The energy produced by burning : -32.92 kJ
<h3>Further explanation</h3>
Delta H reaction (ΔH) is the amount of heat change between the system and its environment
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The enthalpy and heat(energy) can be formulated :
The enthalpy of combustion of naphthalene (MW = 128.17 g/mol) is -5139.6 kJ/mol.
The energy released for 0.8210 g of naphthalene :
Answer is: 13181,7 kJ of energy <span>is released when 10.5 moles of acetylene is burned.
</span>Balanced chemical reaction: C₂H₂ + 5/2O₂ → 2CO₂ + H₂O.
<span>ΔHrxn = sum of
ΔHf (products of reaction) - sum of ΔHf (reactants).</span><span>
Or ΔHrxn = ∑ΔHf (products of reaction)
- ∑ΔHf (reactants).
ΔHrxn - enthalpy change of chemical reaction.
<span>ΔHf - enthalpy of formation of reactants or
products.
</span></span>ΔHrxn = (2·(-393,5) + (-241,8)) - 226,6 · kJ/mol.
ΔHrxn = -1255,4 kJ/mol.
Make proportion: 1 mol (C₂H₂) : -1255,4 kJ = 10,5 mol(C₂H₂) : Q.
Q = 13181,7 kJ.
I think the answer is:
B. Chemical Change.
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
