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dusya [7]
3 years ago
5

El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si

una fábrica de baterías para carro provoca un derrame de material con plomo(II). La concentración máxima (en g/L) de plomo(II) que puede contener el agua marina es:_______________
Kps=1,6x10^5
Chemistry
1 answer:
Alchen [17]3 years ago
7 0

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)

Lo cual hace que la expresión de equilibrio se calcule como:

Ksp=[Pb^{2+}][Cl^-]^2

Y que en términos de la solubilidad molar, s, se resuelve como:

1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L

¡Saludos!

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Answer:

They would produce a repulsive force to another

Explanation:

A positive particle approaching another positive particle will repulse it.

According to coulomb's law "like charges repel one another and unlike charges attract".

A charge is an intrinsic property of any matter.

When like charges e.g positive and positive or negative and negative charges are in the vicinity of one another, they repel each other.

When unlike charges; positive and negative are brought together, they simply attract one another.

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How many moles of n are in 0.163 g of n2o?
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An electron in the n = 3 energy level of the hydrogen atom emits a photon with wavelength 656.27 nm. What is the change in energ
Lady_Fox [76]

Answer:change in energy of the electron=3.052x 10^-19J

which energy level does it move= level 2 , n=2

Explanation:

Using the formulae

1/λ = R (1/n1²- 1/n2²)

Where λ= 656.27 nm

1 nm = 1 x 10^-9 m

656.27 nm = 656.27 x 1 x 10^-9 =6.5726 x 10^-7

R =Rydberg constant = 1.0967 x 10^7m-1

1/λ = R (1/n1²- 1/n2²)

1/6.5726 x 10^-7=1.0967 x 10^7(1/n1²- 1/3²)

1/n1²=(1/6.5726 x 10^-7 x   1/1.0967 x 10^7) + 1/9

1/n1²=1,521,467.9 x 9.118x10^-8 + 0.1111

1/n1² =0.2498

n1²= 1/0.2498 =4

n1= \sqrt{4} = 2

it moves to energy level 2

b) Change in energy =ΔE = Rhc (1/n1²- 1/n2²)

Where R==Rydberg constant = 1.0967 x 10^7m-1

h = Planck constant = 6.626x 10^-34js

c = speed of light = 3.0 x 10^8 x m/s

ΔE = Rhc (1/n1²- 1/n2²)

=1.0967 x 10^7m-1 x6.626x 10^-34js X 3.0 x 10^8 x m/s (1/2² - 1/3²)

=2.18 x 10-18 x ( 1 /4 - 1/9)

=3.052x 10^-19J

7 0
3 years ago
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