Potassium and calcium fluoride are both metals
Answer:
Kc = 3.72 × 10⁶
Explanation:
Let's consider the following reaction:
NH₄HS(g) ⇄ NH₃(g) + H₂S(g)
At equilibrium, we have the following concentrations:
[NH₄HS] = 0.196 M (assuming a 1 L flask)
[NH₃] = 9.56 × 10² M
[H₂S] = 7.62 × 10² M
We can replace this data in the Kc expression.
![Kc=\frac{[NH_{3}] \times [H_{2}S] }{[NH_{4}HS]} =\frac{9.56 \times 10^{2} \times 7.62 \times 10^{2}}{0.196} =3.72 \times 10^{6}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNH_%7B3%7D%5D%20%5Ctimes%20%5BH_%7B2%7DS%5D%20%7D%7B%5BNH_%7B4%7DHS%5D%7D%20%3D%5Cfrac%7B9.56%20%5Ctimes%2010%5E%7B2%7D%20%20%5Ctimes%207.62%20%20%5Ctimes%2010%5E%7B2%7D%7D%7B0.196%7D%20%3D3.72%20%5Ctimes%2010%5E%7B6%7D)
Answer:
1. The dye that absorbs at 530 nm.
Explanation:
The dye will absorb light to promote the transition of an electron from the HOMO to the LUMO orbital.
The higher the gap, the higher the energy of transition. The energy can be calculated by E = hc/λ, in which h and c are constants and λ is the wavelength.
The equation shows that the higher the energy, the higher the gap and the lower the wavelength.
Therefore, the dye with absorption at 530 nm has the higher HOMO-LUMO gap.
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
