Answer:

Explanation:
Molarity is found by dividing the moles of solute by liters of solution.

We are given grams of a compound and milliliters of solution, so we must make 2 conversions.
1. Gram to Moles
We must use the molar mass. First, use the Periodic Table to find the molar masses of the individual elements.
- C: 12.011 g/mol
- H: 1.008 g/mol
- O: 15.999 g/mol
Next, look at the formula and note the subscripts. This tells us the number of atoms in 1 molecule. We multiply the molar mass of each element by its subscript.
6(12.011)+12(1.008)+6(15.999)=180.156 g/mol
Use this number as a ratio.

Multiply by the given number of grams.

Flip the fraction and divide.


2. Milliliters to Liters
There are 1000 milliliters in 1 liter.

Multiply by 2500 mL.


3. Calculate Molarity
Finally, divide the moles by the liters.


The original measurement has 2 significant figures, so our answer must have the same. That is the hundredth place and the 3 tells us to leave the 7.

1 mole per liter is also equal to 1 M.

Answer:
Chemical Change
Explanation:
Physical change normally mean that the change can revert back to its orginal state, which in this case that is not possible therfore it is a chemical change.
This is what i have found i hope this helps
The question is incomplete, complete question is :
Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (
for HF is
.)
[HF] = 0.280 M
Express your answer to two decimal places.
Answer:
The pH of an 0.280 M HF solution is 1.87.
Explanation:3
Initial concentration if HF = c = 0.280 M
Dissociation constant of the HF = 

Initially
c 0 0
At equilibrium :
(c-x) x x
The expression of disassociation constant is given as:
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)


Solving for x, we get:
x = 0.01346 M
So, the concentration of hydrogen ion at equilibrium is :
![[H^+]=x=0.01346 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.01346%20M)
The pH of the solution is ;
![pH=-\log[H^+]=-\log[0.01346 M]=1.87](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.01346%20M%5D%3D1.87)
The pH of an 0.280 M HF solution is 1.87.
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