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katrin [286]
3 years ago
14

Si una solucion 0.0150 M tiene un porcentaje de transmitancia de 35% a una determinada longitud de onda, cual sera el porcentaje

de transmitancia para una solucion 0.0300M de la misma muestra?
Chemistry
1 answer:
tangare [24]3 years ago
4 0

Answer:

17.5% es la transmitancia de la solución problema

Explanation:

Basados en la ley de Beer que dice que:

La transmitancia de una solución es inversamente proporcional a su concentración.

Podemos hallar la transmitancia de una concentración problema sabiendo la concentración de una solución y su transmitancia:

Si una solución 0.0150M tiene una transmitancia de 35%

Y ahora la solución duplica su concentración, su transmitancia se disminuirá en la mitad, esto es:

35% / 2 =

<h3>17.5% es la transmitancia de la solución problema</h3>
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Which of the following would not cause an increase in the pressure of a gaseous system in a container? A The container is made l
ludmilkaskok [199]

Answer:

The container is made larger

Explanation:

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For the reaction: 3 H2(g) + N2(g) &lt;--&gt; 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [
masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.

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5 0
2 years ago
Please i need help ……
nasty-shy [4]

Answer:

H2

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I hope this helped [:

3 0
3 years ago
How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 4.46 atm and 45°C in the reaction shown below?
natka813 [3]

Answer:

8.1433 g of XeF₆  are required.

Explanation:

Balanced chemical equation;

XeF₆ (s) + 3H₂ (g)   →  Xe (g) + 6HF (g)

Given data:

Volume of hydrogen = 0.579 L

Pressure = 4.46 atm

Temperature = 45 °C (45+273= 318 k)

Solution:

First of all we will calculate the moles of hydrogen

PV = nRT

n = PV/ RT

n = 4.46 atm × 0.579 L / 0.0821 atm. dm³. mol⁻¹. K⁻¹ × 318 K

n = 2.6 atm . L / 26.12 atm. dm³. mol⁻¹

n = 0.0995 mol

Mass of hydrogen:

Mass = moles × molar mass

Mass =  0.0995 mol × 2.016 g/mol

Mass =  0.2006 g

Now we will compare the moles of hydrogen with XeF₆ from balance chemical equation.

                                         H₂   :  XeF₆

                                          3    :    1

                                 0.0995   : 1/3× 0.0995 = 0.0332 mol

Now we will calculate the mass of XeF₆.

Mass = moles × molar mass

Mass = 0.0332 mol × 245.28 g/mol

Mass = 8.1433 g

4 0
3 years ago
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