Yes. Everything is made up of mass. If it takes up space, it has mass
Answer:
Fe₃Si₇
Explanation:
In order to determine the empirical formula, we have to follow a series of steps.
Step 1: Determine the percent composition
Fe: 46.01%
Si: 53.99%
Step 2: Divide each percentage by the atomic mass of the element
Fe: 46.01/55.85 = 0.8238
Si: 53.99/28.09 = 1.922
Step 3: Divide all the numbers by the smallest one
Fe: 0.8238/0.8238 = 1
Si: 1.922/0.8238 = 2.33
Step 4: Multiply by numbers that make the coefficients whole.
Fe: 1 × 3 = 3
Si: 2.33 × 3 = 7
The empirical formula is Fe₃Si₇.
Orbital shell notation of fluorine is 2. 7 while that of oxygen s 2. 6. This means that these elements (that follow each other in the periodic table) will have high electronegativity in molecules due to their high atomic number (which causes them to strongly attract electron orbital shell closer to their nucleus). NB: Atomic number of a peroid increased from left to right of the periodic table.
Therefore, in the first molecule, the negative dipole would most likely be located between the F atoms In the second molecule the negative molecule would be most likely located in the between the O and F atoms.
Minerals are solid, naturally occurring, inorganic compounds that possess an orderly internal structure and a regular chemical composition. Minerals should occur naturally. Hope this answers the question. Have a nice day. Feel free to ask more questions.
Answer: 
Explanation:

cM 0 0
So dissociation constant will be:

Given: c = 0.15 M
pH = 1.86
= ?
Putting in the values we get:
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![1.86=-log[H^+]](https://tex.z-dn.net/?f=1.86%3D-log%5BH%5E%2B%5D)
![[H^+]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01)
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)


As ![[H^+]=[ClCH_2COO^-]=0.01](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BClCH_2COO%5E-%5D%3D0.01)

![K_a=1.67\times 10^{-3]](https://tex.z-dn.net/?f=K_a%3D1.67%5Ctimes%2010%5E%7B-3%5D)
Thus the vale of
for the acid is 