Answer:
2CO₂ ⇄ 2CO + O₂.
Explanation:
- <em>The equilibrium constant (Keq)</em> for a reaction is the ratio of the concentration of the products to the reactants, each one is raised to each molar coefficient.
So, for Keq = [CO]²[O₂]/[CO₂]²,
It is clear that the products are CO and O₂, while the reactant is CO₂.
So, the reaction for this is:
<em>2CO₂ ⇄ 2CO + O₂.</em>
Answer:
See explanation
Explanation:
Potassium is a metal. Remember that metals are electropositive in nature. This implies that they give out electrons when they undergo ionic bonding with nonmetals.
Chlorine is a nonmetal. So the bond between chlorine and potassium is ionic. Potassium gives out one electron to chlorine and the both ions now attain a stable octet.
The electronic configuration of potassium is [Ar]4s1. After giving out an electron to chlorine in an ionic bond, its electronic configuration is now [Ar].
Answer: The enthalpy of formation of 
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of
The chemical equation for the combustion of propane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of is -396 kJ/mol
Answer:
0.32M
Explanation:
<u>Step 1:</u> Balance the reaction
K2CO3 + Ba(NO3)2 ⇔ KNO3 + BaCO3
We have a 20 mL 0.2 M K2CO3 and a 30mL 0.4M Ba(NO3)2 solution
SinceK2CO3 is the limiting reactant, there will remain Ba(NO3)2 after it's consumed and produced KNO3 + BaCO3
<u>Step 2: </u>Calculate concentration
To find the concentration of the barium cation we use the following equation:
Concentration = moles of the <u>solute</u> / volumen of the <u>solution</u>
<u />
<u>[Ba2+] </u> = (20 * 10^-3 * 0.2M + 30 * 10^-3 * 0.4M) / ( 20 + 30mL) *10^-3
[Ba2+] = 0.32 M
The concentration of Barium ion in solution is 0.32 M
Answer:
Explanation:
<u>1) Equilibrium equation (given):</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
<u>2) Write the concentration changes when some concentration, A, of CH₂Cl₂ (g) sample is introduced into an evacuated (empty) vessel:</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
A - x x x
<u>3) Replace x with the known (found) equilibrium concentraion of CCl₄ (g) of 0.348 M</u>
- 2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)
A - 0.3485 0.348 0.348
<u>4) Write the equilibrium constant equation, replace the known values and solve for the unknown (A):</u>
- Kc = [ CH₄ (g) ] [ CCl₄ (g) ] / [ CH₂Cl₂ (g) ]²
- A² = 56.0 / 0.348² = 462.
