Yes. bromine and sodium iodide can react to form sodium bromine and free iodine
Answer:
0.508 mole
Explanation:
NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.
The number of mole present in 78.2 g of CCl₄ can be obtained as follow:
Mass of CCl₄ = 78.2 g
Molar mass of CCl₄ = 12 + (35.5×4)
= 12 + 142
= 154 g/mol
Mole of CCl₄ =?
Mole = mass / molar mass
Mole of CCl₄ = 78.2 / 154
Mole of CCl₄ = 0.508 mole
Therefore, 0.508 mole is present in 78.2 g of CCl₄
Explanation:
In this reaction, the reactants are Li and N2. The product is Li3N
So we have;
Li + N2 → Li3N
Upon balancing, we have;
6Li + N2 → 2 Li3N
The sum of the coefficients is 6 + 1 + 2 = 9
