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3 years ago

The molecular mass of a gas is

Physics

1 answer:

bearhunter [10]3 years ago

7 0

Answer:

Explanation:

Then the mass of the gas divided by the moles will give the molar mass. Now divide g by mol to get the molar mass. Since N has a molar mass of 14 g/mol and O has a molar mass of 16 g/mol, the formula N2O would produce the correct molar mass.

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A planet has a surface temperature of 95.0 K and an atmospheric pressure of 1.60 atm. If 4.87 L of atmosphere has a mass of 28.6

mr Goodwill [35]

Answer:

M=28.88 gm/mol

Explanation:

Given that

T= 95 K

P= 1.6 atm

V= 4.87 L

m = 28.6 g

R=0.08206L atm .mol .K

We know that gas equation for ideal gas

P V = n R T

P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant

Now by putting the values

P V = n R T

1.6 x 4.87 = n x 0.08206 x 95

n=0.99 moles

We know that number of moles given as

$n=\dfrac{m}{M}$

M=Molar mass

$n=\dfrac{m}{M}$

$0.99=\dfrac{28.6}{M}$

M=28.88 gm/mol

3 0

3 years ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

3 years ago

Read 2 more answers

1.a. identifying what are the two major motions of earth as it travels through space

Lina20 [59]

Ehh, Earth rotates around its axis and Earth revolves around the sun as it travels through space.

6 0

3 years ago

Read 2 more answers

If, as is typical, each of them breathes about 500 cm3 of air with each breath, approximately what volume of air (in cubic meter

deff fn [24]

Answer:

a) 12614.4 m^3

b) 28.8 m

Explanation:

The complete question is

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm^3 of air with each breath, approximately what volume of air (in cubic meters) do these astronauts breathe in a year? (b) What would the diameter (in meters) of the space station have to be to contain all this air?

The average breathing rate is 12 breaths per minute

there are 60 minutes x 24 hours x 365 days in a year = 525600 minutes in a year

if an average human takes 12 breath per minute, then in a year an average human take 12 x 525600 = 6307200 breath

For the four astronauts, the amount of breath = 4 x 6307200 = 25228800 breath in total.

The volume of air per breath = 500 cm^3

1 cm^3 = 10^-6 m^3

500 cm^3 = $x$ m^3

$x$ = 500 x 10^-6 = 5 x 10^-4 m^3

Therefore in a year, the volume of these astronauts breath in a year = 5 x 10^-4 x 25228800 = 12614.4 m^3

b) If the space station is spherical, the volume will be given as = $\frac{4}{3} \pi r^3$

where r is the radius of the space station

This volume of the space station must be able to contain all the volume of breath produced by the astronauts which is = 12614.4 m^3

Equating, we have

12614.4 = $\frac{4}{3} \pi r^3$

12614.4 = $\frac{4}{3}*3.142*r^3$

12614.4 = 4.1893 $r^{3}$

$r^{3}$ = 12614.4/4.1893 = 3011.1

r = $\sqrt[3]{3011.1}$ = 14.4 m

diameter of the space station = 14.4 m x 2 = 28.8 m

3 0

4 years ago

A wheel of moment of inertia of 5.00 kg∙m2 starts from rest and accelerates under a constant torque of 3.00 N∙m for 8.00 s. What

Strike441 [17]

Answer:

The wheel's rotational kinetic energy is 57.6 J.

Explanation:

Given that,

Moment of inertia = 5.00 kg.m²

Torque = 3.00 N.m

Time = 8.00 s

We need to calculate the angular acceleration

Using formula of the torque act on the wheel

$\tau=I\alpha$

$\alpha=\dfrac{\tau}{I}$

Where, I = moment of inertia

$\alpha$ = angular acceleration

$\tau$ = torque

Put the value into the formula

$\alpha=\dfrac{3.00}{5.00}$

$\alpha=0.6\ rad/s^2$

We need to calculate the final angular velocity

Initially wheel at rest so initial velocity is zero.

Using formula of angular velocity

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}$

$\omega_{f}=\omega_{i}+\alpha t$

Put the value into the formula

$\omega_{f}=0+0.6\times8.00$

$\omega_{f}=4.8\ rad/s$

We need to calculate the rotational kinetic energy of the wheel

Using formula of the rotational kinetic energy

$K.E_{rot}=\dfrac{1}{2}I\omega^2$

$K.E_{rot}=\dfrac{1}{2}\times5.00\times(4.8)^2$

$K.E_{rot}=57.6\ J$

Hence, The wheel's rotational kinetic energy is 57.6 J.

6 0

3 years ago