Answer:
Explanation:
Hello there!
In this case, given the solubilization of cadmium (II) hydroxide:
The solubility product can be set up as follows:
![Ksp=[Cd^{2+}][OH^-]^2](https://tex.z-dn.net/?f=Ksp%3D%5BCd%5E%7B2%2B%7D%5D%5BOH%5E-%5D%5E2)
Now, since we know the concentration of cadmium (II) ions at equilibrium and the mole ratio of these ions to the hydroxide ions is 1:2, we infer that the concentration of the latter at equilibrium is 3.5x10⁻⁵ M. In such a way, the resulting Ksp turns out to be:
Regards!
Answer:
Explanation:
Let the number of moles of oxygen = x
2H2 + O2 --> 2 H2O
x 13.3
Since the balance number for oxygen is 1 and the balance number for water is 2, you must set up a proportion. (Those balance numbers represent the number of moles).
1/x = 2 / 13.3 Cross Multiply
2*x = 13.3 Divide both sides by 2
2x/2 = 13.3/2
x = 6.65
You need 6.65 moles of oxygen.
Answer:
d. 12.3 grams of Al2O3
Explanation:
Based on the reaction:
4Al + 3O2 → 2Al2O3
<em>Where 4 moles of Al reacts in excess of oxygen to produce 2 moles of aluminium oxide.</em>
<em />
To solve this question we must find the moles of Aluminium. With these moles we can find the moles of aluminium oxide using the reaction:
<em>Moles Al -Molar mass: 26.9815g/mol-</em>
6.50g * (1mol / 26.9815g) = 0.241 moles Al
<em>Mass Al₂O₃ -Molar mass: 101.96g/mol-</em>
0.241 moles Al * (2 mol Al2O3 / 4 mol Al) = 0.120 moles Al2O3
0.120 moles Al2O3 * (101.96g / mol) =
12.3g of Al2O3 are produced.
Right answer is:
<h3>d. 12.3 grams of Al2O3
</h3>
C. the denser the plants the better.
Answer:
C
Answer:
A. O=C=O and O≡C−O
Explanation:
Resonance:
When the electron distribution on the molecule become uneven like one molecule have more electron compare to other.Resonance occurs due to overlap of the orbitals.When electron flow through pi system then resonance occurs.
So the option A is correct.
A. O=C=O and O≡C−O
