Answer:
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
Explanation:
Assuming Helium behaves like an ideal gas and temperature is constant.
According to Boyle's law for ideal gases, at constant temperature,
P₁V₁ = P₂V₂
P₁ = 26 atm
V₁ = 19.0 L
P₂ = 1 atm (the balloon is said to expand till the pressure matches the pressure of the atmpsphere; and the pressure of the atmosphere is 1 atm)
V₂ = ?
P₁V₁ = P₂V₂
(26 × 19) = 1 × V₂
V₂ = 494 L (it is assumed the balloon never bursts)
The new volume of the balloon when the pressure equalised with the pressure of the atmosphere = 494 L.
The balloon expands by am additional 475 L.
Hope this Helps!!!
Answer:
The De Broglie wavelength decreases when the momentum increases
Explanation:
The De Broglie wavelength of a particle (or any object) is given by
where
h is the Planck constant
p is the momentum of the object
As we can see, the wavelength is inversely proportional to the momentum of the object: therefore we can say that, if the momentum increases, the De Broglie wavelength will decrease.
Answer: 1.00 M
Explanation:
Moles of Solute/Liters of Solution = molarity. Then you would do 0.500 Moles/ 0.500 L and you get 1.00 M.
I will have to say that this statement is true
