The half reactions as they occur at each electrode
is as follows
at the anode Sn(s) =sn^2+(aq) + 2e -
at the cathode 2 ag^+(aq) + 2e - = 2Ag (s)
net cell reaction = Sn (s) + 2Ag^+(aq) = sn^2+ (aq) + 2 Ag (s)
Answer:
H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)
Explanation:
H₂²⁺(aq) + O₂²⁻(aq) + Mg²⁺(aq) + SO₃²⁻(aq) → Mg²⁺(aq) + SO²⁻₄(aq) + H₂O(l)
A careful observation of the equation above, shows that the equation is already balanced.
To obtain the net ionic equation, we simply cancel Mg²⁺ from both side of the equation as shown below:
H₂²⁺(aq) + O₂²⁻(aq) + SO₃²⁻(aq) → SO²⁻₄(aq) + H₂O(l)
Answer:
increase dramatically(a lot).
Explanation:
The core of the earth is way way way hotter than the surface.
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
<em>4 moles of Al produce 2 moles of Al₂O₃</em>
<em />
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = <em>0.0852 moles of Al₂O₃</em>,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
<h3>8.7g of Al₂O₃ can be produced (Theoretical yield)</h3>
Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
<h3>
78.2% </h3>
