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3 years ago

Mention 5 uses of methane

Chemistry

1 answer:

WARRIOR [948]3 years ago

8 0

Answer:

Used as rocket fuel, is a fertilizer ingredient, used in cooking, used to provide lighting, and used in production of other compounds

Explanation:

Hope this helps!!!

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HELP PLZ ( 10 points )

marta [7]

compressibility - is a measure of the relative volume change of a fluid or solid as a response to a pressure (or mean stress) change. (D)

Flammability - ability of a chemical to burn or ignite, causing fire or combustion. (B)

Heat of Combustion - The amount of heat released per unit mass or unit volume of a substance when the substance is completely burned. (A)

Reactivity - the state or power of being reactive or the degree to which a thing is reactive (C)

Hope this helps!!

4 0

3 years ago

Sound waves are mechanical waves in which the particles in the medium vibrate in a direction parallel to the direction of energy

Marat540 [252]

Longitudinal wave is the another name.

3 0

4 years ago

We would like to estimate how quickly the non-uniformities in gas composition in the alveoli are damped out. Consider an alveolu

Vera_Pavlovna [14]

Answer: It will take 11.775 seconds.

Explanation: As a sphere with a diameter of 0.1 mm, the area of an alveolus is

A = 4.π.r²

r for an alveolus would be: r = 0.00005m or r = 5. $10^{-5}$ m

Finding the area:

A = 4.3.14.(5. $10^{-5}$ )²

A = 3.14. $10^{-8}$ m²

The concentration change is to be 90% of the final, so

c = 0.9.3.14. $10^{-8}$

c = 28.26. $10^{-9}$

The oxygen diffusivity is 2.4. $10^{-9}$ m²/s, that means in 1 second 2.4. $10^{-9}$ of oxygen spread in one alveolus area. So:

1 second = 2.4. $10^{-9}$ m²

t seconds = 28.26. $10^{-9}$ m²

t = $\frac{28.26.10^{-9} }{2.4.10^{-9} }$

t = 11.775s

For a concentration change at the center to be 90%, it will take 11.775s.

4 0

3 years ago

What is the only thing held constant in a combined gas law problem?

vodomira [7]

Only the amount of gas is held constant.

3 0

3 years ago

A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

4 years ago

Mention 5 uses of methane​

Mention 5 uses of methane