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!
The molecular mass of the immunoglobulin G, given the data from the question is 1.53×10⁵ g/mole
<h3>How to determine the molarity</h3>
We'll begin by calculating the molarity of the immunoglobulin G. This is illustrated below:
- Volume = 0.106 L
- Temperature (T) = 25 °C = 25 + 273 = 298 K
- Osmotic pressure (π) = 0.733 mbar = 0.733 × 0.000987 = 0.00072 atm
- Gas constant (R) = 0.0821 atm.L/Kmol
- Van't Hoff factor (i) = 1
- Molarity (M)
π = iMRT
M = π / iRT
M = 0.00072 / (1 × 0.0821 × 298)
M = 0.000029 M
<h3>How to determine the mole of immunoglobulin G</h3>
- Molarity = 0.000029 M
- Volume = 0.106 L
- Mole =?
Mole = Molarity × volume
Mole = 0.000029 × 0.106
Mole = 3.074×10⁻⁶ mole
<h3>How to determine the molar mass of mmunoglobulin G</h3>
- Mole = 3.074×10⁻⁶ mole
- Mass = 0.470 g
- Molar mass =?
Molar mass = mass / mole
Molar mass = 0.47 / 3.074×10⁻⁶
Molar mass = 1.53×10⁵ g/mole
Learn more about Osmotic pressure:
brainly.com/question/5925156
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Answer:
water and oceans
Explanation:
the moons creates tides as it goes around the
earth. This creates a bulge on the side of the earth.
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The emission of the light waves ..................... which means that THE TROUGHS OF ONE WAVE AND THE CREST OF ANOTHER overlap.
When the trough and the crest of wave overlap, it results in constructive interference and the amplitude of the two waves are added together to form a larger amplitude.<span />
