<?xml
version="1.0" encoding="utf-8"?>
<rss version="2.0" 
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:atom="http://www.w3.org/2005/Atom"
>

<channel xml:lang="es">
	<title>EjerciciosFyQ</title>
	<link>https://ejercicios-fyq.com/</link>
	<description>Ejercicios Resueltos, Situaciones de aprendizaje y V&#205;DEOS de F&#237;sica y Qu&#237;mica para Secundaria y Bachillerato</description>
	<language>es</language>
	<generator>SPIP - www.spip.net</generator>
	<atom:link href="https://ejercicios-fyq.com/spip.php?id_rubrique=305&amp;page=backend" rel="self" type="application/rss+xml" />

	<image>
		<title>EjerciciosFyQ</title>
		<url>https://ejercicios-fyq.com/local/cache-vignettes/L144xH25/siteon0-da713.png?1758361862</url>
		<link>https://ejercicios-fyq.com/</link>
		<height>25</height>
		<width>144</width>
	</image>



<item xml:lang="es">
		<title>Concentraciones de los componentes de una mezcla de disoluciones (8460)</title>
		<link>https://ejercicios-fyq.com/Concentraciones-de-los-componentes-de-una-mezcla-de-disoluciones-8460</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Concentraciones-de-los-componentes-de-una-mezcla-de-disoluciones-8460</guid>
		<dc:date>2025-05-16T02:37:32Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Concentraci&#243;n</dc:subject>
		<dc:subject>Molaridad</dc:subject>
		<dc:subject>Molalidad</dc:subject>
		<dc:subject>Fracci&#243;n molar</dc:subject>
		<dc:subject>RESUELTO</dc:subject>

		<description>
&lt;p&gt;Se prepara una disoluci&#243;n mezclando 50.0 g de sulfato de cobre(II) pentahidratado () con 200 mL de una disoluci&#243;n acuosa de 1.50 M, cuya densidad es 1.12 g/mL. Posteriormente, se diluye la mezcla hasta un volumen final de 500 mL, obteniendo una disoluci&#243;n con una densidad de 1.18 g/mL. Calcula: &lt;br class='autobr' /&gt;
a) La molaridad de en la disoluci&#243;n final. &lt;br class='autobr' /&gt;
b) La molalidad de en la disoluci&#243;n final. &lt;br class='autobr' /&gt;
c) El porcentaje en masa de en la disoluci&#243;n final. &lt;br class='autobr' /&gt;
d) La fracci&#243;n molar de agua en la disoluci&#243;n (&#8230;)&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Disoluciones-305" rel="directory"&gt;Disoluciones&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/Concentracion" rel="tag"&gt;Concentraci&#243;n&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Molaridad" rel="tag"&gt;Molaridad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Molalidad" rel="tag"&gt;Molalidad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Fraccion-molar" rel="tag"&gt;Fracci&#243;n molar&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Se prepara una disoluci&#243;n mezclando 50.0 g de sulfato de cobre(II) pentahidratado (&lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L131xH20/207974d82ce07f0864cef87e2dc9ba8b-93ff8.png?1747363085' style='vertical-align:middle;' width='131' height='20' alt=&#034;\ce{CuSO4*5H2O}&#034; title=&#034;\ce{CuSO4*5H2O}&#034; /&gt;) con 200 mL de una disoluci&#243;n acuosa de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L55xH17/7c9ea6c0dca607dbdaa3a969e7564268-ff422.png?1732966467' style='vertical-align:middle;' width='55' height='17' alt=&#034;\ce{H2SO4}&#034; title=&#034;\ce{H2SO4}&#034; /&gt; 1.50 M, cuya densidad es 1.12 g/mL. Posteriormente, se diluye la mezcla hasta un volumen final de 500 mL, obteniendo una disoluci&#243;n con una densidad de 1.18 g/mL. Calcula:&lt;/p&gt;
&lt;p&gt;a) La molaridad de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L47xH15/e5a2fb2832aaadb531f0a90bf4b164bb-8d66e.png?1732971087' style='vertical-align:middle;' width='47' height='15' alt=&#034;\ce{CuSO4}&#034; title=&#034;\ce{CuSO4}&#034; /&gt; en la disoluci&#243;n final.&lt;/p&gt;
&lt;p&gt;b) La molalidad de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L55xH17/7c9ea6c0dca607dbdaa3a969e7564268-ff422.png?1732966467' style='vertical-align:middle;' width='55' height='17' alt=&#034;\ce{H2SO4}&#034; title=&#034;\ce{H2SO4}&#034; /&gt; en la disoluci&#243;n final.&lt;/p&gt;
&lt;p&gt;c) El porcentaje en masa de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L47xH15/e5a2fb2832aaadb531f0a90bf4b164bb-8d66e.png?1732971087' style='vertical-align:middle;' width='47' height='15' alt=&#034;\ce{CuSO4}&#034; title=&#034;\ce{CuSO4}&#034; /&gt; en la disoluci&#243;n final.&lt;/p&gt;
&lt;p&gt;d) La fracci&#243;n molar de agua en la disoluci&#243;n final.&lt;/p&gt;
&lt;p&gt;Datos: Cu = 63.55, S = 32.07, O = 16.00, H = 1.01. Considera que el &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L55xH17/7c9ea6c0dca607dbdaa3a969e7564268-ff422.png?1732966467' style='vertical-align:middle;' width='55' height='17' alt=&#034;\ce{H2SO4}&#034; title=&#034;\ce{H2SO4}&#034; /&gt; se disocia completamente en sus iones.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) Para determinar la molaridad de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e5a2fb2832aaadb531f0a90bf4b164bb.png' style=&#034;vertical-align:middle;&#034; width=&#034;47&#034; height=&#034;15&#034; alt=&#034;\ce{CuSO4}&#034; title=&#034;\ce{CuSO4}&#034; /&gt; en la disoluci&#243;n final debes calcular los moles de la sal pentahidratada que has usado. La masa molecular de la sal es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/ec35b6733748ccc8955461b1051041eb.png' style=&#034;vertical-align:middle;&#034; width=&#034;771&#034; height=&#034;25&#034; alt=&#034;M_{\ce{CuSO4*5H2O}} = 1\cdot 63.55 + 1\cdot 32.07 + 4\cdot 16 + 5\cdot (2\cdot 1.01 + 16) = \color[RGB]{0,112,192}{\bm{249.72\ g\cdot mol^{-1}}}&#034; title=&#034;M_{\ce{CuSO4*5H2O}} = 1\cdot 63.55 + 1\cdot 32.07 + 4\cdot 16 + 5\cdot (2\cdot 1.01 + 16) = \color[RGB]{0,112,192}{\bm{249.72\ g\cdot mol^{-1}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Teniendo en cuenta que cada mol de la sal aporta un mol de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e5a2fb2832aaadb531f0a90bf4b164bb.png' style=&#034;vertical-align:middle;&#034; width=&#034;47&#034; height=&#034;15&#034; alt=&#034;\ce{CuSO4}&#034; title=&#034;\ce{CuSO4}&#034; /&gt;, habr&#225;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e1539d573e85c7d0135e5a3e1a20e678.png' style=&#034;vertical-align:middle;&#034; width=&#034;691&#034; height=&#034;64&#034; alt=&#034;50\ \cancel{g}\ \cdot \frac{1\ \cancel{\ce{mol\ CuSO4*5H2O}}}{249.72\ \cancel{g}}\cdot \frac{1\ \ce{mol CuSO4}}{1\ \cancel{\ce{mol\ CuSO4*5H2O}}} = \fbox{\color[RGB]{0,112,192}{\textbf{0.200 mol \ce{CuSO4}}}}&#034; title=&#034;50\ \cancel{g}\ \cdot \frac{1\ \cancel{\ce{mol\ CuSO4*5H2O}}}{249.72\ \cancel{g}}\cdot \frac{1\ \ce{mol CuSO4}}{1\ \cancel{\ce{mol\ CuSO4*5H2O}}} = \fbox{\color[RGB]{0,112,192}{\textbf{0.200 mol \ce{CuSO4}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La molaridad es el cociente entre los moles calculados y el volumen final de la disoluci&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d9131d0a5f71a1cef951572c56f37042.png' style=&#034;vertical-align:middle;&#034; width=&#034;420&#034; height=&#034;50&#034; alt=&#034;M_{\ce{CuSO4}} = {\color[RGB]{2,112,20}{\bm{\frac{n_{\ce{CuSO4}}}{V_F}}}} = \frac{0.200\ \text{mol}}{0.500\ L} = \fbox{\color[RGB]{192,0,0}{\bf 0.400\ M}}&#034; title=&#034;M_{\ce{CuSO4}} = {\color[RGB]{2,112,20}{\bm{\frac{n_{\ce{CuSO4}}}{V_F}}}} = \frac{0.200\ \text{mol}}{0.500\ L} = \fbox{\color[RGB]{192,0,0}{\bf 0.400\ M}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) Los moles de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7c9ea6c0dca607dbdaa3a969e7564268.png' style=&#034;vertical-align:middle;&#034; width=&#034;55&#034; height=&#034;17&#034; alt=&#034;\ce{H2SO4}&#034; title=&#034;\ce{H2SO4}&#034; /&gt; aportados en la disoluci&#243;n de &#225;cido los tienes que calcular a partir del volumen de esa disoluci&#243;n y su molaridad: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/420490177e88e13770f653eee5df6344.png' style=&#034;vertical-align:middle;&#034; width=&#034;470&#034; height=&#034;48&#034; alt=&#034;n_{\ce{H2SO4}} = M\cdot V = 1.50\ \frac{\text{mol}}{\cancel{L}}\cdot 0.200\ \cancel{L} = \color[RGB]{0,112,192}{\bf 0.300\ mol}&#034; title=&#034;n_{\ce{H2SO4}} = M\cdot V = 1.50\ \frac{\text{mol}}{\cancel{L}}\cdot 0.200\ \cancel{L} = \color[RGB]{0,112,192}{\bf 0.300\ mol}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Como la molalidad viene dada en funci&#243;n de la masa de disolvente, debes saber qu&#233; masa de agua est&#225; contenida en la disoluci&#243;n de &#225;cido. Primero calculas la masa de la disoluci&#243;n de &#225;cido: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/622467b3fcc4ebb0ed9cd09475202ff9.png' style=&#034;vertical-align:middle;&#034; width=&#034;393&#034; height=&#034;42&#034; alt=&#034;m_D = \rho_D\cdot V = 1.12\ \frac{g}{\cancel{{mL}}}\cdot 200\ \cancel{mL} = 224\ g&#034; title=&#034;m_D = \rho_D\cdot V = 1.12\ \frac{g}{\cancel{{mL}}}\cdot 200\ \cancel{mL} = 224\ g&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La masa de agua en la disoluci&#243;n es la diferencia entre la masa total y la masa de &#225;cido: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9999f821c9b92f7af77c91549b019651.png' style=&#034;vertical-align:middle;&#034; width=&#034;579&#034; height=&#034;52&#034; alt=&#034;m_{\ce{H2O}} = 224\ g - \left(0.3\ \cancel{\text{mol}}\cdot \frac{98.09\ g\ \ce{H2SO_4}}{1\ \cancel{\text{mol}}}\right) = \color[RGB]{0,112,192}{\textbf{194.57 g \ce{H2O}}}&#034; title=&#034;m_{\ce{H2O}} = 224\ g - \left(0.3\ \cancel{\text{mol}}\cdot \frac{98.09\ g\ \ce{H2SO_4}}{1\ \cancel{\text{mol}}}\right) = \color[RGB]{0,112,192}{\textbf{194.57 g \ce{H2O}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La sal pentahidratada tambi&#233;n aporta agua a la disoluci&#243;n final, por lo que tienes que determinar la masa de agua contenida en la sal. Por cada mol de sal, se incorporan 5 moles de agua: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d739e3b25803963d12a45eb18063b666.png' style=&#034;vertical-align:middle;&#034; width=&#034;708&#034; height=&#034;59&#034; alt=&#034;0.2\ \cancel{\ce{mol\ CuSO4*5H2O}}\cdot \frac{5\ \cancel{\ce{mol H2O}}}{1\ \cancel{\ce{mol\ CuSO4*5H2O}}}\cdot \frac{18.02\ g}{1\ \cancel{\ce{mol\ H2O}}} = \fbox{\color[RGB]{0,112,192}{\textbf{18.02 g \ce{H2O}}}}&#034; title=&#034;0.2\ \cancel{\ce{mol\ CuSO4*5H2O}}\cdot \frac{5\ \cancel{\ce{mol H2O}}}{1\ \cancel{\ce{mol\ CuSO4*5H2O}}}\cdot \frac{18.02\ g}{1\ \cancel{\ce{mol\ H2O}}} = \fbox{\color[RGB]{0,112,192}{\textbf{18.02 g \ce{H2O}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La masa total de agua, tras la mezcla de las dos disoluciones, es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c9f88237d9387f189c88ca4ea8fbdbc1.png' style=&#034;vertical-align:middle;&#034; width=&#034;525&#034; height=&#034;51&#034; alt=&#034;m_T = (194.57 + 18.02)\ g = 212.59\ \cancel{g}\cdot \frac{1\ kg}{10^3\ \cancel{g}} = \color[RGB]{0,112,192}{\bf 0.2126\ kg}&#034; title=&#034;m_T = (194.57 + 18.02)\ g = 212.59\ \cancel{g}\cdot \frac{1\ kg}{10^3\ \cancel{g}} = \color[RGB]{0,112,192}{\bf 0.2126\ kg}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La molalidad del &#225;cido es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d16d07ec39156566cdd614796552ffff.png' style=&#034;vertical-align:middle;&#034; width=&#034;402&#034; height=&#034;51&#034; alt=&#034;m_{\ce{H2SO4}} = {\color[RGB]{2,112,20}{\bm{\frac{n_{\ce{H2SO4}}}{m_{\ce{H2O}}}}}} = \frac{0.3\ \text{mol}}{0.2126\ kg} = \fbox{\color[RGB]{192,0,0}{\bf 1.41\ m}}&#034; title=&#034;m_{\ce{H2SO4}} = {\color[RGB]{2,112,20}{\bm{\frac{n_{\ce{H2SO4}}}{m_{\ce{H2O}}}}}} = \frac{0.3\ \text{mol}}{0.2126\ kg} = \fbox{\color[RGB]{192,0,0}{\bf 1.41\ m}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; c) Necesitas conocer la masa de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/e5a2fb2832aaadb531f0a90bf4b164bb.png' style=&#034;vertical-align:middle;&#034; width=&#034;47&#034; height=&#034;15&#034; alt=&#034;\ce{CuSO4}&#034; title=&#034;\ce{CuSO4}&#034; /&gt; y la masa total de la disoluci&#243;n final: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a34a27aed83e176f0a3aca0f156fa9c0.png' style=&#034;vertical-align:middle;&#034; width=&#034;599&#034; height=&#034;87&#034; alt=&#034;\left m_{\ce{CuSO4}} = n_{\ce{CuSO4}}\cdot M_{\ce{CuSO4}} = 0.200\ \cancel{\text{mol}}\cdot 159.62\ \dfrac{g}{\cancel{\text{mol}}} = {\color[RGB]{0,112,192}{\bf 31.92\ g}} \atop m_D = V_D\cdot \rho_D = 500\ \cancel{mL}\cdot 1.18\ \dfrac{g}{\cancel{mL}} = {\color[RGB]{0,112,192}{\bf 590\ g}} \right \}&#034; title=&#034;\left m_{\ce{CuSO4}} = n_{\ce{CuSO4}}\cdot M_{\ce{CuSO4}} = 0.200\ \cancel{\text{mol}}\cdot 159.62\ \dfrac{g}{\cancel{\text{mol}}} = {\color[RGB]{0,112,192}{\bf 31.92\ g}} \atop m_D = V_D\cdot \rho_D = 500\ \cancel{mL}\cdot 1.18\ \dfrac{g}{\cancel{mL}} = {\color[RGB]{0,112,192}{\bf 590\ g}} \right \}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El porcentaje en masa es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0f24f385026c5e6c9b90abb1c0aaf274.png' style=&#034;vertical-align:middle;&#034; width=&#034;313&#034; height=&#034;51&#034; alt=&#034;\%\ (m) = \frac{31.92\ \cancel{g}}{590\ \cancel{g}}\cdot 100 = \fbox{\color[RGB]{192,0,0}{\bf 5.41\ \%}}&#034; title=&#034;\%\ (m) = \frac{31.92\ \cancel{g}}{590\ \cancel{g}}\cdot 100 = \fbox{\color[RGB]{192,0,0}{\bf 5.41\ \%}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; d) La fracci&#243;n molar es el cociente de los moles del componente, el agua en este caso, y los moles totales. Los moles agua son: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/3f440ababfc15da958e686912ed7f104.png' style=&#034;vertical-align:middle;&#034; width=&#034;382&#034; height=&#034;55&#034; alt=&#034;n_{\ce{H2O}} = {\color[RGB]{2,112,20}{\bm{\frac{m_{\ce{H2O}T}}{M_{\ce{H2O}}}}}} = \frac{212.59\ \cancel{g}}{18.02\ \frac{\cancel{g}}{\text{mol}}} = \color[RGB]{0,112,192}{\bf 11.8\ mol}}&#034; title=&#034;n_{\ce{H2O}} = {\color[RGB]{2,112,20}{\bm{\frac{m_{\ce{H2O}T}}{M_{\ce{H2O}}}}}} = \frac{212.59\ \cancel{g}}{18.02\ \frac{\cancel{g}}{\text{mol}}} = \color[RGB]{0,112,192}{\bf 11.8\ mol}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Como conoces los moles del soluto, de las dos sustancias que no son el agua, los moles totales ser&#225;n la suma de todos ellos: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/83adf66ddff339e4d73db8b38e289c58.png' style=&#034;vertical-align:middle;&#034; width=&#034;635&#034; height=&#034;53&#034; alt=&#034;x_{\ce{H2O}} = \frac{n_{\ce{H2O}}}{n_{\ce{H2O}} + n_{\ce{CuSO4}} + n_{\ce{H2SO4}}} = \frac{11.80\ \cancel{\text{mol}}}{(11.80 + 0.2 + 0.3)\ \cancel{\text{mol}}} = \fbox{\color[RGB]{192,0,0}{\bf 0.959}}&#034; title=&#034;x_{\ce{H2O}} = \frac{n_{\ce{H2O}}}{n_{\ce{H2O}} + n_{\ce{CuSO4}} + n_{\ce{H2SO4}}} = \frac{11.80\ \cancel{\text{mol}}}{(11.80 + 0.2 + 0.3)\ \cancel{\text{mol}}} = \fbox{\color[RGB]{192,0,0}{\bf 0.959}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Coeficientes de actividad de los iones en una disoluci&#243;n (8231)</title>
		<link>https://ejercicios-fyq.com/Coeficientes-de-actividad-de-los-iones-en-una-disolucion-8231</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Coeficientes-de-actividad-de-los-iones-en-una-disolucion-8231</guid>
		<dc:date>2024-06-19T03:26:25Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Coeficiente de actividad</dc:subject>
		<dc:subject>Actividad</dc:subject>
		<dc:subject>Ecuaci&#243;n Debye-H&#252;ckel</dc:subject>

		<description>
&lt;p&gt;Calcula los coeficientes de actividad de cada uno de los iones presentes en una disoluci&#243;n acuosa que es con respecto al , con respecto al y con respecto al . &lt;br class='autobr' /&gt;
Datos: ; ; ; ; B = 0.328.&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Disoluciones-305" rel="directory"&gt;Disoluciones&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Coeficiente-de-actividad" rel="tag"&gt;Coeficiente de actividad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Actividad" rel="tag"&gt;Actividad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Ecuacion-Debye-Huckel" rel="tag"&gt;Ecuaci&#243;n Debye-H&#252;ckel&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Calcula los coeficientes de actividad de cada uno de los iones presentes en una disoluci&#243;n acuosa que es &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L96xH20/9f8429a64bf536280d0b3de5fdd192ce-86bd3.png?1733033420' style='vertical-align:middle;' width='96' height='20' alt=&#034;2\cdot 10^{-2}\ M&#034; title=&#034;2\cdot 10^{-2}\ M&#034; /&gt; con respecto al &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L70xH20/4d0a18b154a9e22b4e3d12c18485e46c-b8e72.png?1733000569' style='vertical-align:middle;' width='70' height='20' alt=&#034;\ce{Na_2SO4}&#034; title=&#034;\ce{Na_2SO4}&#034; /&gt;, &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L72xH47/5d14bc86595815fe29079d21aad668f9-6805c.png?1732971709' style='vertical-align:middle;' width='72' height='47' alt=&#034;10^{-3}\ M&#034; title=&#034;10^{-3}\ M&#034; /&gt; con respecto al &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L61xH20/97aeba6ec8ef600bc49ea740248a152e-a2208.png?1732953898' style='vertical-align:middle;' width='61' height='20' alt=&#034;\ce{CaSO4}&#034; title=&#034;\ce{CaSO4}&#034; /&gt; y &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L96xH20/2c6e7e65db538c4af2f7df1555b01345-aa525.png?1733033420' style='vertical-align:middle;' width='96' height='20' alt=&#034;3\cdot 10^{-2}\ M&#034; title=&#034;3\cdot 10^{-2}\ M&#034; /&gt; con respecto al &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L90xH23/a02fdf4cffa093934b0df69c61f80a03-886ed.png?1733033420' style='vertical-align:middle;' width='90' height='23' alt=&#034;\ce{Al_2(SO4)3}&#034; title=&#034;\ce{Al_2(SO4)3}&#034; /&gt;.&lt;/p&gt;
&lt;p&gt;Datos: &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L120xH28/c6f8fb400f06149f42daa25dad2a05a8-ce804.png?1733033420' style='vertical-align:middle;' width='120' height='28' alt=&#034;a_{\ce{SO4^{2-}}} = 4\ \mathring{A}&#034; title=&#034;a_{\ce{SO4^{2-}}} = 4\ \mathring{A}&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L110xH27/5b680143177013efa52f6ce2214c5311-0d8a2.png?1733033420' style='vertical-align:middle;' width='110' height='27' alt=&#034;a_{\ce{Ca^{2+}}} = 6\ \mathring{A}&#034; title=&#034;a_{\ce{Ca^{2+}}} = 6\ \mathring{A}&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L108xH26/6dc6358d52d839d4205e618147691f91-929c5.png?1733033420' style='vertical-align:middle;' width='108' height='26' alt=&#034;a_{\ce{Al^{3+}}} = 9\ \mathring{A}&#034; title=&#034;a_{\ce{Al^{3+}}} = 9\ \mathring{A}&#034; /&gt; ; &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L104xH27/70b59db969392c2520c8be079373c2c7-2442a.png?1733033420' style='vertical-align:middle;' width='104' height='27' alt=&#034;a_{\ce{Na^+}} = 4\ \mathring{A}&#034; title=&#034;a_{\ce{Na^+}} = 4\ \mathring{A}&#034; /&gt; ; B = 0.328.&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;Todas las sales son electrolitos fuertes y est&#225;n completamente disociadas en la disoluci&#243;n acuosa. Las concentraciones de cada uno de los iones, teniendo en cuenta la estequiometr&#237;a, son: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/1dd9c12899664241a1d6c93acb190b06.png' style=&#034;vertical-align:middle;&#034; width=&#034;340&#034; height=&#034;25&#034; alt=&#034;[\ce{Na^+}] = 2\cdot 2\cdot 10^{-2}\ M = \color[RGB]{0,112,192}{\bm{4\cdot 10^{-2}\ M}}&#034; title=&#034;[\ce{Na^+}] = 2\cdot 2\cdot 10^{-2}\ M = \color[RGB]{0,112,192}{\bm{4\cdot 10^{-2}\ M}}&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/b9625c9c44d96901f5806008d00edbfd.png' style=&#034;vertical-align:middle;&#034; width=&#034;166&#034; height=&#034;25&#034; alt=&#034;[\ce{Ca^{2+}}] = \color[RGB]{0,112,192}{\bm{10^{-3}\ M}}&#034; title=&#034;[\ce{Ca^{2+}}] = \color[RGB]{0,112,192}{\bm{10^{-3}\ M}}&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6721df9a991e32b4ab1f543e78c746e8.png' style=&#034;vertical-align:middle;&#034; width=&#034;342&#034; height=&#034;25&#034; alt=&#034;[\ce{Al^{3+}}] = 2\cdot 3\cdot 10^{-2}\ M = \color[RGB]{0,112,192}{\bm{6\cdot 10^{-2}\ M}}&#034; title=&#034;[\ce{Al^{3+}}] = 2\cdot 3\cdot 10^{-2}\ M = \color[RGB]{0,112,192}{\bm{6\cdot 10^{-2}\ M}}&#034; /&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/f9f9ac08bb5fcc151460ab0dd55d7e84.png' style=&#034;vertical-align:middle;&#034; width=&#034;509&#034; height=&#034;25&#034; alt=&#034;[\ce{SO4^{2+}}] = (2\cdot 10^{-2} + 10^{-3} + 3\cdot 3\cdot 10^{-2})\ M = \color[RGB]{0,112,192}{\bf 0.111\ M}&#034; title=&#034;[\ce{SO4^{2+}}] = (2\cdot 10^{-2} + 10^{-3} + 3\cdot 3\cdot 10^{-2})\ M = \color[RGB]{0,112,192}{\bf 0.111\ M}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Determinas la fuerza i&#243;nica de la disoluci&#243;n para saber si puedes aplicar la ley l&#237;mite o no: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/cc99067ed30e7bb746f0255b4b1bb9e7.png' style=&#034;vertical-align:middle;&#034; width=&#034;184&#034; height=&#034;60&#034; alt=&#034;{\color[RGB]{2,112,20}{\bm{\mu = \frac{1}{2}\sum_{i=1}^{n} = c_i\cdot z_i^2}}}&#034; title=&#034;{\color[RGB]{2,112,20}{\bm{\mu = \frac{1}{2}\sum_{i=1}^{n} = c_i\cdot z_i^2}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Sustituyes: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/df288013270f97c216578aa9a4d39d38.png' style=&#034;vertical-align:middle;&#034; width=&#034;677&#034; height=&#034;44&#034; alt=&#034;\mu = \frac{1}{2}\left[(0.04\cdot 1^2) + (0.001\cdot 2^2) + (0.06\cdot 3^2) + (0.111\cdot 2^2) \right]\ M = \color[RGB]{0,112,192}{\bf 0.514\ M}&#034; title=&#034;\mu = \frac{1}{2}\left[(0.04\cdot 1^2) + (0.001\cdot 2^2) + (0.06\cdot 3^2) + (0.111\cdot 2^2) \right]\ M = \color[RGB]{0,112,192}{\bf 0.514\ M}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; Este valor te indica que no puedes usar la ley l&#237;mite de Debye-H&#252;ckel porque est&#225; muy por encima de los valores indicados para iones mono, di y polivalentes. Debes emplear la ecuaci&#243;n ampliada de Debye-H&#252;ckel para cada ion. Recuerda que el valor de A es 0.512: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6b6326675d08213a325f96a85c2573d0.png' style=&#034;vertical-align:middle;&#034; width=&#034;265&#034; height=&#034;58&#034; alt=&#034;\color[RGB]{2,112,20}{\bm{-log\ \gamma_i = \frac{A\cdot z_i^2\cdot \sqrt{\mu}}{1 + B\cdot a_i\cdot \sqrt{\mu}}}}&#034; title=&#034;\color[RGB]{2,112,20}{\bm{-log\ \gamma_i = \frac{A\cdot z_i^2\cdot \sqrt{\mu}}{1 + B\cdot a_i\cdot \sqrt{\mu}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; &lt;u&gt;Para el cati&#243;n sodio&lt;/u&gt;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/acd174570542f94c65f270d44a08cbf4.png' style=&#034;vertical-align:middle;&#034; width=&#034;400&#034; height=&#034;54&#034; alt=&#034;-log\ \gamma_{\ce{Na^+}} = \frac{0.512\cdot 1^2\cdot \sqrt{0.514}}{1+ 0.328\cdot 4\cdot \sqrt{0.514}} = \color[RGB]{0,112,192}{\bf 0.189}&#034; title=&#034;-log\ \gamma_{\ce{Na^+}} = \frac{0.512\cdot 1^2\cdot \sqrt{0.514}}{1+ 0.328\cdot 4\cdot \sqrt{0.514}} = \color[RGB]{0,112,192}{\bf 0.189}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El coeficiente de actividad es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/1921ba78847c50025d408af1acfd578e.png' style=&#034;vertical-align:middle;&#034; width=&#034;234&#034; height=&#034;26&#034; alt=&#034;\gamma_{\ce{Na^+}} = 10^{(-0.189)} = \color[RGB]{0,112,192}{\bf 0.647}&#034; title=&#034;\gamma_{\ce{Na^+}} = 10^{(-0.189)} = \color[RGB]{0,112,192}{\bf 0.647}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La actividad del cati&#243;n sodio es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c8a380256a0dbb52eee044b74761fafe.png' style=&#034;vertical-align:middle;&#034; width=&#034;480&#034; height=&#034;30&#034; alt=&#034;a_{\ce{Na^+}} = \gamma_{\ce{Na^+}}\cdot c = 0.647\cdot 0.04\ M = \fbox{\color[RGB]{192,0,0}{\bm{2.59\cdot 10^{-2}\ M}}}&#034; title=&#034;a_{\ce{Na^+}} = \gamma_{\ce{Na^+}}\cdot c = 0.647\cdot 0.04\ M = \fbox{\color[RGB]{192,0,0}{\bm{2.59\cdot 10^{-2}\ M}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;u&gt;Para el cati&#243;n calcio&lt;/u&gt;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/54630bc28b7fe4d002cde1ab02e34aa2.png' style=&#034;vertical-align:middle;&#034; width=&#034;406&#034; height=&#034;54&#034; alt=&#034;-log\ \gamma_{\ce{Ca^{2+}}} = \frac{0.512\cdot 2^2\cdot \sqrt{0.514}}{1+ 0.328\cdot 6\cdot \sqrt{0.514}} = \color[RGB]{0,112,192}{\bf 0.609}&#034; title=&#034;-log\ \gamma_{\ce{Ca^{2+}}} = \frac{0.512\cdot 2^2\cdot \sqrt{0.514}}{1+ 0.328\cdot 6\cdot \sqrt{0.514}} = \color[RGB]{0,112,192}{\bf 0.609}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El coeficiente de actividad es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5c4f6d09aa9b82831578814ce08df25b.png' style=&#034;vertical-align:middle;&#034; width=&#034;239&#034; height=&#034;26&#034; alt=&#034;\gamma_{\ce{Ca^{2+}}} = 10^{(-0.609)} = \color[RGB]{0,112,192}{\bf 0.246}&#034; title=&#034;\gamma_{\ce{Ca^{2+}}} = 10^{(-0.609)} = \color[RGB]{0,112,192}{\bf 0.246}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La actividad del cati&#243;n calcio es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a078d1f4910eda217cae2d97c9c4aea6.png' style=&#034;vertical-align:middle;&#034; width=&#034;497&#034; height=&#034;30&#034; alt=&#034;a_{\ce{Ca^{2+}}} = \gamma_{\ce{Ca^{2+}}}\cdot c = 0.246\cdot 10^{-3}\ M = \fbox{\color[RGB]{192,0,0}{\bm{2.46\cdot 10^{-4}\ M}}}&#034; title=&#034;a_{\ce{Ca^{2+}}} = \gamma_{\ce{Ca^{2+}}}\cdot c = 0.246\cdot 10^{-3}\ M = \fbox{\color[RGB]{192,0,0}{\bm{2.46\cdot 10^{-4}\ M}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;u&gt;Para el cati&#243;n aluminio&lt;/u&gt;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/410b3d5128b940bf8f836d9dd3ad47af.png' style=&#034;vertical-align:middle;&#034; width=&#034;390&#034; height=&#034;54&#034; alt=&#034;-log\ \gamma_{\ce{Al^{3+}}} = \frac{0.512\cdot 3^2\cdot \sqrt{0.514}}{1+ 0.328\cdot 9\cdot \sqrt{0.514}} = \color[RGB]{0,112,192}{\bf 1.06}&#034; title=&#034;-log\ \gamma_{\ce{Al^{3+}}} = \frac{0.512\cdot 3^2\cdot \sqrt{0.514}}{1+ 0.328\cdot 9\cdot \sqrt{0.514}} = \color[RGB]{0,112,192}{\bf 1.06}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El coeficiente de actividad es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/96f47113cf48cda9be3aca8add826cfb.png' style=&#034;vertical-align:middle;&#034; width=&#034;222&#034; height=&#034;25&#034; alt=&#034;\gamma_{\ce{Al^{3+}}} = 10^{(-1.6)} = \color[RGB]{0,112,192}{\bf 0.087}&#034; title=&#034;\gamma_{\ce{Al^{3+}}} = 10^{(-1.6)} = \color[RGB]{0,112,192}{\bf 0.087}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La actividad del cati&#243;n aluminio es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/3b3ee6f6c9c1a88c1c3905cf7dca4d39.png' style=&#034;vertical-align:middle;&#034; width=&#034;487&#034; height=&#034;30&#034; alt=&#034;a_{\ce{Al^{3+}}} = \gamma_{\ce{Al^{3+}}}\cdot c = 0.087\cdot 0.06\ M = \fbox{\color[RGB]{192,0,0}{\bm{5.22\cdot 10^{-3}\ M}}}&#034; title=&#034;a_{\ce{Al^{3+}}} = \gamma_{\ce{Al^{3+}}}\cdot c = 0.087\cdot 0.06\ M = \fbox{\color[RGB]{192,0,0}{\bm{5.22\cdot 10^{-3}\ M}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;u&gt;Para el ani&#243;n sulfato&lt;/u&gt;: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/d257d3837b7e2e9d9b82797fd7a54616.png' style=&#034;vertical-align:middle;&#034; width=&#034;417&#034; height=&#034;54&#034; alt=&#034;-log\ \gamma_{\ce{SO4^{2-}}} = \frac{0.512\cdot 2^1\cdot \sqrt{0.514}}{1+ 0.328\cdot 4\cdot \sqrt{0.514}} = \color[RGB]{0,112,192}{\bf 0.757}&#034; title=&#034;-log\ \gamma_{\ce{SO4^{2-}}} = \frac{0.512\cdot 2^1\cdot \sqrt{0.514}}{1+ 0.328\cdot 4\cdot \sqrt{0.514}} = \color[RGB]{0,112,192}{\bf 0.757}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El coeficiente de actividad es: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/72cab4855b7537f486a83c2085dcd828.png' style=&#034;vertical-align:middle;&#034; width=&#034;249&#034; height=&#034;27&#034; alt=&#034;\gamma_{\ce{SO4^{2-}}} = 10^{(-0.757)} = \color[RGB]{0,112,192}{\bf 0.175}&#034; title=&#034;\gamma_{\ce{SO4^{2-}}} = 10^{(-0.757)} = \color[RGB]{0,112,192}{\bf 0.175}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; La actividad del ani&#243;n sulfato es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a3e30188e590c9171cf35bae4c188ffd.png' style=&#034;vertical-align:middle;&#034; width=&#034;522&#034; height=&#034;31&#034; alt=&#034;a_{\ce{SO4^{2-}}} = \gamma_{\ce{SO4^{2-}}}\cdot c = 0.175\cdot 0.111\ M = \fbox{\color[RGB]{192,0,0}{\bm{1.94\cdot 10^{-2}\ M}}}&#034; title=&#034;a_{\ce{SO4^{2-}}} = \gamma_{\ce{SO4^{2-}}}\cdot c = 0.175\cdot 0.111\ M = \fbox{\color[RGB]{192,0,0}{\bm{1.94\cdot 10^{-2}\ M}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Concentraci&#243;n de cati&#243;n calcio en una muestra de suero sangu&#237;neo (7786)</title>
		<link>https://ejercicios-fyq.com/Concentracion-de-cation-calcio-en-una-muestra-de-suero-sanguineo-7786</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Concentracion-de-cation-calcio-en-una-muestra-de-suero-sanguineo-7786</guid>
		<dc:date>2022-11-25T06:05:05Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Molaridad</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Normalidad</dc:subject>
		<dc:subject>Partes por mill&#243;n</dc:subject>

		<description>
&lt;p&gt;Al analizar una muestra de suero sangu&#237;neo se encuentra que contiene de de suero. Si la densidad del suero es 1.053 g/mL y el peso at&#243;mico del calcio es 40.08, cu&#225;l es la concentraci&#243;n de expresada en: a) molaridad b) ; c) ppm de en peso?&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Disoluciones-305" rel="directory"&gt;Disoluciones&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/Molaridad" rel="tag"&gt;Molaridad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Normalidad" rel="tag"&gt;Normalidad&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Partes-por-millon" rel="tag"&gt;Partes por mill&#243;n&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Al analizar una muestra de suero sangu&#237;neo se encuentra que contiene &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L59xH15/34b31c33b17616e2fdf66a215bd19b64-66a65.png?1732993652' style='vertical-align:middle;' width='59' height='15' alt=&#034;102.5\ \mu g&#034; title=&#034;102.5\ \mu g&#034; /&gt; de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L28xH22/ac64bf7da4597c8e5c9ef634000858b7-93db5.png?1732993652' style='vertical-align:middle;' width='28' height='22' alt=&#034;\textstyle{\ce{Ca^2+}\over \text{mL}}&#034; title=&#034;\textstyle{\ce{Ca^2+}\over \text{mL}}&#034; /&gt; de suero. Si la densidad del suero es 1.053 g/mL y el peso at&#243;mico del calcio es 40.08, cu&#225;l es la concentraci&#243;n de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L34xH16/1b67c0de4cacf2e095bcfa8751aa32cc-bba08.png?1732976434' style='vertical-align:middle;' width='34' height='16' alt=&#034;\ce{Ca^2+}&#034; title=&#034;\ce{Ca^2+}&#034; /&gt; expresada en: a) molaridad b) &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L55xH23/30a6fb0ac68d9fed14acaa24141abd07-378e9.png?1732993652' style='vertical-align:middle;' width='55' height='23' alt=&#034;\textstyle{\ce{meq\ Ca^2+}\over \text{L\ de\ suero}}&#034; title=&#034;\textstyle{\ce{meq\ Ca^2+}\over \text{L\ de\ suero}}&#034; /&gt;; c) ppm de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L34xH16/1b67c0de4cacf2e095bcfa8751aa32cc-bba08.png?1732976434' style='vertical-align:middle;' width='34' height='16' alt=&#034;\ce{Ca^2+}&#034; title=&#034;\ce{Ca^2+}&#034; /&gt; en peso?&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) Para calcular la molaridad debes llevar el soluto a moles y la disoluci&#243;n a litro: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/a6770555f5b918033ba14d48b79c2ea9.png' style=&#034;vertical-align:middle;&#034; width=&#034;459&#034; height=&#034;43&#034; alt=&#034;102.5\ \frac{\cancel{\mu g}\ \ce{Ca^2+}}{\cancel{mL}\ \text{suero}}\cdot \frac{1\ \cancel{g}}{10^6\ \cancel{\mu g}}\cdot \frac{1\ \text{mol}}{40.08\ \cancel{g}}\cdot \frac{10^3\ \cancel{mL}}{1\ L} = \fbox{\color[RGB]{192,0,0}{\bm{2.557\cdot 10^{-3}\ M}}}&#034; title=&#034;102.5\ \frac{\cancel{\mu g}\ \ce{Ca^2+}}{\cancel{mL}\ \text{suero}}\cdot \frac{1\ \cancel{g}}{10^6\ \cancel{\mu g}}\cdot \frac{1\ \text{mol}}{40.08\ \cancel{g}}\cdot \frac{10^3\ \cancel{mL}}{1\ L} = \fbox{\color[RGB]{192,0,0}{\bm{2.557\cdot 10^{-3}\ M}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) Cada mol de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/1b67c0de4cacf2e095bcfa8751aa32cc.png' style=&#034;vertical-align:middle;&#034; width=&#034;34&#034; height=&#034;16&#034; alt=&#034;\ce{Ca^2+}&#034; title=&#034;\ce{Ca^2+}&#034; /&gt; contiene dos equivalente del cati&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/44fce54eca9759879a1da469310c4d00.png' style=&#034;vertical-align:middle;&#034; width=&#034;391&#034; height=&#034;43&#034; alt=&#034;2.557\ \frac{\cancel{mol}\ \ce{Ca^2+}}{L}\cdot \frac{2\ \cancel{eq}}{1\ \cancel{mol}}\cdot \frac{10^3\ \text{meq}}{1\ \cancel{eq}} = \fbox{\color[RGB]{192,0,0}{\bm{5.114\cdot 10^3\ \frac{meq}{L}}}}&#034; title=&#034;2.557\ \frac{\cancel{mol}\ \ce{Ca^2+}}{L}\cdot \frac{2\ \cancel{eq}}{1\ \cancel{mol}}\cdot \frac{10^3\ \text{meq}}{1\ \cancel{eq}} = \fbox{\color[RGB]{192,0,0}{\bm{5.114\cdot 10^3\ \frac{meq}{L}}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; c) Para hacer las partes por mill&#243;n puedes calcular la masa del suero y compararla con la masa del cati&#243;n: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/c45a47db54c29a3f6cb745342d5ebce0.png' style=&#034;vertical-align:middle;&#034; width=&#034;332&#034; height=&#034;39&#034; alt=&#034;102.5\ \frac{\mu g}{\cancel{mL\ suero}}\cdot \frac{1\ \cancel{mL\ suero}}{1.053\ g} = \fbox{\color[RGB]{192,0,0}{\bf 97.34\ ppm}}&#034; title=&#034;102.5\ \frac{\mu g}{\cancel{mL\ suero}}\cdot \frac{1\ \cancel{mL\ suero}}{1.053\ g} = \fbox{\color[RGB]{192,0,0}{\bf 97.34\ ppm}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Volumen de una disoluci&#243;n de agua oxigenada para preparar otra m&#225;s diluida (7240)</title>
		<link>https://ejercicios-fyq.com/Volumen-de-una-disolucion-de-agua-oxigenada-para-preparar-otra-mas-diluida-7240</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Volumen-de-una-disolucion-de-agua-oxigenada-para-preparar-otra-mas-diluida-7240</guid>
		<dc:date>2021-06-23T08:52:28Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Concentraci&#243;n</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>Porcentaje (masa/volumen)</dc:subject>
		<dc:subject>EDICO</dc:subject>

		<description>
&lt;p&gt;El perhidrol es una soluci&#243;n de agua oxigenada de 100 vol. Sup&#243;n que tienes una soluci&#243;n de perhidrol desde hace varios a&#241;os y la valoras para verificar su t&#237;tulo, obteniendo un valor de 85 vol. &#191;C&#243;mo preparar&#237;as 10 L de una soluci&#243;n de agua oxigenada 1 vol. a partir de ese perhidrol? &#191;Qu&#233; concentraci&#243;n en tendr&#237;a la soluci&#243;n preparada?&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Disoluciones-305" rel="directory"&gt;Disoluciones&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/Concentracion" rel="tag"&gt;Concentraci&#243;n&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Porcentaje-masa-volumen" rel="tag"&gt;Porcentaje (masa/volumen)&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/EDICO" rel="tag"&gt;EDICO&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;El perhidrol es una soluci&#243;n de agua oxigenada de &lt;i&gt;100 vol&lt;/i&gt;. Sup&#243;n que tienes una soluci&#243;n de perhidrol desde hace varios a&#241;os y la valoras para verificar su t&#237;tulo, obteniendo un valor de &lt;i&gt;85 vol&lt;/i&gt;. &#191;C&#243;mo preparar&#237;as 10 L de una soluci&#243;n de agua oxigenada &lt;i&gt;1 vol.&lt;/i&gt; a partir de ese perhidrol? &#191;Qu&#233; concentraci&#243;n en &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L58xH18/4b4bef89f2a6760cb5007f14280f864e-f3c5d.png?1733076207' style='vertical-align:middle;' width='58' height='18' alt=&#034;\%\ (p/V)&#034; title=&#034;\%\ (p/V)&#034; /&gt; tendr&#237;a la soluci&#243;n preparada?&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;La reacci&#243;n de descomposici&#243;n del agua oxigenada es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/7579979f193a94fb3b1402eb58724748.png' style=&#034;vertical-align:middle;&#034; width=&#034;183&#034; height=&#034;34&#034; alt=&#034;\color[RGB]{2,112,20}{\textbf{\ce{H2O2 -&gt; H2O + 1/2 O2}}}&#034; title=&#034;\color[RGB]{2,112,20}{\textbf{\ce{H2O2 -&gt; H2O + 1/2 O2}}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; Observa que cada mol de per&#243;xido de hidr&#243;geno produce medio mol de ox&#237;geno. Si defines una concentraci&#243;n de &lt;i&gt;1 vol&lt;/i&gt; como el volumen de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0c78f1c6ccd624f193dcd7229489125c.png' style=&#034;vertical-align:middle;&#034; width=&#034;38&#034; height=&#034;15&#034; alt=&#034;\ce{H2O2}&#034; title=&#034;\ce{H2O2}&#034; /&gt; necesario para que se produzca 1 L de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/67d6e5f3cdb441ecb11258efaeae6139.png' style=&#034;vertical-align:middle;&#034; width=&#034;17&#034; height=&#034;15&#034; alt=&#034;\ce{O_2}&#034; title=&#034;\ce{O_2}&#034; /&gt;, en condiciones normales, puedes ver que los moles necesarios de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0c78f1c6ccd624f193dcd7229489125c.png' style=&#034;vertical-align:middle;&#034; width=&#034;38&#034; height=&#034;15&#034; alt=&#034;\ce{H2O2}&#034; title=&#034;\ce{H2O2}&#034; /&gt; son: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/6d92db66a3121e0a4b8a123851550646.png' style=&#034;vertical-align:middle;&#034; width=&#034;408&#034; height=&#034;38&#034; alt=&#034;1\ vol:\ \frac{1\ \cancel{\ce{mol\ O2}}}{22.4\ \cancel{L}}\cdot \frac{2\ \ce{mol\ H2O2}}{1\ \cancel{\ce{mol\ O2}}} = \color[RGB]{0,112,192}{\bm{8.93\cdot 10^{-2}}\ \textbf{\ce{mol\ H2O2}}}}&#034; title=&#034;1\ vol:\ \frac{1\ \cancel{\ce{mol\ O2}}}{22.4\ \cancel{L}}\cdot \frac{2\ \ce{mol\ H2O2}}{1\ \cancel{\ce{mol\ O2}}} = \color[RGB]{0,112,192}{\bm{8.93\cdot 10^{-2}}\ \textbf{\ce{mol\ H2O2}}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El volumen que necesitas de disoluci&#243;n titulada es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/41179e90189bf20cd26fd67bc6475fdb.png' style=&#034;vertical-align:middle;&#034; width=&#034;413&#034; height=&#034;38&#034; alt=&#034;c_1\cdot V_1 = c_2\cdot V_2\ \to\ V_1 = \frac{c_2\cdot V_2}{c_1} = \frac{1\ \cancel{vol}\cdot 10\ L}{85\ \cancel{vol}} = \fbox{\color[RGB]{192,0,0}{\bf 0.12\ L}}&#034; title=&#034;c_1\cdot V_1 = c_2\cdot V_2\ \to\ V_1 = \frac{c_2\cdot V_2}{c_1} = \frac{1\ \cancel{vol}\cdot 10\ L}{85\ \cancel{vol}} = \fbox{\color[RGB]{192,0,0}{\bf 0.12\ L}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; &lt;b&gt;Bastar&#237;a con tomar 0.12 L de la disoluci&#243;n titulada y disolverla hasta los 10 L de disoluci&#243;n final&lt;/b&gt;. &lt;br/&gt; &lt;br/&gt; Al ser de una concentraci&#243;n &lt;i&gt;1 vol.&lt;/i&gt; quiere decir que contiene &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/93cebe1fa9435f7abbc6083b2bcb05cd.png' style=&#034;vertical-align:middle;&#034; width=&#034;105&#034; height=&#034;16&#034; alt=&#034;8.93\cdot 10^{-2}\ mol&#034; title=&#034;8.93\cdot 10^{-2}\ mol&#034; /&gt; de &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/0c78f1c6ccd624f193dcd7229489125c.png' style=&#034;vertical-align:middle;&#034; width=&#034;38&#034; height=&#034;15&#034; alt=&#034;\ce{H2O2}&#034; title=&#034;\ce{H2O2}&#034; /&gt;. Si lo conviertes en masa: &lt;br/&gt; &lt;br/&gt; &lt;img src='https://ejercicios-fyq.com/local/cache-TeX/2490a397ac29d7b1bb910dffaa9cd2d6.png' style=&#034;vertical-align:middle;&#034; width=&#034;324&#034; height=&#034;38&#034; alt=&#034;8.93\cdot 10^{-2}\ \cancel{mol}\ \ce{H2O2}\cdot \frac{34\ g}{1\ \cancel{mol}} = \color[RGB]{0,112,192}{\textbf{3.04\ g\ \ce{H2O2}}}&#034; title=&#034;8.93\cdot 10^{-2}\ \cancel{mol}\ \ce{H2O2}\cdot \frac{34\ g}{1\ \cancel{mol}} = \color[RGB]{0,112,192}{\textbf{3.04\ g\ \ce{H2O2}}}&#034; /&gt; &lt;br/&gt; &lt;br/&gt; El porcentaje (m/V) est&#225; referido a un volumen de 100 mL de disoluci&#243;n en lugar de los 10 L: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/5294547520fcd1429206181f299b7ac7.png' style=&#034;vertical-align:middle;&#034; width=&#034;404&#034; height=&#034;26&#034; alt=&#034;\%\ (\textstyle{m\overV}) = \frac{m_{\ce{H2O2}}}{V_D}\cdot 100 = \frac{3.04\ g\ \ce{H2O2}}{10^4\ mL\ D}\cdot 100 = \fbox{\color[RGB]{192,0,0}{\bm{3.04\cdot 10^{-2} \%}}}&#034; title=&#034;\%\ (\textstyle{m\overV}) = \frac{m_{\ce{H2O2}}}{V_D}\cdot 100 = \frac{3.04\ g\ \ce{H2O2}}{10^4\ mL\ D}\cdot 100 = \fbox{\color[RGB]{192,0,0}{\bm{3.04\cdot 10^{-2} \%}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;
&lt;p&gt; &lt;br/&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Descarga el enunciado y la resoluci&#243;n del problema en formato EDICO si lo necesitas&lt;/b&gt;.&lt;/p&gt;
&lt;div class='spip_document_1684 spip_document spip_documents spip_document_file spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt;
&lt;a href=&#034;https://ejercicios-fyq.com/apuntes/descarga.php?file=Ej_7240.edi&#034; class=&#034; spip_doc_lien&#034; title='Zip - ' type=&#034;application/zip&#034;&gt;&lt;img src='https://ejercicios-fyq.com/plugins-dist/medias/prive/vignettes/zip.svg?1772792240' width='64' height='64' alt='' /&gt;&lt;/a&gt;
&lt;/figure&gt;
&lt;/div&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>
<item xml:lang="es">
		<title>Factor de Van't Hoff, grado de disociaci&#243;n y aumento ebullosc&#243;pico de una disoluci&#243;n de una sal (7232)</title>
		<link>https://ejercicios-fyq.com/Factor-de-Van-t-Hoff-grado-de-disociacion-y-aumento-ebulloscopico-de-una</link>
		<guid isPermaLink="true">https://ejercicios-fyq.com/Factor-de-Van-t-Hoff-grado-de-disociacion-y-aumento-ebulloscopico-de-una</guid>
		<dc:date>2021-06-20T21:41:09Z</dc:date>
		<dc:format>text/html</dc:format>
		<dc:language>es</dc:language>
		<dc:creator>F_y_Q</dc:creator>


		<dc:subject>Descenso criosc&#243;pico</dc:subject>
		<dc:subject>Aumento ebullosc&#243;pico</dc:subject>
		<dc:subject>RESUELTO</dc:subject>
		<dc:subject>EDICO</dc:subject>

		<description>
&lt;p&gt;Una soluci&#243;n acuosa de nitrato de magnesio 0.200 m, presenta un descenso criosc&#243;pico de . Las constantes criosc&#243;pica y ebullosc&#243;pica del agua son, 1.86 y 0.520 respectivamente. Calcula: &lt;br class='autobr' /&gt;
a) El valor del factor de Van't Hoff. &lt;br class='autobr' /&gt;
b) El grado de disociaci&#243;n. &lt;br class='autobr' /&gt;
c) &#191;Cu&#225;l ser&#237;a el aumento ebullosc&#243;pico de la disoluci&#243;n?&lt;/p&gt;


-
&lt;a href="https://ejercicios-fyq.com/Disoluciones-305" rel="directory"&gt;Disoluciones&lt;/a&gt;

/ 
&lt;a href="https://ejercicios-fyq.com/Descenso-crioscopico" rel="tag"&gt;Descenso criosc&#243;pico&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/Aumento-ebulloscopico" rel="tag"&gt;Aumento ebullosc&#243;pico&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/RESUELTO" rel="tag"&gt;RESUELTO&lt;/a&gt;, 
&lt;a href="https://ejercicios-fyq.com/EDICO" rel="tag"&gt;EDICO&lt;/a&gt;

		</description>


 <content:encoded>&lt;div class='rss_texte'&gt;&lt;p&gt;Una soluci&#243;n acuosa de nitrato de magnesio 0.200 m, presenta un descenso criosc&#243;pico de &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L62xH13/916a0bd19f2c60378b55b950a4d8b251-5a815.png?1732977384' style='vertical-align:middle;' width='62' height='13' alt=&#034;0.960\ ^oC&#034; title=&#034;0.960\ ^oC&#034; /&gt;. Las constantes criosc&#243;pica y ebullosc&#243;pica del agua son, 1.86 y 0.520 &lt;img src='https://ejercicios-fyq.com/local/cache-vignettes/L31xH21/0d23061eddac069b131626d0689d413d-f3af3.png?1732977384' style='vertical-align:middle;' width='31' height='21' alt=&#034;\textstyle{^oC\cdot kg\over mol}&#034; title=&#034;\textstyle{^oC\cdot kg\over mol}&#034; /&gt; respectivamente. Calcula:&lt;/p&gt;
&lt;p&gt;a) El valor del factor de Van't Hoff.&lt;/p&gt;
&lt;p&gt;b) El grado de disociaci&#243;n.&lt;/p&gt;
&lt;p&gt;c) &#191;Cu&#225;l ser&#237;a el aumento ebullosc&#243;pico de la disoluci&#243;n?&lt;/math&gt;&lt;/p&gt;&lt;/div&gt;
		&lt;hr /&gt;
		&lt;div &lt;div class='rss_ps'&gt;&lt;p&gt;a) Para calcular el factor de Van't Hoff solo tienes que partir de la ecuaci&#243;n del descenso criosc&#243;pico, que es el dato que conoces en el enunciado: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/77f1748d7c590d4f54fa069ef3741868.png' style=&#034;vertical-align:middle;&#034; width=&#034;449&#034; height=&#034;47&#034; alt=&#034;\Delta T = i\cdot k_c\cdot m\ \to\ i = \frac{\Delta T}{k_c\cdot m} = \frac{0.960\ \cancel{^oC}}{1.86\ \frac{\cancel{^oC}\cdot \cancel{kg}}{\cancel{mol}}\cdot 0.200\ \frac{\cancel{mol}}{\cancel{kg}}} = \fbox{\color[RGB]{192,0,0}{\bf 2.58}}&#034; title=&#034;\Delta T = i\cdot k_c\cdot m\ \to\ i = \frac{\Delta T}{k_c\cdot m} = \frac{0.960\ \cancel{^oC}}{1.86\ \frac{\cancel{^oC}\cdot \cancel{kg}}{\cancel{mol}}\cdot 0.200\ \frac{\cancel{mol}}{\cancel{kg}}} = \fbox{\color[RGB]{192,0,0}{\bf 2.58}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; b) El grado de disociaci&#243;n, teniendo en cuenta que es un electrolito fuerte y que se disocia en 3 iones, un cati&#243;n y dos aniones, es: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/f9b8ae460268be9a96bc394afbc0a9d4.png' style=&#034;vertical-align:middle;&#034; width=&#034;395&#034; height=&#034;39&#034; alt=&#034;i = 1 + \alpha(n - 1)\ \to\ \alpha = \frac{i - 1}{(3 -1)} = \frac{2.58 - 1}{2} = \fbox{\color[RGB]{192,0,0}{\bf 0.790}}&#034; title=&#034;i = 1 + \alpha(n - 1)\ \to\ \alpha = \frac{i - 1}{(3 -1)} = \frac{2.58 - 1}{2} = \fbox{\color[RGB]{192,0,0}{\bf 0.790}}&#034; /&gt;&lt;/p&gt; &lt;br/&gt; c) Para determinar el aumento ebullosc&#243;pico debes considerar la constante ebullosc&#243;pica del agua: &lt;br/&gt; &lt;br/&gt; &lt;p class=&#034;spip&#034; style=&#034;text-align: center;&#034;&gt;&lt;img src='https://ejercicios-fyq.com/local/cache-TeX/9ae0e799ca07170de7732c40e19ace35.png' style=&#034;vertical-align:middle;&#034; width=&#034;455&#034; height=&#034;45&#034; alt=&#034;\Delta T = i\cdot k_{eb}\cdot m = 2.58\cdot 0.520\ \frac{^oC\cdot \cancel{kg}}{\cancel{mol}}\cdot 0.200\ \frac{\cancel{mol}}{\cancel{kg}} = \fbox{\color[RGB]{192,0,0}{\bm{0.268\ ^oC}}}&#034; title=&#034;\Delta T = i\cdot k_{eb}\cdot m = 2.58\cdot 0.520\ \frac{^oC\cdot \cancel{kg}}{\cancel{mol}}\cdot 0.200\ \frac{\cancel{mol}}{\cancel{kg}} = \fbox{\color[RGB]{192,0,0}{\bm{0.268\ ^oC}}}&#034; /&gt;&lt;/p&gt;
&lt;/math&gt;&lt;/p&gt;
&lt;p&gt; &lt;br/&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Descarga el enunciado y la resoluci&#243;n del problema en formato EDICO si lo necesitas&lt;/b&gt;.&lt;/p&gt;
&lt;div class='spip_document_1685 spip_document spip_documents spip_document_file spip_documents_center spip_document_center'&gt;
&lt;figure class=&#034;spip_doc_inner&#034;&gt;
&lt;a href=&#034;https://ejercicios-fyq.com/apuntes/descarga.php?file=Ej_7232.edi&#034; class=&#034; spip_doc_lien&#034; title='Zip - ' type=&#034;application/zip&#034;&gt;&lt;img src='https://ejercicios-fyq.com/plugins-dist/medias/prive/vignettes/zip.svg?1772792240' width='64' height='64' alt='' /&gt;&lt;/a&gt;
&lt;/figure&gt;
&lt;/div&gt;&lt;/div&gt;
		
		</content:encoded>


		

	</item>



</channel>

</rss>
