6.02x10^23; This represents the number of molecules in 1 mole of a substance.
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:
83.20 g of Na3PO4
Explanation:
1 mole of Na3PO4 contains 3 moles of Na+.
Mole of Na ion to be prepared = Molarity x volume
= 0.700 x 725/1000
= 0.5075 mole
If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:
0.5075/3 x 1 = 0.1692 mole of Na3PO4
mole of Na3PO4 = mass/molar mass = 0.1692
Hence, mass of Na3PO4 = 0.1692 x molar mass
= 0.1692 x 163.94
= 83.20 g.
83.20 g of Na3PO4 will be needed.