Answer:
In the electrolysis of dilute sulfuric acid, which electrolysis in aqueous solution to form hydrogen ions, H⁺, and sulfur IV ions SO₄²⁻ in the presence of H⁺ and OH⁻ ions from the water molecules
At the anode
The anode, positive electrode, attracts the negative OH⁻ and SO₄²⁻ ions where the OH⁻ gives up electrons to form water molecules and oxygen as follows;
4OH⁻ → 2H₂O + O₂ + 4e⁻
At the cathode
The positive H⁺ ions from the water molecules and the acid are attracted to the cathode where they combine with 2 electrons to form hydrogen gas as follows;
2e⁻ + 2H⁺ → H₂ (gas)
Explanation:
Answer:
evaporation, vaporisation, boiling if change occurs at boiling point
Explanation:
Answer:
Explanation:
<u>1) Equilibrium equation (given)</u>:
- H₂ (g) + I₂ (g) ⇄ 2HI (g)
<u>2) Equilibrium constant expression</u>:
<u>3) Determine the conentrations from the stoichiometry</u>:
a) <u>Number of moles</u>:
H₂ (g) + I₂ (g) ⇄ 2HI (g)
Start 0.4 1.55 0
Change - 0.4×0.6 = - 0.24 - 0.24 + 0.24×2 = + 0.48
--------------------------- ------------- --------------------------
End 0.16 1.31 0.48
b) <u>Concentrations</u>:
- [ H₂ (g) ] = 0.16 mole / 2 liter = 0.08 M
- [ I₂ (g) ] = 1.31 mole / 2 liter = 0.655 M
- [HI (g) ] = 0.48 mole / 2 liter = 0.24 M
<u>4) Compute Kc</u>:
- Kc = (0.24 M)² / (0.08 M × 0.655 M) = 1.099 ≈ 1.1 ← answer
Mass of Nd₂(SO₄)₃ : 2.653 g
<h3>Further explanation</h3>
Reaction
Nd₂Fe₁₄B + 17H₂SO₄ → Nd₂(SO₄)₃ + 14FeSO₄ + B + 17H₂
5 g Nd₂Fe₁₄B
MW = 2. 144,242 + 14. 55.845 + 10.811
MW = 288.484 + 781.83+10.811=1081.125 g/mol
mol :
mol Nd₂Fe₁₄B = mol Nd₂(SO₄)₃=0.0046
so mass of Nd₂(SO₄)₃ :
MW Nd₂(SO₄)₃ = 2.144.242+3.32,065+12.15,999
MW = 288.484+96.195+191.988=576.667 g/mol
66.2% yield
The balanced equation for the reaction is:
2C6H10 + 17O2 ==> 12CO2 + 10H2O
So for every 2 moles of C6H10 consumed, we should get 12 moles of CO2. Or to simplify, for each mole of C6H10, we should get 6 moles of CO2. Now let's calculate the molar mass of C6H10 and CO2 and then determine how many moles of each we really have.
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass C6H10 = 6*12.0107 + 10*1.00794 = 82.1436 g/mol
Molar mass CO2 = 12.0107 + 2*15.999 = 44.0087 g/mol
Moles C6H10 = 8.88 g / 82.1436 g/mol = 0.10810337 mol
Moles CO2 = 18.9 g / 44.0087 g/mol = 0.429460538 mol
Since we had 0.10810337 moles of C6H10, we should have gotten
6*0.10810337 = 0.648620221 moles of CO2, but only got 0.429460538 moles. So let's divide the actual yield by the theoretical yield to get the percentage yield.
0.429460538 / 0.648620221 = 0.662114014 = 66.2%