Answer:
76.0%
Explanation:
Let's consider the following reaction.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
At equilibrium, the equilibrium constant Kp is:
Kp = 1.16 = pCO₂ ⇒ pCO₂ = 1.16 atm
We can calculate the moles of CO₂ at equilibrium using the ideal gas equation.
![P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1.16atm\times 14.4 L}{(0.08206atm.L/mol.K)\times 1073K} =0.190mol](https://tex.z-dn.net/?f=P.V%3Dn.R.T%5C%5Cn%3D%5Cfrac%7BP.V%7D%7BR.T%7D%20%3D%5Cfrac%7B1.16atm%5Ctimes%2014.4%20L%7D%7B%280.08206atm.L%2Fmol.K%29%5Ctimes%201073K%7D%20%3D0.190mol)
From the balanced equation, we know that 1 mole of CO₂ is produced by 1 mole of CaCO₃. Taking into account that the molar mass of CaCO₃ is 100.09 g/mol, the mass of CaCO₃ that reacted is:
![0.190molCO_{2}.\frac{1molCaCO_{3}}{1molCO_{2}} .\frac{100.09gCaCO_{3}}{1molCaCO_{3}} =19.0gCaCO_{3}](https://tex.z-dn.net/?f=0.190molCO_%7B2%7D.%5Cfrac%7B1molCaCO_%7B3%7D%7D%7B1molCO_%7B2%7D%7D%20.%5Cfrac%7B100.09gCaCO_%7B3%7D%7D%7B1molCaCO_%7B3%7D%7D%20%3D19.0gCaCO_%7B3%7D)
The percentage by mass of the CaCO₃ that reacted to reach equilibrium is:
![\frac{19.0g}{25.0g} \times 100\%=76.0\%](https://tex.z-dn.net/?f=%5Cfrac%7B19.0g%7D%7B25.0g%7D%20%5Ctimes%20100%5C%25%3D76.0%5C%25)
Explanation:
About which substance you are talking?
Answer:
.
Start color: yellowish-green.
End color: dark purple.
Assumption: no other ion in the solution is colored.
Explanation:
In this reaction, chlorine gas
oxidizes iodine ions
to elemental iodide
. At the same time, the chlorine atoms are converted to chloride ions
.
Fluorine, chlorine, bromine, and iodine are all halogens. They are all found in the 17th column of the periodic table from the left. One similarity is that their anions are not colored. However, their elemental forms are typically colored. Besides, moving down the halogen column, the color becomes darker for each element.
Among the reactants of this reaction,
is colorless. If there's no other colored ion, only the yellowish-green hue of
would be visible. Hence the initial color of the reaction would be the yellowish-green color of
.
Similarly, among the products of this reaction,
is colorless. If there's no other colored ion, only the dark purple hue of
would be visible. Hence the initial color of the reaction would be the dark purple color of
.