An intensive property is a property that does not change depending on how much mass of it you are considered. An example of an intensive property is density. No matter how much water you examine, the density of the sample will be 1g/cm³.
Answer:
2.67 × 10⁻²
Explanation:
Equation for the reaction is expressed as:
CaCrO₄(s) ⇄ Ca₂⁺(aq) + CrO₂⁻⁴(aq)
Given that:
Kc=7.1×10⁻⁴
Kc= ![[Ca^{2+}][CrO^{2-}_4]](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%5D%5BCrO%5E%7B2-%7D_4%5D)
Kc= [x][x]
Kc= [x²]
7.1×10⁻⁴ = [x²]
x = 
x = 0.0267
x = 
Answer:
0.11%
Explanation:
Without mincing words, let us dive straight into the solution to the question/problem. The first step to solve this question is to write out the chemical reaction, that is the reaction showing the dissociation of acetic acid.
CH3COOH <=======================================> CH3COO⁻ + H⁺
Initially, the amount present in the acetic acid which is = 12M, the concentration for CH3COO⁻ and H⁺ is 0 respectively.
At equilibrium, the amount present in the acetic acid which is = 12 - x, the concentration for CH3COO⁻ = x and H⁺ = x respectively. Note that the ka for acetic acid = 1.8 × 10⁻⁵.
1.8 × 10⁻⁵ = x²/ 14 - x. Therefore, x = 0.0158 M.
The next thing to do is to calculate for the percentage of dissociation, this can be done as given below:
percentage of dissociation = x/14 × 100. Recall that the value that we got for x = 0.0158 M. Hence, the percentage of dissociation = 0.0158 M/ 14m × 100 = 0.11%