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serg [7]
3 years ago
9

A decrease in the magnitude of velocity is called

Chemistry
1 answer:
IceJOKER [234]3 years ago
3 0
<span>The answer is deceleration. Acceleration is the general term to refer to the change in velocity. Acceleration = change in velocity / change in time. When you want to highlight the fact that the change in velocity is a decrease in the magnitude, you can use the term deceleration, which means that the acceleration is negative.</span>
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90 POINTSSSS!!!
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Answer:

\large \boxed{\text{-2043.96 kJ/mol}}

Explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn:    C₃H₈(g) +   O₂(g) ⟶   CO₂(g) +   H₂O(g)

Balanced eqn:    C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_{3}$H$_{8}$(g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H$_{2}$O(g)} & -241.82\\\end{array}

(b)Total enthalpies of reactants and products

\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\=  \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}

ΔᵣH° is negative, so the reaction is exothermic.

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