Answer:
There is more space between gas particles than the size of the particles.
Explanation:
This scenario can be understand by taking a very simple example. As we know that 1 mole of any gas at standard temperature and pressure occupy 22.4 liters of volume. Lets take Hydrogen gas and Oxygen gas, 1 mole of each gas will occupy same volume. Why it is so? Why same volume although Oxygen is 16 times more heavier? This is because the space between gas molecules is very large. Approximately the distance between gas molecules is 300 times greater than their own diameter from its neighbor molecules.
Answer:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
Explanation:
The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:
<u>Where:</u>
<em>μ (l): is the chemical potential of 2-propanol in solution </em>
<em>μ° (l): is the chemical potential of pure 2-propanol </em>
<em>R: is the gas constant = 8.314 J K⁻¹ mol⁻¹ </em>
<em>T: is the temperature = 82.3 °C = 355.3 K </em>
<em>x: is the mole fraction of 2-propanol = 0.41 </em>
Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.
I hope it helps you!
Answer:
<h2>number of moles of solute dissolved in one liter of solution. </h2>
1. 8.28
Explanation:
I'm not sure but I hope it's help
The rate constant of the reaction K we can get it from this formula:
K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B
So when after 10 s and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) = 16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14
