Answer:
a) .66
Explanation:
from (0,4) to (3,6)
the slope would be
[tex] \frac{2}{3} [/tex]
when diving the fraction the half slope is .66666...
What is the average velocity of atoms in 2.00 mol of neon (a monatomic gas)
at 308 K? Use the equation: -mv2
2
For m, use 0.02000 kg. Remember that R = 8.31 J/(mol-K).
= ER
3
2
nRT
A. 1540 m/s
B. 876 m/s
C. 87.6 m/s
O D. 15,400 m/s
Answer:
v = 876 m/s
Explanation:
It is given that,
Number of mol of Neon is 2 mol
Temperature, T = 308 K
Mass, m = 0.02 kg
Value of R - 8.31 J/mol-K
We need to find the average velocity of atoms in 2.00 mol of neon. Neon is a monoatomic gas. Let v is the velocity. So,
[tex]\dfrac{1}{2}mv^2=\dfrac{3}{2}nRT\\\\v=\sqrt{\dfrac{3nRT}{m}} \\\\v=\sqrt{\dfrac{3\times 2\times 8.314\times 308}{0.02}} \\\\v=876.47\ m/s[/tex]
So, the correct option is (B).