Answer:
It is necessary because Trans-cinnamic is a limiting reagent in the mixture reaction while Bromine is the excess reagent
Explanation:
It is necessary to maintain excess bromine in the reaction mixture because Bromine is the excess reagent in the reaction mixture and if it's quantity is less it would consume the limiting reagent ( Trans-cinnamic ) completely . hence Bromine should maintain excess quantity in the reaction mixture
Burning paper because it can’t be changed back
<span>Total mass = 2.75g
Mass % of Mg = 1.0g x 100/2.75g = 36.36 %
Mass % of O = 1.75g x 100/2.75g = 63.64 %
Mol of Mg = 36.36/24 = 1.515
Mol of O = 63.64/16 = 3.977
Ratio of Mol of Mg and O in the substance = (1.515 : 3.977) x 2 = 3 : 8
The empirical formula of substance is Mg3O8</span>
Answer:
[HI] = 0.264M
Explanation:
Based on the equilibrium:
2HI(g) ⇄ H₂(g) + I₂(g)
It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:
<em>Kc = 0.0156 = [H₂] [I₂] / [HI]²</em>
<em />
As initial concentration of HI is 0.660mol / 2.00L = <em>0.330M, </em>the equlibrium concentrations will be:
[HI] = 0.330M - 2X
[H₂] = X
[I₂] = X
<em>Where X is reaction coefficient.</em>
<em />
Replacing in Kc:
0.0156 = [X] [X] / [0.330M - 2X]²
0.0156 = X² / [0.1089 - 1.32X + 4X²
]
0.00169884 - 0.020592 X + 0.0624 X² = X²
0.00169884 - 0.020592 X - 0.9376 X² = 0
Solving for X:
X = - 0.055 → False solution, there is no negative concentrations
X = 0.0330 → Right solution.
Replacing in HI formula:
[HI] = 0.330M - 2×0.033M
<h3>[HI] = 0.264M</h3>
