Answer: a. 17.7 KJ/Mol
b. T=210K
Explanation:
Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressure is 35 torr at – 111.95°c and 253 torr at – 83.6°c. using these data calculate.
the question isnt completely originally, but we could look at the likely derivation from the questions
(a) the standard enthalpy of vaporization
using the clausius clapeyron equation
In (PT1vap / PT2vap) = delta H (vap) / R ( (1/T1) - (1/T2) )
In (35Torr/253Torr) = delta H (vap) / 8.3145 ( (1/189.55) - (1/161.2) )
Therefore, Delta H (vap) = 17.7 KJ/Mol
b. Also the boiling point
What is the normal boiling point of arsine?
At the boiling point Pvap = atmospheric pressure = 1 atm=760 torr
substitution into the equation as stated in question 1
ln(760/253)=17700/8.314(1/189.55-1/T)
T=210K
Answer:
v = 31.3 m / s
Explanation:
The law of the conservation of stable energy that if there are no frictional forces mechanical energy is conserved throughout the point.
Let's look for mechanical energy at two points, the highest where the body is at rest and the lowest where at the bottom of the plane
Highest point
Em₀ = U = m g y
Lowest point
= K = ½ m v²
As there is no friction, mechanical energy is conserved
Em₀ =
m g y = ½ m v²
v = √ 2 g y
Where we can use trigonometry to find and
sin 30 = y / L
y = L sin 30
Let's replace
v = RA (2 g L sin 30)
Let's calculate
v = RA (2 9.8 100.0 sin30)
v = 31.3 m / s
Answer:
(A) The speed just as it left the ground is 30.25 m/s
(B) The maximum height of the rock is 46.69 m
Explanation:
Given;
weight of rock, w = mg = 20 N
speed of the rock at 14.8 m, u = 25 m/s
(a) Apply work energy theorem to find its speed just as it left the ground
work = Δ kinetic energy
F x d = ¹/₂mv² - ¹/₂mu²
mg x d = ¹/₂m(v² - u²)
g x d = ¹/₂(v² - u²)
gd = ¹/₂(v² - u²)
2gd = v² - u²
v² = 2gd + u²
v² = 2(9.8)(14.8) + (25)²
v² = 915.05
v = √915.05
v = 30.25 m/s
B) Use the work-energy theorem to find its maximum height
the initial velocity of the rock = 30.25 m/s
at maximum height, the final velocity = 0
- mg x H = ¹/₂mv² - ¹/₂mu²
- mg x H = ¹/₂m(0) - ¹/₂mu²
- mg x H = - ¹/₂mu²
2g x H = u²
H = u² / 2g
H = (30.25)² / 2(9.8)
H = 46.69 m
The visible spectrum is composed of red, orange,yellow, green, blue, violet, indigo.
<h3>What is visible spectrum?</h3>
The visible spectrum refers to the portion of the electromagnetic spectrum that can be seen with the eyes. All other portions of the electromagnetic spectrum are invisible.
The question is incomplete as the details are missing. The visible spectrum is composed of red, orange,yellow, green, blue, violet, indigo.
Learn more about the visble spectrum: brainly.com/question/1596783
Answer:
given
y=6.0sin(0.020px + 4.0pt)
the general wave equation moving in the positive directionis
y(x,t) = ymsin(kx -?t)
a) the amplitude is
ym = 6.0cm
b)
we have the angular wave number as
k = 2p /?
or
? = 2p / 0.020p
=1.0*102cm
c)
the frequency is
f = ?/2p
= 4p/2p
= 2.0 Hz
d)
the wave speed is
v = f?
= (100cm)(2.0Hz)
= 2.0*102cm/s
e)
since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction
f)
the maximum transverse speed is
umax =2pfym
= 2p(2.0Hz)(6.0cm)
= 75cm/s
g)
we have
y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]
= -2.0cm
