Answer:
Well they didn't transfer any energy when they weren’t touching and it did t produce any energy if it didn’t move. Since they are on top of each other they are causing momentum on each other creating kinetic energy
Explanation:
<h3>Answer:</h3>
Platoic Acid
<h3>Explanation:</h3>
While naming Carboxylic Acids we know that when the Carboxylic Acid looses proton it is converted into corresponding conjugate base called as Carboxylate.
Examples:
HCOOH → HCOO⁻ + H⁺
Formic acid Formate Ion
H₃CCOOH → H₃CCOO⁻ + H⁺
Acetic acid Acetate Ion
H₅C₂COOH → H₅C₂COO⁻ + H⁺
Propanoic acid Propanoate Ion
Therefore, if the conjugate base is Platoate then the corresponding acid will be Platoic Acid means we will replace the -ate by -ic acid <em>i.e.</em>
RCOO⁻ + H⁺ → RCOOH
Platoate Ion Platoic Acid
Answer:
VP as function of time => VP(Ar) > VP(Ne) > VP(He).
Explanation:
Effusion rate of the lighter particles will be higher than the heavier particles. That is, the lighter particles will leave the container faster than the heavier particles. Over time, the vapor pressure of the greater number of heavier particles will be higher than the vapor pressure of the lighter particles.
=> VP as function of time => VP(Ar) > VP(Ne) > VP(He).
Review Graham's Law => Effusion Rate ∝ 1/√formula mass.
<span>The steps of solubility of water in N-butanol is as follows:1. N-butanol molecules are attracted to the surface of the water, 2. N-butanol molecules surround water molecules, 3. Butanol mixes with water and 4. Water molecules are carried into N-butanol.</span>
Answer:
The new molar concentration of CO at equilibrium will be :[CO]=1.16 M.
Explanation:
Equilibrium concentration of all reactant and product:
![[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M](https://tex.z-dn.net/?f=%5BCO_2%5D%20%3D%200.24%20M%2C%20%5BH_2%5D%20%3D%200.24%20M%2C%20%5BH_2O%5D%20%3D%200.48%20M%2C%20%5BCO%5D%20%3D%200.48%20M)
Equilibrium constant of the reaction :
![K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BH_2O%5D%5BCO%5D%7D%7B%5BCO_2%5D%5BH_2%5D%7D%3D%5Cfrac%7B0.48%20M%5Ctimes%200.48%20M%7D%7B0.24%20M%5Ctimes%200.24%20M%7D)
K = 4
Concentration at eq'm:
0.24 M 0.24 M 0.48 M 0.48 M
After addition of 0.34 moles per liter of 
and 
are added.
(0.24+0.34) M (0.24+0.34) M (0.48+x)M (0.48+x)M
Equilibrium constant of the reaction after addition of more carbon dioxide and water:
Solving for x: x = 0.68
The new molar concentration of CO at equilibrium will be:
[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M
