<u>Answer:</u> The above reaction is non-spontaneous.
<u>Explanation:</u>
For the given chemical reaction:
Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.
We know that:
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the 
of the reaction, we use the equation:
Relationship between standard Gibbs free energy and standard electrode potential follows:
As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.
Hence, the above reaction is non-spontaneous.
Q = mΔT(Cp)
where Q = heat energy in J (joules),
m = mass in g, ΔT = change in temper. (°C),
Cp = heat capacity in J/(g°C)
Water has a higher heat capacity, meaning that once heat energy is absorbed, it holds that heat longer than bread. Also though, a higher heat capacity of water means that it takes more energy to heat it up.
I don't see any specific data listed for this lab??
Answer:
Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
Explanation:
The equilibriums that take place are:
Cu⁺² + 2OH⁻ ↔ Cu(OH)₂(s) ksp = 2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
Co⁺² + 2OH⁻ ↔ Co(OH)₂(s) ksp = 1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
Keep in mind that <em>the concentration of each ion is halved </em>because of the dilution when mixing the solutions.
For Cu⁺²:
2.2x10⁻²⁰ = [Cu⁺²]*[OH⁻]²
2.2x10⁻²⁰ = 0.25 M*[OH⁻]²
[OH⁻] = 2.97x10⁻¹⁰ M
For Co⁺²:
1.3x10⁻¹⁵ = [Co⁺²]*[OH⁻]²
1.3x10⁻¹⁵ = 0.25 M*[OH⁻]²
[OH⁻] = 7.21x10⁻⁸ M
<u>Because Copper requires less concentration of OH⁻ than Cobalt</u>, Cu(OH)₂ will precipitate first, with [OH⁻] = 2.97x10⁻¹⁰ M
Pure water does
not have enough ions to conduct electricity. A mixture of metals such as iron,
zinc and copper in the wet soil can trigger electrolysis that requires excess
energy in the form of over potential to conduct electricity. The excess energy
is needed due to limited self-ionization of water. The wet soil then can
conduct current when positive and negative ions are present. The water ions begin
to flow from anode (positive electrode) to cathode (negative electrode) to be oxidize
and produce electricity.
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Answer : The correct option is, (a) paramagnetic with two unpaired electrons.
Explanation :
According to the molecular orbital theory, the general molecular orbital configuration will be,
![(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)](https://tex.z-dn.net/?f=%28%5Csigma_%7B1s%7D%29%2C%28%5Csigma_%7B1s%7D%5E%2A%29%2C%28%5Csigma_%7B2s%7D%29%2C%28%5Csigma_%7B2s%7D%5E%2A%29%2C%28%5Csigma_%7B2p_z%7D%29%2C%5B%28%5Cpi_%7B2p_x%7D%29%3D%28%5Cpi_%7B2p_y%7D%29%5D%2C%5B%28%5Cpi_%7B2p_x%7D%5E%2A%29%3D%28%5Cpi_%7B2p_y%7D%5E%2A%29%5D%2C%28%5Csigma_%7B2p_z%7D%5E%2A%29)
As there are 14 electrons present in the given configuration.
The molecular orbital configuration of molecule will be,
![(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^1=(\pi_{2p_y})^1],[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0](https://tex.z-dn.net/?f=%28%5Csigma_%7B1s%7D%29%5E2%2C%28%5Csigma_%7B1s%7D%5E%2A%29%5E2%2C%28%5Csigma_%7B2s%7D%29%5E2%2C%28%5Csigma_%7B2s%7D%5E%2A%29%5E2%2C%28%5Csigma_%7B2p_z%7D%29%5E2%2C%5B%28%5Cpi_%7B2p_x%7D%29%5E1%3D%28%5Cpi_%7B2p_y%7D%29%5E1%5D%2C%5B%28%5Cpi_%7B2p_x%7D%5E%2A%29%5E0%3D%28%5Cpi_%7B2p_y%7D%5E%2A%29%5E0%5D%2C%28%5Csigma_%7B2p_z%7D%5E%2A%29%5E0)
The number of unpaired electron in the given configuration is, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic.
Hence, the correct option is, (a) paramagnetic with two unpaired electrons.
