When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant.
Let's do the ICE analysis
2 NH₃ ⇄ N₂ + 3 H₂
I 0 1.3 1.65
C +2x -x -3x
-------------------------------------
E 0.1 ? ?
The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:
Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
Therefore,
Equilibrium H₂ = 1.65 - 3(0.05) = 1.5 molEquilibrium N₂ = 1..3 - 0.05 = 1.25 mol
For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is
![K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}](https://tex.z-dn.net/?f=%20K_%7BC%7D%20%3D%20%5Cfrac%7B%5B%20C%5E%7Bc%7D%20%5D%7D%7B%5B%20A%5E%7Ba%7D%20%5D%5B%20B%5E%7Bb%7D%20%5D%7D%20)
Hence, Kc could vary depending on the stoichiometric coefficients of the reaction.
I think its C i don't know if i am right
When water reaches its boiling point and turns into water vapor, the molecular structure of water remains the same. It is only the state of the substance that is changed in this process. I hope this helps you
I'd go for D here. It also fits in with the idea of thermal expansion - as something is heated up, molecules vibrate and maybe collide. they vibrate with bigger amplitudes, so taking up more space, so expanding. maybe
The chemical reaction would be as follows:
<span>2Na + S → Na2S
We are given the amount of the reactants to be used in the reaction. We use these to calculate the amount of product. We do as follows:
45.3 g Na ( 1 mol / 22.99 g ) = 1.97 mol Na
105 g S ( 1 mol / 32.06 g ) = 3.28 mol S
The limiting reactant would be Na. We calculate as follows:
1.97 mol Na ( 1 mol Na2S / 2 mol Na ) (78.04 g / mol ) = 76.87 g Na2S produced</span>