All categories

3 years ago

Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 31.5°c.

1 answer:

8 0

The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant = 8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec

You might be interested in

The path a star takes during its lifetime depends on

Svetlanka [38]

Answer:

c.) Mass (is the answer)

7 0

3 years ago

I need help fast. what is the definition of Kinetic molecular model

seraphim [82]

It state that the average kinetic energy from a gas particle depends only on the temperature of the gas

8 0

3 years ago

Good evening! Can someone please answer this, ill give you brainliest and your earning 50 points. Would be very appreciated.

BARSIC [14]

Answer:

Gases that are very good at absorbing long wave photons of infrared light

Explanation:

Green house gases examples are Carbon monoxide, nitric oxide etc.
These gases absorb long wave photons of infrared light of sun's rays
They melts the glaciers and increase the sea level .

7 0

2 years ago

What are the characteristics of image when object is between f1 and 2f1 for concave lense?

k0ka [10]

Answer:

The required diagram is shown in the figure. When an object is placed in front of the convex lens, i.e., between 2F

and F

, its image is formed beyond 2F

on the other side of the lens. The image is real, inverted and enlarged.

solution

5 0

3 years ago

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$