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%
Astronomers use chemical signatures to determine<span> the age and .... </span>Gas<span> mixtures that contain more than </span>4<span>% hydrogen in </span>air<span> are potentially explosive. ... water vapor, </span>carbon dioxide<span>, and several other </span><span>gases</span>
Answer:
Noble gases.
Good Luck & Have a good day!
(.❛ ᴗ ❛.)
You first need to write the balanced chemical reaction for what is going on.
Ca(OH)₂+2HCl→2H₂O+CaCl₂
After you make the balanced chemical reaction, First you find the moles of HCl used. To do this multiply 0.0375L by 0.124M to get 0.00465mol HCl. Then you multiply 0.00465mol HCl by (1mol Ca(OH)₂)/(2mol HCl) to get 0.002325mol Ca(OH)₂. Finally to find concentration of Ca(OH)₂ used you divide 0.002325mol by 0.020L to get 0.116M Ca(OH)₂.
Therefore the concentration of the unknown solution of Ca(OH)₂ was 0.116M.
I hope this helps. Let me know if anything is unclear.
