For 2HCl(g) ==> H2(g) + Cl2(g) the Keq = [H2][Cl2]/[HCl]^2
The equilibrium constant, K_eq, for the reaction 2HCl(g) ⇌ H2(g) + Cl2(g) is equal to the concentration of H2 and Cl2 divided by the concentration of HCl squared, as products are in the numerator and reactants in the denominator.
The correct form for the equilibrium constant expression for this reaction is Kc = [HF]^2 / ([H2] * [F2]), where the square brackets denote molar concentrations of each species at equilibrium.
In the reaction ( \text{Mg} + 2 \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 + \text{H}_2 ), the oxidation state of each hydrogen atom in ( \text{H}_2 ) (denoted as ( \text{H}_2(g) )) is 0. This is because in diatomic molecules like ( \text{H}_2 ), the atoms are in their elemental form, and their oxidation state is defined as zero.
The reaction represented by H₂O → H₂(g) + O₂(g) is a decomposition reaction. In this reaction, water (H₂O) is broken down into its constituent elements, hydrogen gas (H₂) and oxygen gas (O₂), typically through the process of electrolysis or thermolysis. Decomposition reactions involve the separation of a compound into simpler substances.
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The equilibrium constant, K_eq, for the reaction 2HCl(g) ⇌ H2(g) + Cl2(g) is equal to the concentration of H2 and Cl2 divided by the concentration of HCl squared, as products are in the numerator and reactants in the denominator.
Keq = [H2O][CO] [H2][CO2]
Its 0
HI will be consumed. The reaction will proceed to the left. More I2 will form.
The correct form for the equilibrium constant expression for this reaction is Kc = [HF]^2 / ([H2] * [F2]), where the square brackets denote molar concentrations of each species at equilibrium.
The sample with the greatest mass at STP would be the one with the highest molar mass, as 1 mole of any substance at STP occupies the same volume (22.4 L). Among the given options, the sample with Cl2 gas (molar mass = 70.9 g/mol) would have the greatest mass.
In:(H2)g oxidation state: 0 In:(O2)g oxidation state: 0 In:(H2O)l oxidation state: H: +1 and O: -2
Yes, the equation obeys the law of conservation of matter. The number of atoms for each element is the same on both sides of the equation, indicating that no atoms are created or destroyed during the reaction.
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A. Of the choices listed, the helium atom is the smallest in both size and mass, and so its behavior is more "ideal." 1.He(g) Of the choices listed, the helium atom is the smallest in both size and mass, and so its behavior is more "ideal."
The oxidation half-reaction is: Fe => Fe+3 + 3e-, and the reduction half-reaction is: F2 + 2e- => 2 F-1. For a complete equation, the oxidation half-reaction as written must be multiplied by 2 and added to the reduction half-reaction as written multiplied by 3 to result in an overall reaction of 2 Fe + 3 F2 = 2 FeF3.