The asthenosphere is directly below the lithosphere so the answer to your question is the asthenosphere, because the outer core is towards the center of Earth.
Answer:
20.27 mol
Explanation:
454 L x (1 mol/22.4 L) = 20.27 mol
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:
Lo cual hace que la expresión de equilibrio se calcule como:
![Ksp=[Pb^{2+}][Cl^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BPb%5E%7B2%2B%7D%5D%5BCl%5E-%5D%5E2)
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](https://tex.z-dn.net/?f=1.6x10%5E%7B-5%7D%3Ds%282s%29%5E2%5C%5C%5C%5C1.6x10%5E%7B-5%7D%3D4s%5E3%5C%5C%5C%5Cs%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1.6x10%5E%7B-5%7D%7D%7B4%7D%20%7D%20%5C%5C%5C%5Cs%3D0.0159molPbCl_2%2FL)
Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):
¡Saludos!
We can set up an ICE table for the reaction:
HClO H+ ClO-
Initial 0.0375 0 0
Change -x +x +x
Equilibrium 0.0375-x x x
We calculate [H+] from Ka:
Ka = 3.0x10^-8 = [H+][ClO-]/[HClO] = (x)(x)/(0.0375-x)
Approximating that x is negligible compared to 0.0375 simplifies the equation to
3.0x10^-8 = (x)(x)/0.0375
3.0x10^-8 = x2/0.0375
x2 = (3.0x10^-8)(0.0375) = 1.125x10^-9
x = sqrt(1.125x10^-9) = 0.0000335 = 3.35x10^-5 = [H+]
in which 0.0000335 is indeed negligible compared to 0.0375.
We can now calculate pH:
pH = -log [H+] = - log (3.35 x 10^-5) = 4.47
