I put
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please don't plagiarise tho, re-word it.
Answer: Option D) 298 g/mol is the correct answer
Explanation:
Given that;
Mass of sample m = 13.7 g
pressure P = 2.01 atm
Volume V = 0.750 L
Temperature T = 399 K
Now taking a look at the ideal gas equation
PV = nRT
we solve for n
n = PV/RT
now we substitute
n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K
)
= 1.5075 / 32.7579
= 0.04601 mol
we know that
molar mass of the compound = mass / moles
so
Molar Mass = 13.7 g / 0.04601 mol
= 297.7 g/mol ≈ 298 g/mol
Therefore Option D) 298 g/mol is the correct answer
Answer:
Explanation:
Mean temperature is given by

Tmean = (Ti + T∞)/2

Tmean = 107.5⁰C
Tmean = 107.5 + 273 = 380.5K
Properties of air at mean temperature
v = 24.2689 × 10⁻⁶m²/s
α = 35.024 × 10⁻⁶m²/s
= 221.6 × 10⁻⁷N.s/m²
= 0.0323 W/m.K
Cp = 1012 J/kg.K
Pr = v/α = 24.2689 × 10⁻⁶/35.024 × 10⁻⁶
= 0.693
Reynold's number, Re
Pv = 4m/πD²
where Re = (Pv * D)/
Substituting for Pv
Re = 4m/(πD
)
= (4 x 0.003)/( π × 6 ×10⁻³ × 221.6 × 10⁻⁷)
= 28728.3
Since Re > 2000, the flow is turbulent
For turbulent flows, Use
Dittus - Doeltr correlation with n = 0.03
Nu = 0.023Re⁰⁸Pr⁰³ = (h₁D)/k
(h₁ × 0.006)/0.0323 = 0.023(28728.3)⁰⁸(0.693)⁰³
(h₁ × 0.006)/0.0323 = 75.962
h₁ = (75.962 × 0.0323)/0.006
h₁ = 408.93 W/m².K
Answer:
infinite
Explanation:
The tangent of the angle the arm makes with the vertical will be the ratio of the centripetal acceleration to the acceleration due to gravity on the center of mass of the arm. The angle can only be 90° (fully-open arms) if that ratio is infinite.
The speed must be infinite for the arms to be fully open.