Answer:
Hints
concentration=density×%by mass
Molarity= concentration/molar mass
Molality=no of moles/mass of solvent in kg
Mass of solvent is of water given by=density of water×volume of container
Solved
Beta particles is the smallest type of radiation as its just electron/s. Alpha particles and Gamma rays are larger than beta particles. Alpha particles is a helium nuclei and we know that iodine-131 is an example of gamma rays. SO beta particles are smaller than a helium nuclei and a Iodine atom.
Atoms and void would be the answer
Answer:oxygen is the central atom in water,it has two lone pairs of electrons, the bond angle is 104.27° The ideal H-O-H bond angle ought to have been 109.28°
Explanation:
The presence of lone pairs causes slight distortion in the bond angles of molecules. This distortion increases with the number of lone pairs present. Water has two lone pairs hence a greater distortion compared to ammonia.
Answer:
37 %.
Explanation:
¡Hola!
En este caso, para el problema descrito, conocemos la corriente de entrada y la de salida del agua, por lo que podemos obtener el flujo de la corriente que contiene el zumo a la salida una vez el agua fue evaporada:
![F_{sol}=11500kg/dia-3000kg/dia=8500 kg/dia](https://tex.z-dn.net/?f=F_%7Bsol%7D%3D11500kg%2Fdia-3000kg%2Fdia%3D8500%20kg%2Fdia)
Luego, por medio de un balance de zumo de limón en el evaporador en el cual la cantidad que entra es igual a la que sale con sus respectivas concentraciones:
![x_z^{entra}*11500kg/dia=x_z^{sale}*8500kg/dia](https://tex.z-dn.net/?f=x_z%5E%7Bentra%7D%2A11500kg%2Fdia%3Dx_z%5E%7Bsale%7D%2A8500kg%2Fdia)
Como la concentración del zumo a la salida es del 50 % (0.50), la de entrada es:
![x_z^{entra}=\frac{x_z^{sale}*8500kg/dia}{11500kg/dia} =\frac{0.50*8500kg/dia}{11500 kg/dia}\\ \\x_z^{entra}=0.37](https://tex.z-dn.net/?f=x_z%5E%7Bentra%7D%3D%5Cfrac%7Bx_z%5E%7Bsale%7D%2A8500kg%2Fdia%7D%7B11500kg%2Fdia%7D%20%3D%5Cfrac%7B0.50%2A8500kg%2Fdia%7D%7B11500%20kg%2Fdia%7D%5C%5C%20%5C%5Cx_z%5E%7Bentra%7D%3D0.37)
Que es igual al 37%.
¡Saludos!