I think the answer is tenfold
hope this helps :)
It is called exothermic reaction because it releases heat and light and it is called combustion reaction because it is reacting and is being oxidised by O2 to MgO.
It can also be called as oxidation reaction since Mg is oxidised to MgO.
Answer:
1. Ionic bonding
2. Covalent bonding
3. Metallic bonding
Explanation:
Ionic bonding also referred to as electrovalent bonding is a kind of chemical bonding that involves the transfer of electrons between the valence shells of two elements with a large electronegativity difference usually a metal and a nonmetal.
For example an ionic bonding scenario might play out between a group one metal and a group seven halogen. While group one metals have one electron hindering their stability, group seven halogens need that one electron that could make them achieve this stability. It is this that causes them to come together in a way where the electron is transferred completely from the valence shell of the group 1 atom and accepted into the valence shell of the group 7 halogen.
Covalent bonding involves the sharing of electrons between atoms of comparable electronegativities. The electro negativity difference is not large enough to permit the total movement of the electrons and hence the electrons are then controlled by the nuclei of the two atoms
Between two metals, what we have is called the metallic bonding
Answer:
The pH of the solution is 11.48.
Explanation:
The reaction between NaOH and HCl is:
NaOH + HCl → H₂O + NaCl
From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:
Now, we need to find the concentration of the OH⁻ ions.
![[OH^{-}] = \frac{n_{NaOH}}{V}](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20)
Where V is the volume of the solution = 1.00 L
![[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L](https://tex.z-dn.net/?f=%20%5BOH%5E%7B-%7D%5D%20%3D%20%5Cfrac%7Bn_%7BNaOH%7D%7D%7BV%7D%20%3D%20%5Cfrac%7B3.01%20%5Ccdot%2010%5E%7B-3%7D%20moles%7D%7B1.00%20L%7D%20%3D%203.01%20%5Ccdot%2010%5E%7B-3%7D%20mol%2FL%20)
Finally, we can calculate the pH of the solution as follows:
![pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52](https://tex.z-dn.net/?f=%20pOH%20%3D%20-log%28%5BOH%5E%7B-%7D%5D%29%20%3D%20-log%283.01%20%5Ccdot%2010%5E%7B-3%7D%29%20%3D%202.52%20)
Therefore, the pH of the solution is 11.48.
I hope it helps you!
