A fluorine atom has 7 valence electrons in its second shell.

Does the sign of the charge (positive or negative) affect how that charge is dissipated? Explain.

labwork [276]

This is an add proceeded by

5 0

3 years ago

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

7 0

4 years ago

A 20 kg block rests on a rough horizontal table. A rope is attached to the block and is pulled with a force of 80 N to the left

statuscvo [17]

Explanation:

it will move to thr left?

4 0

3 years ago

An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45° relative to edge of the sidewalk. If it has

Marat540 [252]

Answer:14 s

Explanation:

Given

Velocity of ant is 0.1 m/s in a direction $45^{\circ}$

if it has traveled 1 m perpendicular to the edge of the sidewalk

i.e. from diagram

$\sin 45=\frac{1}{L}$

$L=\sqrt{2}$

$time=\frac{Distance}{speed}$

$t=\frac{\sqrt{2}}{0.1}$

$t=14.14 s$

3 0

3 years ago

A 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal exp

Gnom [1K]

Answer:

a.) 1567.2 m/s

b.) 149.4 m/s

Explanation:

Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.

The x-component of the third part can be calculated by assuming that it moves in a positive x axis.

The third mass = 26 - ( 7.8 + 8.8)

The third mass = 26 - 16.6

The third mass = 9.4kg

since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion

26 x 350 = -8.8 x 640 + 9.4V

9100 = -5632 + 9.4V

9.4V = 9100 + 5632

9.4V = 14732

V = 14732/9.4

V = 1567.2 m/s

(b) y-component of the velocity of the third part will be

7.8 x 180 = 9.4 V

1404 = 9.4V

V = 1404/9.4

V = 149.4 m/s

7 0

3 years ago