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1 year ago

The normal boiling point of cyclohexane is 81.0 oC. What is the vapor pressure of cyclohexane at 81.0 oC?

Physics

1 answer:

Aloiza [94]1 year ago

6 0

Answer:

The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.

Explanation:

Given that,

Boiling point = 81.0°C

Atmospheric pressure :

Atmospheric pressure is the force per unit area exerted by the weight of the atmosphere.

The value of atmospheric pressure is

$P=101325\ Pa$

Vapor pressure :

Vapor pressure is equal to the atmospheric pressure.

Hence, The vapor pressure of cyclohexane at 81.0°C is 101325 Pa.

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by comparing the given solution to other familiar solutions containing phenolphthalein

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1 year ago

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A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

ankoles [38]

Given required solution

M=10kg W=? W=Fd
v=5.0m/s F=mg
t=2.40s =10*10=100N
S=VT
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1 year ago

Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 c

tatuchka [14]

Answer:

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F=K*q₁*q₂/d² Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)

d: distance between the charges in meters

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm= 2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

$\frac{k*q_{2} *q_{3} }{d_{2}^{2} } -\frac{k*q_{1} *q_{3} }{d_{1}^{2} }=0$

We eliminate k*q₃ of the equation

$\frac{q_{1} }{d_{1}^{2} } = \frac{q_{2} }{d_{2}^{2} }$

$q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2} }$

$q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4} }{5^{2}*10^{-4} }$

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC

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1 year ago

A person walks 25 m west and then 45 m at the angle of 60 degrees north of east what is the magnitude of the total displacement?

expeople1 [14]

To solve this question, we need to use the component method and split our displacements into their x and y vectors. We will assign north and east as the positive directions.

The first movement of 25m west is already split. x = -25m, y = 0m.

The second movement of 45m [E60N] needs to be split using trig.
x = 45cos60 = 22.5m
y = 45sin60 = 39.0m

Then, we add the two x and two y displacements to get the total displacement in each direction.

x = -25m + 22.5m = -2.5m
y = 0m + 39.0m

We can use Pythagorean theorem to find the total displacement.
d² = x² + y²
d = √(-2.5² + 39²)
d = 39.08m

And then we can use tan to find the angle.
inversetan(y/x) = angle
inversetan(39/2.5) = 86.3

Therefore, the total displacement is 39.08m [W86.3N]

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2 years ago

As a youngster, you drive a nail in the trunk of a young tree that is 3 meters tall. The nail is about 1.5 meters from the groun

Lera25 [3.4K]

Answer:

15m

Explanation:

Hello! first to solve this problem we must find that so much that the tree grew in the 15 years this is achieved by dividing the height of the tree before and after

$\frac{30m}{3m} =10 times$

the tree grew 10 times its initial length in 15 years, so to find how tall the nail is, we multiply this factor by 1.5m

X=(1.5m)(10)=15m

the nail is 15 meters above ground level

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1 year ago