<span>3.5 moles CaCO3 (1 mole carbon/1 mole CaCO3) = 3.5 moles</span>
Atomic radii increase when going down a group and decreases when going towards the anion periods. So A and D.
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Answer:
<span>The mole concept is important in chemistry because, "</span>Atoms and molecules are very small and the mole concept allows us to count atoms and molecules by weighing macroscopic amounts of material".
Explanation:
To understand this question lets take an example of Hydrogen atom. Let suppose you need to react Hydrogen with Oxygen. You need exactly Two Hydrogen atoms and one Oxygen atom to form one water molecule.
The mass of 1 hydrogen atom is 1.76 × 10⁻²⁴ grams. How will you count the Hydrogen atoms??? How can you measure exactly for 1 Million Hydrogen Atoms???
Answer to these questions and Calculations lies in Mole. It is found that 1 Mole of Hydrogen weights exactly 1.008 gram and contains 6.022 × 10²³ atoms. Now, having this reference in hand you can calculate for any number of Hydrogen atoms.
Result:
So the Mole helps us to zoom a microscopic level to a macroscopic level. :)
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)](https://tex.z-dn.net/?f=PbCl_2%28s%29%5Crightleftharpoons%202Cl%5E-%28aq%29%2BPb%5E%7B2%2B%7D%28aq%29)
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):
![s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L](https://tex.z-dn.net/?f=s%3D0.0159molPbCl_2%2FL%2A%5Cfrac%7B278.1gmolPbCl_2%7D%7B1molmolPbCl_2%7D%20%5C%5C%5C%5Cs%3D4.41g%2FL)
¡Saludos!