Answer:
Kc = 77.9
Explanation:
To solve this equilibrium problem we will use an ICE Chart. We recognise 3 stages: Initial (I), Change (C) and Equilibrium (E). We complete each row with the <em>concentration or change of concentration in that stage</em>. Since the container is of 1.00 L, the initial concentrations are [NO] = 0.103 M and [Br₂] = 9.75 × 10⁻² M and the equilibrium concentration of Br₂ is 6.21 × 10⁻² M. Then,
2 NO(g) + Br₂(g) ⇄ 2 NOBr(g)
I 0.103 9.75 × 10⁻² 0
C -2x -x +2x
E 0.103 -2x 9.75 × 10⁻² - x 2x
Also, we know that
[Br₂]eq = 9.75 × 10⁻² - x = 6.21 × 10⁻² ⇒ x = 3.54 × 10⁻² M
We can use the value of x to find the concentrations at equilibrium:
[NO] = 0.103 -2x = 0.103 - 2 . 3.54 × 10⁻² = 3.22 × 10⁻² M
[Br₂] = 6.21 × 10⁻² M
[NOBr] = 2x = 2 . 3.54 × 10⁻² = 7.08 × 10⁻² M
We can use these concentrations in the equilibrium constant (Kc) expression.
![Kc=\frac{[NOBr]^{2} }{[NO]^{2}.[Br_{2}] } =\frac{(7.08 \times 10^{-2} )^{2} }{(3.22 \times 10^{-2})^{2}.6.21 \times 10^{-2} } =77.9](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%20%7D%7B%5BNO%5D%5E%7B2%7D.%5BBr_%7B2%7D%5D%20%7D%20%3D%5Cfrac%7B%287.08%20%5Ctimes%2010%5E%7B-2%7D%20%20%29%5E%7B2%7D%20%7D%7B%283.22%20%5Ctimes%2010%5E%7B-2%7D%29%5E%7B2%7D.6.21%20%5Ctimes%2010%5E%7B-2%7D%20%7D%20%3D77.9)
Answer:
A full moon occurs when the Moon appears as a complete circle in the sky.
Explanation:
- We see it as a full orb because the whole of the side of the Moon facing the Earth is lit up by the Sun's rays.
- The moon shows no visible light of its own, so we only see the parts of the moon that are lit up by other objects.
Answer:
d
Explanation:
Isomers are molecules that have the same molecular formula(the same number of each atom), but different structural formulas
in this case both have the formula C2H5O
D.) Atoms are always in motion