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2 years ago

A competitive high school swimmer takes 56.7 seconds to swim 100. yards. What is his rate in m/min?

Chemistry

1 answer:

lilavasa [31]2 years ago

7 0

Answer:

96.72 m/min

Explanation:

The following data were obtained from the question:

Time (t) = 56.7 s

Distance (d) = 100 yard

Rate (R) =?

Next, we shall convert 56.7 s to minutes. This can be obtained as follow:

60 s = 1 min

Therefore,

56.7 s = 56.7 s × 1 min /60 s

56.7 s = 0.945 min

Next, we shall convert 100 yard to metre (m). This can be obtained as follow:

1 yard = 0.914 m

Therefore,

100 yard = 100 yard × 0.914 m /1 yard

100 yard = 91.4 m

Finally, we shall determine the rate of the swimmer as follow:

Time (t) = 0.945 min

Distance (d) = 91.4 m

Rate (R) =?

R = d/t

R = 91.4/0.945

R = 96.72 m/min

Thus the rate of the swimmer is 96.72 m/min

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$\Delta_{\text{rxn}}H^{\circ} = \sum \left( \Delta_{\text{f}} H^{\circ} \text{products}\right) - \sum \left (\Delta_{\text{f}}H^{\circ} \text{reactants} \right)$

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$\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_{2}$O} & -241.8\\\text{CO$_{2}$(g)} & -393.5 \\\text{H$_{2}$(g)} & 0 \\\end{array}$

(b) Total enthalpies of reactants and products

$\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_{2}$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_{2}(g) & -393.5&-393.5 \\\text{H}_{2} & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}$

(c) Enthalpy of reaction $\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{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}$

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