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vodka [1.7K]
3 years ago
13

Si tienes una disolución al 30 % en masa, ¿Cuál sería la nueva concentración si añades agua suficiente hasta duplicar su volumen

?
Chemistry
1 answer:
Alika [10]3 years ago
4 0

Answer:

17.6% en masa será la nueva concentración de la solución

Explanation:

Una disolución al 30% en masa contiene 30g de soluto en 100g de solución (Solvente + soluto). Así, la masa de solvente (Agua) es:

100g - 30g = 70g

Dado que la densidad del agua es de 1g/mL, 70g de agua ocupan 70mL.

Si el volumen se duplica, habrán 140mL de agua = 140g de solvente.

Así, la masa total de la solución será 30g + 140g = 170g y el porcentaje en masa será:

30g soluto / 170g solución × 100 =

<h3>17.6% en masa será la nueva concentración de la solución</h3>
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Which of the following ions is least likely to form colored compounds?
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2 years ago
A 41.1 g sample of solid CO2 (dry ice) is added to a container at a temperature of 100 K with a volume of 3.4 L.A. If the contai
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Answer:

Approximately 6.81 × 10⁵ Pa.

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  • O: 15.999.

Calculate the molar mass of carbon dioxide \rm CO_2:

M\!\left(\mathrm{CO_2}\right) = 12.011 + 2\times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

Find the number of moles of molecules in that 41.1\;\rm g sample of \rm CO_2:

n = \dfrac{m}{M} = \dfrac{41.1}{44.009} \approx 0.933900\; \rm mol.

If carbon dioxide behaves like an ideal gas, it should satisfy the ideal gas equation when it is inside a container:

P \cdot V = n \cdot R \cdot T,

where

  • P is the pressure inside the container.
  • V is the volume of the container.
  • n is the number of moles of particles (molecules, or atoms in case of noble gases) in the gas.
  • R is the ideal gas constant.
  • T is the absolute temperature of the gas.

Rearrange the equation to find an expression for P, the pressure inside the container.

\displaystyle P = \frac{n \cdot R \cdot T}{V}.

Look up the ideal gas constant in the appropriate units.

R = 8.314 \times 10^3\; \rm L \cdot Pa \cdot K^{-1} \cdot mol^{-1}.

Evaluate the expression for P:

\begin{aligned} P &=\rm \frac{0.933900\; mol \times 8.314 \times 10^3 \; L \cdot Pa \cdot K^{-1} \cdot mol^{-1} \times 298\; K}{3.4\; L} \cr &\approx \rm 6.81\times 10^5\; Pa \end{aligned}.

Apply dimensional analysis to verify the unit of pressure.

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Answer:

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