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otez555 [7]
3 years ago
15

Here is the balanced chemical equation for the combustion of isopropyl alcohol: 2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(g) Multiply

ing 42 by 2/3, or 0.667, would be most useful for which calculation?
A. Calculating the moles of O2 needed to produce 42 moles of CO2.

B. Calculating the moles of CO2 that will form from 42 moles of C3H7OH.

C. Calculating the moles of O2 that will react completely with 42 moles of C3H7OH.

D. Calculating the moles of CO2 that will form from 42 moles of O2.
Chemistry
2 answers:
Whitepunk [10]3 years ago
5 0
I believe the correct answer from the choices listed above is option A. Multiplying 42 by 2/3, or 0.667, would be most useful for c<span>alculating the moles of O2 needed to produce 42 moles of CO2. Hope this answers the question.</span>
Lelechka [254]3 years ago
3 0
Hello there.

<span>Here is the balanced chemical equation for the combustion of isopropyl alcohol: 2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(g) Multiplying 42 by 2/3, or 0.667, would be most useful for which calculation?

</span><span>A. Calculating the moles of O2 needed to produce 42 moles of CO2.
</span>
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How many moles of NaCl are present in 2.50 L of a 0.070 M solution?
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Volume of NaCl solution 2.5 L .

Molarity of NaCl solution is 0.070 M .

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Let, n be the number of moles.

We know, molarity is given by :

M = \dfrac{n}{V(in\ L)}

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n = M \times V\\\\n = 0.070\times 2.5 \\\\n = 0.175\ moles

Therefore, number of moles of NaCl is 0.175 moles.

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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
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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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