Photon has no mass at all
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.
Half-life is the length of time it takes for half of the radioactive atoms of a specific radionuclide to decay. A good rule of thumb is that, after seven half-lives, you will have less than one percent of the original amount of radiation.
<h3>What do you mean by half-life?</h3>
half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive.
<h3>What affects the half-life of an isotope?</h3>
Since the chemical bonding between atoms involves the deformation of atomic electron wavefunctions, the radioactive half-life of an atom can depend on how it is bonded to other atoms. Simply by changing the neighboring atoms that are bonded to a radioactive isotope, we can change its half-life.
Learn more about half life of an isotope here:
<h3>
brainly.com/question/13979590</h3><h3 /><h3>#SPJ4</h3>
Answer:
Covalent Bond
Explanation:
In the diagram Carbon and each of the 4 Hydrogens are sharing electrons. They are also both non metals. Both of these are characteristics of Covalent Bonds.
We want to solve Q = mcΔT for the liquid water; its change in temperature will tell us the amount of thermal energy that flowed out of the reaction. The specific heat, c, of water is 4.184 J/g °C.
Q = (72.0 g)(4.184 J/g °C)(100 °C - 25 °C) = 22593.6 J
Q ≈ 2.26 × 10⁴ J or 22.6 kJ (three significant figures).
