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.
10 is mixture of elements and compounds
11 is compound
12-15 are mixture of compounds
Hello!
When the final velocity is less than the initial velocity, this is deceleration
<h2>
Why?</h2>
Acceleration is defined as the physical magnitude that measures the change in velocity with time. The units to express acceleration are speed over time.
The equation for acceleration is: 
Where: a=acceleration, v=final velocity, vo=initial velocity, t=final time, to=initial time.
If the final velocity is less than the initial velocity, then the acceleration is negative, and that is called deceleration. An example of this is when a car brakes.
Have a nice day!
Sulfur forms compounds in oxidation states −2 (sulfide, S2−), +4 (sulfite, SO32−), and +6 (sulfate, SO42−). I don't know what type of ion but hope this helps!! :)
The reaction described above is the formation of an acetal. The initial starting material has a central carbonyl and two terminal alcohol functional groups. In the presence of acid, the carbonyl will become protonated, making the carbon of the carbonyl susceptible to nucleophilic attack from one of the alcohols. The alcohol substitutes onto the carbon of the carbonyl to provide us with the intermediate shown.
The intermediate will continue to react in the presence of acid and the -OH that was once the carbonyl will become protonated, turning it into a good leaving group. The protonated alcohol leaves and is substituted by the other terminal alcohol to give the final acetal product. The end result of the overall reaction is the loss of water from the original molecule to give the spiroacetal shown in the image provided.
