Answer: The answer is 167
Explanation: This is because that was right on edg. so yea heart this tho plsss
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:
2.25g of NaF are needed to prepare the buffer of pH = 3.2
Explanation:
The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:
pH = pKa + log [A-] / [HA]
<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>
The moles of HF are:
500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF
Replacing:
3.2 = 3.17 + log [A-] / [0.0500moles]
0.03 = log [A-] / [0.0500moles]
1.017152 = [A-] / [0.0500moles]
[A-] = 0.0500mol * 1.017152
[A-] = 0.0536 moles NaF
The mass could be obtained using the molar mass of NaF (41.99g/mol):
0.0536 moles NaF * (41.99g/mol) =
<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
Boyle's law states that pressure is inversely proportional to volume of gas at constant temperature
PV = k
where P - pressure , V - volume and k - constant
P1V1 = P2V2
where parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting these values in the equation
1.25 atm x 0.75 L = P x 1.1 L
P = 0.85 atm
final pressure is B) 0.85 atm
