Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
Answer:
The oxidation state of the carbon is +4.
Explanation:
Calcium is in group 2 of the periodic table, therefore, its oxidation state is +2.
The oxidation state of the oxygen is -2.
As the compound is neutral, the sum of the oxidation states of all atoms must be 0.
Oxidation State Ca + Oxidation State C + (Oxidation State O)×3 = 0
+2 + x + (-2)×3 = 0
2 + x - 6 = 0
x = 6 -2
x = 4
Hence, the oxidation state of the carbon is +4.
Answer:
2 C₄H₁₀(l) + 13 O₂(g) ⇄ 8 CO₂(g) + 10 H₂O(g)
Explanation:
When a substance burns we talk about a combustion reaction. When combustion is complete the products are carbon dioxide and water, like in this case. The equation is:
C₄H₁₀(l) + O₂(g) ⇄ CO₂(g) + H₂O(g)
First, we balance the element with the largest stoichiometric coefficient (C).
C₄H₁₀(l) + O₂(g) ⇄ 4 CO₂(g) + H₂O(g)
Then, we balance H because it is in just 1 compound on each side.
C₄H₁₀(l) + O₂(g) ⇄ 4 CO₂(g) + 5 H₂O(g)
Finally, we balance O.
C₄H₁₀(l) + 6.5 O₂(g) ⇄ 4 CO₂(g) + 5 H₂O(g)
Since we want the smallest whole numbers, we multiply all coefficients by 2.
2 C₄H₁₀(l) + 13 O₂(g) ⇄ 8 CO₂(g) + 10 H₂O(g)
Heat energy released here
Q = mass x specific heat capacity of water x deltaT
= 25.0 x 4.184 x (-10.0)
= - 1046 Joules
1046 Joules of heat energy is released in this process.
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