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:
11.39
Explanation:
Given that:
Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:
Thus,
Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)
Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
![K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20BH%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BB%5D%7D)
x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>
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Answer:
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