In a combustion reaction, one of the reactants is always oxygen.
Answer:
0.189 g.
Explanation:
- This problem is an application on <em>Henry's law.</em>
- Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
- Solubility of the gas ∝ partial pressure
- If we have different solubility at different pressures, we can express Henry's law as:
<em>S₁/P₁ = S₂/P₂,</em>
S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm
S₂ = ??? g/L and P₂ = 5.73 atm
- So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.
<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>
<em></em>
Answer:
The correct option is False
Explanation:
Ionization energy is the <em>minimum amount of energy required to remove a valence electron from one mole of an atom in it's gaseous state</em>. Ionization energy requires the removal of an electron from a gaseous atom. The definition in the question is that of electronegativity.
Electronegativity is the <u>ability of an atom to attract electrons towards itself in a chemical bond.</u>
Answer:
Explanation:
Hello,
For the given chemical reaction:
We first must identify the limiting reactant by computing the reacting moles of Al2S3:
Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:
Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:
Finally, we compute the percent yield with the obtained 2.10 g:
Best regards.
Answer:
0.001 M OH-
Explanation:
[OH-] = 10^-pOH, so
pOH + pH = 14 and 14 - pH = pOH
14 - 11 = 3
[OH⁻] = 10⁻³ ; [OH-] = 0.001 M OH-