Answer:
The center of the atom contains the nucleus so yes
Explanation:
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.
The balanced chemical reaction is:
N2 + 3H2 = 2NH3
We are given the amount of ammonia formed
from the reaction. This is where we start our calculations.
0.575 g NH3 (1 mol NH3 / 17.03 g NH3) (3 mol
H2 / 2 mol NH3) ( 2.02 g H2 / 1 mol H2) = 0.10 g H2
This one is an oxidation-rdcution equation:
<span>H2 + F2 → 2HF
How can you tell?
If the oxidation states of the atoms in the reactans are different from the oxidation states of the same atoms in the products then it is an oxidation-reduction reaction.
Both atoms H and F in the reactants have oxidation states 0.
That is a basic rule: any atom alone or bonded to the same kind of atom has oxidation state 0.
The oxidation states in HF are: H: +1, and F: -1.
So, the H increased its oxidation state, which is that ii is oxydized ; while F reduced its oxidation state so it is reduced.
Answer: H2 + F2 ----> 2HF
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