They gain energy and move faster
Answer:
When the big power lines are down, electricity is going through the ground and up you bike to the handles, so when yo touch it, you are electrocuted
Explanation:
Answer:
The atomic mass is 14, which is the combined mass of protons and neutrons. The atomic number is the number of protons. SO if we have a total mass of 14 and we know 7 of that is protons, 14-7=7
the nitrogen has 7 neutrons
Answer:
1.5 atm is the pressure of dinitrogen tetroxide after equilibrium is reached the second time.
Explanation:
![2NO_2(g)\rightleftharpoons N_2O_4(g)](https://tex.z-dn.net/?f=2NO_2%28g%29%5Crightleftharpoons%20N_2O_4%28g%29)
initially
4.9 atm 0
At equilbrium:
(4.9-2p) p
Pressure of nitrogen dioxide gas at equilibrium = 2.7 atm
So, (4.9-2p) = 2.7 atm
p = 1.1 atm
The value of equilibrium constant
:
![K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7Bp_%7BN_2O_4%7D%7D%7B%28p_%7BNO_2%7D%29%5E2%7D)
![K_p=\frac{p}{(4.9-2p)^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7Bp%7D%7B%284.9-2p%29%5E2%7D)
![K_p=\frac{1.1 atm}{(4.9-2\times 1.1)^2}=0.1509](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B1.1%20atm%7D%7B%284.9-2%5Ctimes%201.1%29%5E2%7D%3D0.1509)
After addition of 1.2 atm of nitrogen dioxide gas more to the flask,will the reestablish an equilibrium in flask.
![2NO_2(g)\rightleftharpoons N_2O_4(g)](https://tex.z-dn.net/?f=2NO_2%28g%29%5Crightleftharpoons%20N_2O_4%28g%29)
(2.7+1.2) atm 1.1 atm
At second equilibrium
(2.7+1.2-2x) atm (1.1+x) atm
The expression of
:
![K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7Bp_%7BN_2O_4%7D%7D%7B%28p_%7BNO_2%7D%29%5E2%7D)
![0.1509=\frac{(1.1+x)}{(2.7+1.2-2x)^2}](https://tex.z-dn.net/?f=0.1509%3D%5Cfrac%7B%281.1%2Bx%29%7D%7B%282.7%2B1.2-2x%29%5E2%7D)
Solving for x:
x = 0.3827 atm
The pressure of dinitrogen tetroxide after equilibrium is reached the second time:
(1.1+x) atm = (1.1+0.3827) atm =1.4827 atm ≈ 1.5 atm
1.5 atm is the pressure of dinitrogen tetroxide after equilibrium is reached the second time.
Doubling the mass doubles the energy, while doubling the velocity quadruples it. Your question is basically about order of operations; the exponent only applies to the variable it's immediately on. As you note, you'd have to put parentheses to make it cover the
m as well. We say that energy is linear in mass, but quadratic in velocity.