To determine the empirical formula and the molecular formula of the compound, we assume a basis of the compound of 100 g. We do as follows:
Mass Moles
K 52.10 52.10/39.10 = 1.33 1.33/1.32 ≈ 1
C 15.8 15.8/12 = 1.32 1.32/1.32 ≈ 1
O 32.1 32.1 / 16 = 2.01 2.01/1.32 ≈ 1.5
The empirical formula would most likely be KCO.
The molecular formula would be K2C2O3.
No, within a crystal like structure or ionic lattice of sodium chloride, there are ions of Na and Cl. Na+ and Cl- respectively that are attracted to each other due to their opposite charges. Many of these ions form a crystal structure.
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!
Answer:
B-Sucrose molecules are too large to conduct electricity in once dissolved in water.
D-Salts, like NaCL, have ionic bonds and are considered to be electrolytes:when dissolved in water, salts dissociate and form ions.
Explanation: