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miskamm [114]
3 years ago
14

Calculate the number of molecules/m3 in an ideal gas at stp.

Physics
1 answer:
vivado [14]3 years ago
4 0
STP (Standard Temperature and Pressure) has the following conditions:

Temperature = 273.15 K = 0°C
Pressure = 101325 Pa = 101.325 KPa = 1 atm

We also know that 1 mole = 6.022x10^23 molecules

Using the ideal gas equation: PV=nRT

n/V = P/RT

molecules/V = P*6.022x10^23/RT
molecules/V = 101325 Pa (6.022x10^23 molecules/mole)/ (8.314 Pa-m3/mol-K)(273.15K)

molecules/V = 7.339x10^27 molecules/m^3 - Final answer
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HEY CAN ANYONE ANSWER DIS RQ!!!!!!
emmasim [6.3K]

Answer:

Density is the mass to volume ratio of an object.

> It tells you how compact the mass is.

> Density = mass/volume

The density of water is 1 g/ml or 1 kg/L or 1000 kg/m3

• If an object has a density less than that of water, it

will float.

• If an object has a density more than that of water, it

will sink

Explanation:

hope this helps some

7 0
3 years ago
Read 2 more answers
7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back
saw5 [17]

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

K.E =\dfrac{1}{2}mv^2

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

K.E=\dfrac{1}{2}\times0.5\times(4)^2

K.E = 4\ J

The average force exerted by the floor on the ball = change in kinetic energy over collision time

F = \dfrac{4}{0.05}

F=80\ N

Hence, The force exerted by the floor is 80 N.

4 0
3 years ago
For an object that travels at a fixed speed along a circular path, the acceleration of the object is
lidiya [134]
For an object that travels at a fixed speed along a circular path, the acceleration of the object is LARGER IN MAGNITUDE THE SMALLER THE RADIUS OF CIRCLE.
8 0
3 years ago
Read 2 more answers
What is the kinetic energy of a 100 kg object that is moving with a speed of 12.5m/s
Doss [256]

The kinetic energy of any moving object is

                           (1/2) (mass) (speed²) .

For the object you described, that's

                            (1/2) (100 kg) (12.5 m/s)²

                         =      (50 kg)  (156.25 m²/s²)

                         =              7,812.5 joules  
______________________________

Your attachment is way out of focus, and impossible to read.

7 0
3 years ago
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