Answer:
Explanation:
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In this case, since the pH of the given metal is 10.15, we can compute the pOH as shown below:
Now, we compute the concentration of hydroxyl ions in solution:
![[OH^-]=10^{-pOH}=10^{-3.95}=1.41x10^{-4}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-pOH%7D%3D10%5E%7B-3.95%7D%3D1.41x10%5E%7B-4%7DM)
Now, since this hydroxide has the form MOH, we infer the concentration of OH- equals the concentration of M^+ at equilibrium, assuming the following ionization reaction:
Whose equilibrium expression is:
![Ksp=[M^+][OH^-]](https://tex.z-dn.net/?f=Ksp%3D%5BM%5E%2B%5D%5BOH%5E-%5D)
Therefore, the Ksp for the saturated solution turns out:
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The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
C, they could work together to make things from the top down.
Answer:
Al,Ga,B
Explanation:
Now since i helped you can you help me with this plz
Matteo took 5 math quizzes. The mean of the 5 quizzes was 8.2. Here are four of his quiz scores 7, 7, 8, 10. What is the 5th quiz score? Show work.
