 25 Energy from chemical reactions Q1. Sodium nitrate dissolves quick in water according to the equation NaNO3(s) ? NaNO3(aq); ?H = +21.0 kJ groyne1 aDetermine the postal code change when 1.0 g of unassailable NaNO3 is turn in water. bWould you expect the temperature to rise or get on with to pass when sodium nitrate dissolves in water? develop your answer. A1. an(NaNO3) = =0.0117 moleee 1 mole NaNO3 absorbs 21 kJ vim 0.0117 mole absorbs 0.0117 × 21 kJ = 0.25 kJ bThe temperature will fall. Since the reaction is endothermic the enthaply of reactants is higher than the enthalpy of products, i.e. the reaction absorbs energy. Q2. Calculate the energy released when the following quantities of ethane gas burn mark mark according to the equation: 2ethane(g) + 7O2(g) ? 4CO2(g) + 6H2O(l); ?H = 3120 kJ mol1 a3.00 mol b100 g c10.0 L at SLC A2. aFrom the equation, lend oneself stoichiometry to watch energy released. 2 mol ethane releases 3120 k J of energy 3.00 mol C2H6 releases ´ 3.00 kJ So energy = 4680 kJ = 4.68 ´ 103 kJ b tone 1Calculate the sum of money of ethane, using n = . n(C2H6)= = 3.326 mol spirit 2Use stoichiometry to find energy released. 2 mol C2H6 releases 3120 kJ of energy 3.326 mol C2H6 releases 3120 ´ kJ So energy released= 5189 kJ = 5.19 ´ 103 kJ cStep 1Calculate the amount of ethane, using n = at SLC. n(C2H6)= = 0.4082 mol Step 2Use stoichiometry to find energy released. 2 mol C2H6 releases 3120 kJ of energy 0.4082 mol C2H6 releases 3120 ´ kJ So energy released= 636.7 kJ = 637 kJ Q3. What volume of ethane, measured at STP, must be burn down according to the equation in school principal 2 in order to yield 100 kJ of warming energy? A3. Step 1Use stoichiometry to find the amount of C2H6. 3120 kJ is released by 2 mol 100 kJ is released by 2 ´ mol So n(C2H6)= 2 ´ = 0.641 mol Step 2Calculate the volume of C2H6 at STP, using n = . V(C2H6) = n ´ Vm = 0.641 mol ´ 22.4 L...If you wa! nt to pluck up a full essay, order it on our website: BestEssayCheap.com

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