3.mixture
4. element, or an compound (its in order)
6. solvent
I hope these answers help you!
We calculate the entropy of an ideal gas follows:
<span>For an isothermal compression, change in internal energy is equal to zero.</span>
<span>Thus, the heat added to the gas is equal to the work done on the gas which is given as 1750 J.</span>
<span>Entropy would be 1750/301 = 5.81 J/K </span>
Answer:
The
expression for the weak base equilibrium is:
![K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
Explanation:

The expression of the equilibrium constant of base
can be given as:
![K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%5BH_2O%5D%7D)
]![K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%20%5BH_2O%5D%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
As we know, water is pure solvent, we can put ![[H_2O]=1](https://tex.z-dn.net/?f=%5BH_2O%5D%3D1)
![K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%201%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
So, the the
expression for the weak base equilibrium is:
![K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
Answer:
bleh
Explanation:
Neutralization reaction between magnesium hydroxide and hydrochloric acid Mg(OH)2(s) + 2HCl(aq) → 2H2O(l) + MgCl2(aq) b.
so uh pretty sure Mg(OH)2(s) + 2HCl(aq) → 2H2O(l) + MgCl2(aq) b.
correct me if im wrong <3
Answer: The given statement is true.
Explanation:
A catalyst is defined as the substance that lowers the activation energy of a chemical reaction without itself getting consumed so that there will be increase in rate of reaction.
As catalyst lowers the activation energy so, reactant molecules with lower energy also participates in the reaction. Hence, more collisions occur due to which there is rapid formation of products take place through another path.
Therefore, we can conclude that the statement a catalyst increases the rate of a reaction without being consumed. It accomplishes this by providing another mechanism that has a lower activation energy, is true.