Answer:
False
Step-by-step explanation:
The trough-to-trough distance is the <em>wavelength</em> (λ) of a wave.
The <em>amplitude</em> (a) of a wave is half the peak-to-trough distance. It is also the distance from the trough to the rest position.
Molar mass O₂ = 16 x 2 = 32.0 g/mol
1 mol O₂--------------------------- 32.0 g
30.5 moles O₂-------------------- ? ( mass O₂)
mass O₂ = ( 30.5 ) x 32.0 / 1
mass O₂ = 976.0 / 1
= 976.0 grams of O₂
hope this helps!
a gas behaves more like an ideal gas at higher temperature and lower pressure, as the potential energy due to intermolecular forces becomes less significant compared with the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them.
Part 1 (non-ideal behavior ):
we will use Van der Waals formula:
P = (nRT/(V-nb)) - (n^2 a/V2)
when n (moles of H2) = 18 g / 2 = 9 moles
R (constant) = 0.0821
T(tempreature in kelvin = 20 +273 = 293 K
V(Volume) = 1 L
a (constant for H2) = 0.2476
b(constant for H2) = 0.02661
So by substitution:
∴P = (9*0.0821*293/(1-(9*0.02661))) - (9^2*0.2476/1)
∴P ≈264.6 atm
Part 2 (ideal behavior):
we will use the ideal gas formula :
PV = nRT
when we have n = 9
R= 0.0821
T=293 K
V= 1 L
∴P = (9*0.0821*293)/1 L
∴P = 216.5 atm