no...the atoms will not behave the same
as when temperature is increased, the atoms vibration and kinetic energy will also be increased....they come in excited state...
where as when temperature is reduced ,atoms kinetic energy slows down....
Answer:
https://www.getriddles.com/science-riddles/
Explanation:
go to this link, you will find some answers
hope it helps!
Warmer air is less dense than cold air.As air warm it rises while the cold air sink. Warmer air masses forces the cooler air to move which causes wind. These is illustrated when you open a hot oven the hotter air inside the oven rises into cooler air outside the oven.
Answer:
1) Oil is less dense than water so when oil spills, it spreads across the entire water surface.
2) The oil spreads very quickly with lighter oils such as gasoline.
3) Wind, Currents, and Warm Temperatures will cause Oil to spread quicker.
Answer:
189.2 KJ
Explanation:
Data Given
wavelength of the light = 632.8 nm
Convert nm to m
1 nm = 1 x 10⁻⁹
632.8 nm = 632.8 x 1 x 10⁻⁹ = 6.328 x 10⁻⁷m
Energy of 1 mole of photon = ?
Solution
Formula used
E = hc/λ
where
E = energy of photon
h = Planck's Constant
Planck's Constant = 6.626 x 10⁻³⁴ Js
c = speed of light
speed of light = 3 × 10⁸ ms⁻¹
λ = wavelength of light
Put values in above equation
E = hc/λ
E = 6.626 x 10⁻³⁴ Js ( 3 × 10⁸ ms⁻¹ / 6.328 x 10⁻⁷m)
E = 6.626 x 10⁻³⁴ Js (4.741 x 10¹⁴s⁻¹)
E = 3.141 x 10⁻¹⁹J
3.141 x 10⁻¹⁹J is energy for one photon
Now we have to find energy of 1 mole of photon
As we know that
1 mole consists of 6.022 x10²³ numbers of photons
So,
Energy for one mole photons = 3.141 x 10⁻¹⁹J x 6.022 x10²³
Energy for one mole photons = 1.89 x 10⁵ J
Now convert J to KJ
1000 J = 1 KJ
1.89 x 10⁵ J = 1.89 x 10⁵ /1000 = 189.2 KJ
So,
energy of one mole of photons = 189.2 KJ
