Answer:
A
Explanation:
CHEMICALS WEATHER IN CHEMICAL WEATHERING
Answer:

Explanation:
We are asked to find how many moles are in 4.8 × 10²³ fluorine atoms. We convert atoms to moles using Avogadro's Number or 6.022 × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this case, the particles are atoms of fluorine.
We will convert using dimensional analysis and set up a ratio using Avogadro's Number.

We are converting 4.8 × 10²³ fluorine atoms to moles, so we multiply the ratio by this number.

Flip the ratio so the units of atoms of fluorine cancel each other out.


Condense into 1 fraction.

Divide.

The original measurement of atoms has 2 significant figures, so our answer must have the same. For the number we found, that is the hundredths place. The 7 in the thousandths tells us to round the 9 in the hundredths place up to a 0. Then, we also have to round the 7 in the tenths place up to an 8.

4.8 × 10²³ fluorine atoms are equal to <u>0.80 moles of fluorine.</u>
It would be 35.8 Calories or calories. Not sure about that part. Hope this helps though.
Answer : The equilibrium concentration of
in the solution is, 
Explanation :
The dissociation of acid reaction is:

Initial conc. c 0 0
At eqm. c-x x x
Given:
c = 

The expression of dissociation constant of acid is:
![K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_6H_5COO%5E-%5D%7D%7B%5BC_6H_5COOH%5D%7D)

Now put all the given values in this expression, we get:
![6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}](https://tex.z-dn.net/?f=6.3%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%5B%287.0%5Ctimes%2010%5E%7B-2%7D%29-x%5D%7D)

Thus, the equilibrium concentration of
in the solution is, 