Answers
Answer:
The vapor pressure at 60.0°C is 2.416 atm
Explanation:
To solve this problem, we use Clausius-Clapeyron equation
where;
Initial pressure P₁ = 0.703 atm
Initial Temperature T₁ = 25+273 = 298K
Final temperature T₂ = 60+273 = 333K
Change in enthalpy of vaporization ΔH = 29.1 KJ/mol = 29100J/mol
R is Boltzman constant = 8.314 J/K.mol
⇒
P₂ = P₁ (3.43663) = (0.703 atm)(3.43663) = 2.416 atm
P₂ = 2.416 atm
Therefore, the vapor pressure at 60.0°C is 2.416 atm.
Related Questions
The presence of dwarf galaxies around the Milky Way supports what picture of our galaxy’s formation?
As the Moon revolves around the Earth, it also rotates on its axis. Why is it that the same side of the Moon is always visible from Earth?
Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)
Which of the following wouldhave low electromagnetic energy? A. X-rays B. ultraviolet waves C. radio waves
Answers
Answer:
The final temperature is 50.8degrees celcius
Explanation:
Pls refer to attached handwritten document
Answer: 50.63° C
Explanation:
Given
Length of heater, L = 200 mm = 0.2 m
Diameter of heater, D = 15 mm = 0.015 m
Thermal conductivity, k = 5 W/m.K
Power of the heater, q = 25 W
Temperature of the block, = 35° C
T1 = T2 + (q/kS)
S can be gotten from the relationship
S = 2πL/In(4L/D)
On substituting we have
S = (2 * 3.142 * 0.2) / In (4 * 0.2 / 0.015)
S = 1.2568 / In 53.33
S = 1.2568 / 3.98
S = 0.32 m
Proceeding to substitute into the main equation, we have
T1 = T2 + (q/kS)
T1 = 35 + (25 / 5 * 0.32)
T1 = 35 + (25 / 1.6)
T1 = 35 + 15.625
T1 = 50.63° C
Answers
Answer:
1.26812 N
Explanation:
= Mass of toy =
= Length of string =
= Period of rotation =
Time period is given by
The rotational velocity is 1.3792 m/s
The tension in the rope will be the centripetal force acting on the toy
The tension in the string is 1.26812 N.
Answers
Answer:TL;DR: 3.535 cm
Explanation:
Xcm = ΣxMoments/ΣMasses = (10*0 + 10*5)/(10+10) = 50/20 = 2.5 cm
by symmetry,
Ycm = 2.5 cm
The distance D from the point Xcm,Ycm to the origin is D = √(2.5²+2.5²) = 3.535 cm
Final answer:
The center of mass of the bent wire is approximately 11.18 cm from the bend.
Explanation:
In order to find the center of mass of the bent wire, we need to divide it into two segments: the horizontal segment and the vertical segment. The length of each segment is half of the total length of the wire, which is 20 cm, so each segment is 10 cm long.
The center of mass of the horizontal segment is located exactly at its middle point, which is 5 cm from the corner. The center of mass of the vertical segment is also located at its middle point, which is 10 cm from the corner. Since the horizontal and vertical segments are orthogonal, the distance from the bend to the center of mass of the bent wire is the hypotenuse of a right triangle with legs of length 5 cm and 10 cm. Using the Pythagorean theorem, we can calculate the distance:
d = sqrt(5^2 + 10^2) = sqrt(25 + 100) = sqrt(125) = 11.18 cm
Therefore, the center of mass of the bent wire is approximately 11.18 cm from the bend.
Answers
Answers
Answer:
ratio of the piccolo's length to the flute's length is 0.4916
Explanation:
given data
frequency of piccolo = 522.5 Hz
frequency of flute = 256.9 Hz
to find out
ratio of the piccolo's length to the flute's length
solution
we get here length of tube that is express as
length of tube = velocity of sound ÷ fundamental frequency .......................1
so here ratio of Piccolo length to flute that is
= 0.4916
so ratio of the piccolo's length to the flute's length is 0.4916
b. Determine the value of the coefficient of friction between the tires and the road.
Answers
(a) It's the force of (static) friction that keeps the car on the road and prevents it from skidding, and this friction is directed toward the center of the curve.
Recall that centripetal acceleration has a magnitude a of
a = v ² / R
where
v = tangential speed
R = radius of the curve
so that
a = (35 m/s)² / (215 m) ≈ 5.69767 m/s² ≈ 5.7 m/s²
Parallel to the road, the only force acting on the car is friction. So by Newton's second law, we have
∑ F = Fs = ma
where
Fs = magnitude of static friction
m = mass of the car
Then
Fs = (950 kg) (5.7 m/s²) ≈ 5412.79 N ≈ 5400 N
(b) Perpendicular to the road, the car is in equilbrium, so its weight and the normal force of the road on the car are equal in magnitude. By Newton's second law,
N - W = 0
where
N = magnitude of normal force
W = weight
so that
N = W = m g = (950 kg) (9.8 m/s²) = 9310 N
Friction is proportional to the normal force by a factor of µ, the coefficient of static friction:
Fs = µN
Assuming 35 m/s is the maximum speed the car can travel without skidding, we find
µ = Fs / N = (5400 N) / (9310 N) ≈ 0.581395 ≈ 0.58