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Anna [14]
2 years ago
13

. ¿Qué cantidad de HNO3 concentrado y de solvente se utilizará en la preparación de 500 mLal 5% (v/v)?

Chemistry
1 answer:
Viefleur [7K]2 years ago
6 0

Answer:

464.29mL de solvente y 35.71mL de ácido nítrico concentrado deben agregarse.

Explanation:

El ácido nítrico concentrado viene al 70% v/v por temas de estabilidad. El volumen de ácido nítrico que se debe agregar si se quieren hace 500mL al 5% de HNO3 es:

500mL * (5mL / 100mL) = 25mL de ácido nítrico se deben agregar.

Como el ácido nítrico está al 70%:

25mL ácido nítrico * (100mL / 70mL ácido nítrico) = 35.71mL de ácido nítrico concentrado deben agregarse.

Y el volumen de solvente debe ser:

500mL - 35.71mL = 464.29mL

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Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T
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Explanation:

The given data is as follows.

          \Delta H = 286 kJ = 286 kJ \times \frac{1000 J}{1 kJ}

                            = 286000 J

 S_{H_{2}O} = 70 J/^{o}K,      S_{H_{2}} = 131 J/^{o}K

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Hence, formula to calculate entropy change of the reaction is as follows.

          \Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)

                     = [(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]

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             \Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}

                            = 286000 J - (163.5 J/K \times 298 K)

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Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

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Answer:

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Explanation:

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Explanation:

i hope that helps

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