This is a tricky question. All that matters are ratios of percentages, not percentages themselves. So no one should directly compare 27.2 with 42.9. We must and shall compare the ratios (27.2 to 72.8) and (42.9 to 57.1).
Take them both down to 1 to and see what happens.
Working out the formulas knowing atomic masses is a bit beside the point; this is how people first DISCOVERED the idea of atomic mass.
A
Carbon Oxygen
27.2g 72.8g (100-27.2)
Moles 27.2/12 72.8/16
2.27 4.55
Ratio 1 2
Do the same with the other
Answer:
Explanation:
An element whose atomic number is 68 (number Z), has 68 electrons and 68 protons. According to the periodic table it corresponds to the element Erbium (Er), whose mass number (A) is 167. The number of neutrons is 167-68 = 99.
This statement is true in fact to mount Everest atmospheric pressure sit at about one third of that at sea level.
Answer:
fundamental frequency in helium = 729.8 Hz
Explanation:
Fundamental frequency of an ope tube/pipe = v/2L
where v is velocity of sound in air = 340 m/s; λ is wave length of wave = 2L ; L is length of the pipe
To find the length of the pipe,
frequency = velocity of sound / 2L
272 = 340 / 2 L
L = 0.625 m
If the pipe is filled with helium at the same temperature, the velocity of sound will change as well as the frequency of note produced since velocity is directly proportional to frequency of sound.
Also, the velocity of sound is inversely proportional to square root of molar mass of gas; v ∝ 1/√m
v₁/v₂ = √m₂/m₁
v₁ = velocity of sound in air, v₂ = velocity of sound in helium, m₁ = molar mass of air, m₂ = molar mass of helium
340 / v = √4 / 28.8
v₂ = 340 / 0. 3727
v₂ = 912.26 m /s
fundamental frequency in helium = v₂ / 2L
fundamental frequency in helium = 912.26 / (2 x 0.625)
fundamental frequency in helium = 729.8 Hz
