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GuDViN [60]
3 years ago
6

Torrey's neighbor told her that a marble rolling down a hill increases in velocity as it rolls down, but does not increase in ki

netic energy because energy can neither be created nor destroyed, so it always has the same amount kinetic energy is Torrey's neighbor correct ? If they are correct, explain how you know . If they are not correct would you explain to them the fault in their thinking ?
Chemistry
1 answer:
ziro4ka [17]3 years ago
6 0

Answer:

Torrey's neighbour is incorrect because increase in kinetic energy is proportional to velocity.  If the velocity increases so will the object's kinetic energy.  Because the mass is constant, if the velocity increases, so does the kinetic energy.

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The volume of a gas is the same as its CONTAINER.

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3 years ago
What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

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