Answer:
the molecular formula for the gas is NO₂
Explanation:
since it contains
Nitrogen = n → 30.45%
Oxygen = o → 69.55%
and 30.45%+69.55% = 100% , then the gas only contains nitrogen and oxygen
Also we know that the proportion of oxygen over nitrogen is
proportion of oxygen over nitrogen = moles of oxygen / moles of nitrogen
since
moles = mass / molecular weight
then for a sample of 100 gr of the unknown gas
mass of oxygen = 69.55%*100 gr = 69.55 gr
mass of Nitrogen = 30.45%*100 gr = 30.45 gr
proportion of oxygen over nitrogen = (mass of oxygen/ molecular weight)/(mass of nitrogen / molecular weight of nitrogen ) = (69.55 gr/ 16 gr/mol) /( 30.45 gr /14 gr/mol) = 1.998 mol of O/ mol of N≈ 2 mol of O/ mol of N
therefore there are 2 atoms of oxygen per atom of nitrogen
thus the molecular formula for the gas is:
NO₂
12.5 times 14 and convert to meters its 1.75 meters per second
Because it reverses an image there for making the objects appear on opposite side
wavelength of the EM wave produced by your iclicker is 0.33 m.
<h3>What makes an EM wave?</h3>
- When an electric field (illustrated in red arrows) combines with a magnetic field, electromagnetic waves are generated (which is shown in blue arrows). An electromagnetic wave's magnetic and electric fields are perpendicular to each other and to the wave's direction.
- A changing magnetic field causes a changing electric field, and vice versa—the two are inextricably related. Electromagnetic waves are created by changing fields. Electromagnetic waves, unlike mechanical waves, do not require a medium to propagate.
The clicker emits EM (electromagnetic) wave which travels at the speed of light, that is
v = 3 x 10⁸ m/s
The frequency is
f = 900mHz = 9 x 10⁸ Hz
velocity = frequency * wavelength, the wavelength, λ, is given by
fλ = v
λ = v/f
= (3 x 10⁸ m/s) / (9 x 10⁸ 1/s)
= 1/3 m = 0.333 m
To learn more about electromagnetic waves refer,
brainly.com/question/25847009
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To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.
Mathematically the conservation of these two energies can be given through
Where,
W = Work
Final gravitational Potential energy
Initial gravitational Potential energy
When the spacecraft of mass m is on the surface of the earth then the energy possessed by it
Where
M = mass of earth
m = Mass of spacecraft
R = Radius of earth
Let the spacecraft is now in an orbit whose attitude is 
then the energy possessed by the spacecraft is
Work needed to put it in orbit is the difference between the above two
Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is
