Answer:
2Mg + O₂ ⟶ 2MgO
Explanation:
Step 1. Start with the most complicated-looking formula (O₂?).
Put a 1 in front of it.
Mg + 1O₂ ⟶ MgO
Step 2. Balance O.
We have fixed 2 O on the left. We need 2O on the right. Put a 2 in front of MgO.
Mg + 1O₂ ⟶ 2MgO
Step 3. Balance Mg.
We have fixed 2 Mg on the right-hand side. We need 2 Mg atoms on the left. Put a 2 in front of Mg.
2Mg + 1O₂ ⟶ 2MgO
Every formula now has a coefficient. The equation should be balanced. Let’s check.
<u>Atom</u> <u>On the left</u> <u>On the righ</u>t
Mg 2 2
O 2 2
All atoms are balanced.
The balanced equation is
2Mg + O₂ ⟶ 2MgO
<u>Answer:</u> The rate law for the reaction is ![\text{Rate}=k[NO_3][CO]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BNO_3%5D%5BCO%5D)
<u>Explanation:</u>
Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.
In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.
For the given chemical reaction:

The intermediate reaction of the mechanism follows:
Step 1: 
Step 2: 
As, step 2 is the slow step. It is the rate determining step
Rate law for the reaction follows:
![\text{Rate}=k[NO_3][CO]](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3Dk%5BNO_3%5D%5BCO%5D)
Hence, the rate law for the reaction is written above.
Phosphorus has 5 valence
valence electrons is located to their group#
yes because of TAVE available valence electrons