Answer:
The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.
Explanation:
Ionic compounds are compound formed from the transfer of electron(s). One atom of the element loses electron(s) while the other atom gains electron(s).
The compound Magnesium chloride is an ionic compound . The bond between an atom of magnesium and 2 atoms of chlorine is an ionic bonding.
The valency electron of magnesium is 2 electron , for the atom of magnesium to attain octet rule, it will easily lose it 2 electrons to the chlorine atoms.
The chlorine atom on the other hand has 7 valency electrons, to attain octet configuration it will most likely gain 1 electron to become stable.
The magnesium atom loses 2 electron to the 2 atoms of chlorine. The 7 valency electrons of each chlorine atom will now be 8 to attain stable configuration. The final compound is written as MgCl2.
Answer:
2
Explanation:
Half-life
Concentration
![{[A]_0}_A=1.2\ \text{M}](https://tex.z-dn.net/?f=%7B%5BA%5D_0%7D_A%3D1.2%5C%20%5Ctext%7BM%7D)
![{[A]_0}_B=0.6\ \text{M}](https://tex.z-dn.net/?f=%7B%5BA%5D_0%7D_B%3D0.6%5C%20%5Ctext%7BM%7D)
We have the relation
![t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%5Cpropto%20%5Cdfrac%7B1%7D%7B%5BA%5D_0%5E%7Bn-1%7D%7D)
So
![\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}](https://tex.z-dn.net/?f=%5Cdfrac%7B%7Bt_%7B1%2F2%7D%7D_A%7D%7B%7Bt_%7B1%2F2%7D%7D_B%7D%3D%5Cleft%28%5Cdfrac%7B%7B%5BA%5D_0%7D_B%7D%7B%7B%5BA%5D_0%7D_A%7D%5Cright%29%5E%7Bn-1%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B2%7D%7B4%7D%3D%5Cleft%28%5Cdfrac%7B0.6%7D%7B1.2%7D%5Cright%29%5E%7Bn-1%7D%5C%5C%5CRightarrow%20%5Cdfrac%7B1%7D%7B2%7D%3D%5Cleft%28%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5E%7Bn-1%7D)
Comparing the exponents we get
The order of the reaction is 2.
![t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA%5D_0%5E%7Bn-1%7D%7D%5C%5C%5CRightarrow%20k%3D%5Cdfrac%7B1%7D%7Bt_%7B1%2F2%7D%5BA%5D_0%5E%7Bn-1%7D%7D%5C%5C%5CRightarrow%20k%3D%5Cdfrac%7B1%7D%7B2%5Ctimes%201.2%5E%7B2-1%7D%7D%5C%5C%5CRightarrow%20k%3D0.4167%5C%20%5Ctext%7BM%7D%5E%7B-1%7D%5Ctext%7Bmin%7D%5E%7B-1%7D)
The rate constant is
The products are on the right side of the equation. For this one it would be 2AlPO4 + 3CaSO4
Answer:
-0.050 kJ/mol.K
Explanation:
- A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. K
- The reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. K
All in all, ΔG° = 0 at 400. K.
We can find ΔS° using the following expression.
ΔG° = ΔH° - T.ΔS°
0 = -20 kJ/mol - 400. K .ΔS°
ΔS° = -0.050 kJ/mol.K
Answer:
1. denser
2. subduction
3. divergence
4. hot spots
5. earthquakes
6. An underwater hot-spot volcano occasionally erupts, and the lava solidifies into an island. However, the plate that lies above the hot spot continues to move due to slab pull and ridge push, so the stationary volcano will be in a different location when it erupts again. This forms a new island, and the process may continue.
