I would say if you would add colors it would be the best
Answer:- Oxidation number of Cl does not change as it is -1 on both sides.
Explanations:- oxidation number of Mg on reactant side is 0 as it is in its elemental form(not combined with another element).
Oxidation number of hydrogen in its compounds is +1, so if H is +1 in HCl the oxidation number of Cl is -1 as the sum has to be zero.
On product side, Mg oxidation number is +2 as the oxidation number of alkaline earth metals in their compounds is +2.
Two Cl are present in magnesium chloride, so if Mg is +2 then Cl is -1.
Oxidation number of H on product side is 0 as it is present in its elemental for,
,
So, it is only chlorine(Cl) whose oxidation number does not change for the given equation.
Answer:
0.49 mol
Explanation:
Step 1: Write the balanced equation
Mg + 2 HCI ⇒ MgCl₂ + H₂
Step 2: Calculate the moles corresponding to 12 g of Mg
The molar mass of Mg is 24.31 g/mol.

Step 3: Calculate the moles of H₂ produced by 0.49 moles of Mg
The molar ratio of Mg to H₂ is 1:1. The moles of H₂ produced are 1/1 × 0.49 mol = 0.49 mol.
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M