Answer:
82.59 m/s or 297.324 km/h
Explanation:
From the question,
Applying
V = √[2(P'/ρ)].................. Equation 1 ( From
Where V = Speed of the aircraft, Differential Pressure of the air craft, ρ = Density of air at an altitude of 3000 m.
Given: P' = 3100 N/m², ρ = 0.909 kg/m³
Substitute into equation 1
V = √[2(3100/0.909)]
V = √(2×3410.34)
V = √(6820.68)
V = 82.59 m/s
V = 297.324 km/h
Hence the speed of the aircraft is 82.59 m/s or 297.324 km/h
Answer: The pH of an aqueous solution of .25M acetic acid is 2.7
Explanation:
cM 0 0
So dissociation constant will be:
Give c= 0.25 M and 
= ?
Putting in the values we get:
![[H^+]=c\times \alpha](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dc%5Ctimes%20%5Calpha)
![[H^+]=0.25\times 0.0084=0.0021](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.25%5Ctimes%200.0084%3D0.0021)
Also ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![pH=-log[0.0021]=2.7](https://tex.z-dn.net/?f=pH%3D-log%5B0.0021%5D%3D2.7)
Thus pH is 2.7
The formula for pH given the pKa and the concentrations
are:
pH = pKa + log [a–]/[ha]
<span>
Therefore calculating:</span>
3.75 = 3.75 + log [a–]/[ha]
log [a–]/[ha] = 0
[a–]/[ha] = 10^0
<span>[a–]/[ha] = 1</span>
