Answer:
[CO] = 0.078M
[Cl2] = 0.078M
[COCl2] = 0.477M
Explanation:
Based on the reaction:
CO(g) Cl2(g) ⇄ COCl2(g)
<em>Where equilibrium constant, kc, is:</em>
kc = 77.5 = [COCl2] / [CO] [Cl2]
[] represents the equilibrium concentration of each gas. The initial concentration of each gas is:
[CO] = 0.555mol/1.00L = 0.555M
[Cl2] = 0.555M
And equilibrium concentrations are:
[CO] = 0.555M - x
[Cl2] = 0.555M - x
[COCl2] = x
<em>Where x is reaction coordinate</em>
Replacing in kc expression:
77.5 = [x] / [0.555M - x] [0.555M - x]
77.5 = x / 0.308025 - 1.11 x + x²
23.8719 - 86.025 x + 77.5 x² = x
23.8719 - 87.025 x + 77.5 x² = 0
x = 0.477M. Right answer
x = 0.646M. False answer. Produce negative concentrations
Replacing:
<h3>[CO] = 0.555M - 0.477M = 0.078M</h3><h3>[Cl2] = 0.078M</h3><h3>[COCl2] = 0.477M</h3>
And those concentrations are the equilibrium concentrations
Im not 100% sure, but I think the answer is C. If not, Im sorry for bothering you.
Answer:
Sodium hydroxide solution.
Explanation:
because it is a strong base. Since carbonic acid is a weak acid, a solution formed resists change in pH on addition hence buffer formed.
You have to calculate the oxidation estates of the atoms in each compound.
I will start with K2Cr2O7 because I believe that Cr is the best candidate to reduce its oxidation number in 3 units.
In K2Cr2O7:
- K has oxidation state of 1+, then K2 has a charge of 2* (1+) = 2+.
- O has oxidation state of 2*, then O7 has a charge of 7* (2-) = 14-.
That makes that Cr2 has charge of 14 - 2 = +12, so each Cr has +12/2 = +6 oxidation state.
In Cr2O3:
- O has oxidation state of 2-, then O3 has charge 3 * (2-) = - 6
- Then, Cr2 has charge 6+, and each Cr has charge 6+ / 2 = 3+.
So, we have seen that Cr reduced its oxidation state in 3 units, from 6+ to 3+.
Answer: Cr has a change in oxidation number of - 3.
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
