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Vilka [71]
4 years ago
10

4.39 moles of gas in a box has a pressure of 2.25 atm at temperature of 385K. What is the volume of the box?

Physics
2 answers:
stira [4]4 years ago
6 0

Answer:

Volume of the container = 0.0618 m³

Explanation:

Assuming that the gas is ideal, we can apply ideal gas equation:

PV=nRT

where P = pressure of the gas

          V = volume occupied by the gas

          n =  number of moles of gas

          R = Universal gas constant = 8.314 J/mol-K

          T = Temperature of the gas

Here we have to find volume.

Given: P=2.25 atm T=385 K n=4.39 moles

Putting the above values in the ideal gas equation, we get:

V=\frac{nRT}{P}=\frac{4.39\times8.314\times385}{2.25\times1.01\times10^{5}}=0.0618\ m^3

Volume of the gas = volume of the container in which it is filled = 0.0618 m³

bazaltina [42]4 years ago
5 0

Answer:

0.0619 m^3

Explanation

number of moles = n = 4.39 mol

pressure = P = 2.25 atm =2.25×1.01×10^5 Pa= 2.27×10^5 Pa

Molar gas constant =R = 8.31 J/(mol K)

Temperature T= 385K

volume of gas = V =?

BY GENERAL GAS LAW WE HAVE

PV = nRT

or V = nRT/P

or V = (4.39×8.31×385)/(2.27×10^5)

V = 0.0618728

V =  0.0619 m^3

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A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

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W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
3 years ago
There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl
kaheart [24]

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28  × 10⁵km

Orbital period  T = 28.9days

T = 28.9  × 24 × 60 × 60

   = 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s

7 0
3 years ago
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goldenfox [79]

Answer:

B. 120

Explanation:

24/.2 = 120

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Degger [83]

Answer:Nazi skinhead code for Heil Hitler. H being the 8th letter of the alphabet, therefore HH=88.

Explanation:

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faltersainse [42]

Answer:

v= 1.911×10^8m/s

Explanation:

n=c/v

v=c/n

v= 3.0×10^8/1.57

v= 1.911×10^8m/s

7 0
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