Answer : The equilibrium concentration of 
in the solution is, 
Explanation :
The dissociation of acid reaction is:
Initial conc. c 0 0
At eqm. c-x x x
Given:
c = 
The expression of dissociation constant of acid is:
![K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_6H_5COO%5E-%5D%7D%7B%5BC_6H_5COOH%5D%7D)
Now put all the given values in this expression, we get:
![6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}](https://tex.z-dn.net/?f=6.3%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%5B%287.0%5Ctimes%2010%5E%7B-2%7D%29-x%5D%7D)
Thus, the equilibrium concentration of in the solution is,
In an experiement things that are changing are called variables.
Answer:
58.94 mL
Explanation:
V1 = 48.3 mL V2 = v mL
T1 = 22 degree celsius OR 295 k T2 = 87 degree celsius OR 360 k
We will use the gas equation:
PV = nRT
Since the Pressure (p) , number of moles (n) and the universal gas constant(R) are all constants in this given scenario,
we can say that
V / T = k , (where k is a constant)
Since this is the first case,
V1 / T1 = k --------------------(1)
For case 2:
Since we have the same constants, the equation will be the same
V / T = k (where k is the same constant from before)
V2 / T2 = k (Since this is the second case) ------------------(2)
From (1) and (2):
V1 / T1 = V2 / T2
Now, replacing the variables with the given values
48.3 / 295 = v / 360
v = 48.3*360 / 295
v = 58.94 mL
Therefore, the final volume of the gas is 58.94 mL
Answer:
3 moles
Explanation:
3 moles atoms are needed to react with 50 grams of O2 gas.
