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GrogVix [38]
3 years ago
13

A piston in a heat engine does 500 joules of work, and 1400 joules of heat are added to the system. Determine the change in inte

rnal energy and explain how this example demonstrates the conservation of energy.
Chemistry
1 answer:
san4es73 [151]3 years ago
8 0

Answer : The change in internal energy is, 900 Joules.

Solution : Given,

Heat given to the system = +1400 J

Work done by the system = -500 J

Change in internal energy is equal to the sum of heat energy and work done.

Formula used :

\Delta U=q+w

where,

\Delta U = change in internal energy

q = heat energy

w = work done

As per question, heat is added to the system that means, q is positive and work done by the system that means, w is negative.

Now put all the given values in the above formula, we get

\Delta U=(+1400J)+(-500J)=900J

Therefore, the change in internal energy is 900 J.

The change in internal energy depends on the heat energy and work done. As we will change in the heat energy and work done, then changes  will occur in the internal energy. Hence, the energy is conserved.

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Solution X is a strong base with a pH of 12. Solution X is mixed with solution Y,and the pH of the resulting mixture is 8.Based
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The strong Base with a pH of 12 is reduced by 4 units upon being added with solution Y. If you added a strong acid to the strong base, all ions are present in the solution, yes? So every OH- is neutralised by every H+ for example, meaning the resultant pH should be 7. The resultant pH is only 8 however, so solution Y must be a <em>weak acid </em>only!

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3 years ago
A sample of a pure compound that weighs 60.3 g contains 20.7 g Sb (antimony) and 39.6 g F (fluorine). What is the percent compos
Volgvan

Answer:

The percent composition of fluorine is 65.67%

Explanation:

Percent Composition is a measure of the amount of mass an element occupies in a compound. It is measured in percentage of mass.

That is, the percentage composition is the percentage by mass of each of the elements present in a compound.

The calculation of the percentage composition of an element is made by:

percent composition element A=\frac{total mass of element A}{mass of compound} *100

In this case, the percent composition of fluorine is:

percent composition of fluorine=\frac{39.6 g}{60.3 g} *100

percent composition of fluorine= 65.67%

<u><em>The percent composition of fluorine is 65.67%</em></u>

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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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