Identify the literary devices used in the poem and explain their effect.

English
Identify the literary devices used in the poem and explain their effect.

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Answer

\textbf{-890 kJ mol^{-1}}

Step 1: The standard enthalpy change of reaction is given by Hess's law using enthalpies of formation.

ΔHr∘xn=∑nΔHf∘(products)−∑nΔHf∘(reactants)\Delta H^\circ_rxn = \sum n \Delta H^\circ_f (products) - \sum n \Delta H^\circ_f (reactants)

where nn is the stoichiometric coefficient.

Step 2: Identify enthalpies of formation with units (all in kJ mol−1^{-1} at 298 K):

ΔHf∘(\ceCO2(g))=−393.5\Delta H^\circ_f (\ce{CO2(g)}) = -393.5

ΔHf∘(\ceH2O(l))=−285.8\Delta H^\circ_f (\ce{H2O(l)}) = -285.8

ΔHf∘(\ceCH4(g))=−74.8\Delta H^\circ_f (\ce{CH4(g)}) = -74.8

ΔHf∘(\ceO2(g))=0\Delta H^\circ_f (\ce{O2(g)}) = 0

Step 3: Calculate sum for products.

Products: \ce1⋅CO2(g)+2⋅H2O(l)\ce{1 \cdot CO2(g) + 2 \cdot H2O(l)}

∑nΔHf∘(products)=[1×(−393.5)]+[2×(−285.8)] kJ mol−1\sum n \Delta H^\circ_f (products) = [1 \times (-393.5)] + [2 \times (-285.8)] \, kJ \, mol^{-1}

=−393.5+2×(−285.8) kJ mol−1= -393.5 + 2 \times (-285.8) \, kJ \, mol^{-1}

=−393.5−571.6 kJ mol−1= -393.5 - 571.6 \, kJ \, mol^{-1}

=−965.1 kJ mol−1= -965.1 \, kJ \, mol^{-1}

Step 4: Calculate sum for reactants.

Reactants: \ce1⋅CH4(g)+2⋅O2(g)\ce{1 \cdot CH4(g) + 2 \cdot O2(g)}

∑nΔHf∘(reactants)=[1×(−74.8)]+[2×0] kJ mol−1\sum n \Delta H^\circ_f (reactants) = [1 \times (-74.8)] + [2 \times 0] \, kJ \, mol^{-1}

=−74.8+0 kJ mol−1= -74.8 + 0 \, kJ \, mol^{-1}

=−74.8 kJ mol−1= -74.8 \, kJ \, mol^{-1}

Step 5: Subtract sums.

ΔHr∘xn=[−965.1]−[−74.8] kJ mol−1\Delta H^\circ_rxn = [-965.1] - [-74.8] \, kJ \, mol^{-1}

=−965.1+74.8 kJ mol−1= -965.1 + 74.8 \, kJ \, mol^{-1}

=−890.3 kJ mol−1= -890.3 \, kJ \, mol^{-1}

\textbf{-890 kJ mol^{-1}}

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The standard enthalpy change of reaction is given by Hess's law using enthalpies of formation.

Identify the literary devices used in the poem and explain their effect.
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This English question involves literary analysis, grammar, or writing skills. The detailed response below provides a well-structured answer with supporting evidence and clear explanations.

ScanSolve AI Answer
Step 1: The standard enthalpy change of reaction is given by Hess's law using enthalpies of formation. H^_rxn = n H^_f (products) - n H^_f (reactants) where n is the stoichiometric coefficient. Step 2: Identify enthalpies of formation with units (all in kJ mol^-1 at 298 K): H^_f (CO2(g)) = -393.5 H^_f (H2O(l)) = -285.8 H^_f (CH4(g)) = -74.8 H^_f (O2(g)) = 0 Step 3: Calculate sum for products. Products: 1 · CO2(g) + 2 · H2O(l) n H^_f (products) = [1 × (-393.5)] + [2 × (-285.8)] \, kJ \, mol^-1 = -393.5 + 2 × (-285.8) \, kJ \, mol^-1 = -393.5 - 571.6 \, kJ \, mol^-1 = -965.1 \, kJ \, mol^-1 Step 4: Calculate sum for reactants. Reactants: 1 · CH4(g) + 2 · O2(g) n H^_f (reactants) = [1 × (-74.8)] + [2 × 0] \, kJ \, mol^-1 = -74.8 + 0 \, kJ \, mol^-1 = -74.8 \, kJ \, mol^-1 Step 5: Subtract sums. H^_rxn = [-965.1] - [-74.8] \, kJ \, mol^-1 = -965.1 + 74.8 \, kJ \, mol^-1 = -890.3 \, kJ \, mol^-1 -890 kJ mol^-1