In exothermic reactions, there is a release heat and the replacement of weak bonds with stronger ones.
Answer:
Electrons, with a charge of -1.
Explanation:
The nucleus of the atom is composed of protons and neutrons.
The nucleus is where the bulk of the mass of an atom is found.
Electrons orbit the nucleus in various energy levels.
Protons have a charge of +1, neutrons have no charge, and electrons have a charge of -1.
I think you mean:
".. reacts with oxygen gas to produce lead (II) oxide and 'sulfur' dioxide..."
The balanced chemical reaction would be
2PbS + 3O2 --> 2PbO + 2SO2
The stiochiometric ratio is this: 3 moles of O2 produces 2 moles of lead (II) oxide. So,
0.500 moles (2/3) = 0.333 moles
Therefore, 0.333 moles of PbO is produced when 0.500 moles of O2 reacts with PbS.
3.1 is not the same as 2.97
Answer:
![[H_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.183M)
![[I_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.183M)
![[HI]_{eq}=0.025M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D0.025M)
Explanation:
Hello.
In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

Whereas the law of mass action (equilibrium expression) is:
![Kc=\frac{[H_2][I_2]}{[HI]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2%5D%5BI_2%5D%7D%7B%5BHI%5D%5E2%7D)
That in terms of initial concentrations and reaction extent or change
turns out:
![Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7Bx%2Ax%7D%7B%28%5BHI%5D_0-2x%29%5E2%7D%5C%5C%5C%5C54.3%3D%5Cfrac%7Bx%5E2%7D%7B%280.391M-2x%29%5E2%7D)
And the solution via solver or quadratic equation is:

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:
![[H_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.183M)
![[I_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.183M)
![[HI]_{eq}=0.391M-2*0.183M=0.025M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D0.391M-2%2A0.183M%3D0.025M)
Regards.