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.
Answer:
Percent yield = 90.5%
Explanation:
Given data:
Mass of carbon dioxide = 500 g
Mass of water = excess
Actual yield of carbonic acid = 640 g
Percent yield = ?
Solution:
Balanced chemical equation:
CO₂ + H₂O → H₂CO₃
Number of moles of carbon dioxide
Number of moles = Mass / molar mass
Number of moles = 500 g/ 44 g/mol
Number of moles = 11.4 mol
Now we will compare the moles of H₂CO₃ with CO₂.
CO₂ : H₂CO₃
1 : 1
11.4 : 11.4
Mass of carbonic acid:
Mass = number of moles × molar mass
Mass = 11.4 mol × 62.03 g/mol
Mass = 707.14 g
Percent yield:
Percent yield = actual yield / theoretical yield × 100
Percent yield = 640 g/ 707.14 g × 100
Percent yield = 90.5%
They don't change what the substance really is unlike chemical change. They chemical formula of the substance stays the same even though the substance can go under shape change.
Recycling reduces the demand for new wood is correct