Solution:
1. Calculate the molar mass of SO4:
- Sulfur (S) has a molar mass of approximately 32.07 g/mol.
- Oxygen (O) has a molar mass of approximately 16.00 g/mol.
- Since there are four oxygen atoms in SO4, multiply the molar mass of oxygen by 4: 16.00 g/mol * 4 = 64.00 g/mol.
- Now, add the molar mass of sulfur and the molar mass of oxygen: 32.07 g/mol + 64.00 g/mol = 96.07 g/mol.

2. Calculate the molar mass of Na2SO4:
- Sodium (Na) has a molar mass of approximately 22.99 g/mol.
- Since there are two sodium atoms in Na2SO4, multiply the molar mass of sodium by 2: 22.99 g/mol * 2 = 45.98 g/mol.
- Add this to the molar mass of SO4 calculated earlier: 45.98 g/mol + 96.07 g/mol = 142.05 g/mol.

3. Calculate the percent composition of SO4 in Na2SO4:
- Divide the molar mass of SO4 (96.07 g/mol) by the molar mass of Na2SO4 (142.05 g/mol) and multiply by 100 to get the percentage: (96.07 g/mol / 142.05 g/mol) * 100 67.64%.

So, the percent composition of SO4 in Na2SO4 is approximately 67.64%.

Number of sodium ions = Number of chloride ions = 0.949 moles x 6.022 x 10²³ ions/mole
Number of sodium ions = 0.949 moles x 6.022 x 10²³ ions/mole 5.70 x 10²³ sodium ions
There are approximately 5.70 x 10²³ sodium ions in 0.949 moles of sodium chloride (NaCl).

1 mol of any substance = 6.022 × 10^23 particles
Use the formula
Number of Na+ ions = moles of NaCl × 6.022 × 10^23
Given:
Moles of NaCl = 0.949
Calculate
Number of Na+ ions = 0.949 × 6.022 × 10^23
Number of Na+ ions = 5.715 × 10^23
Answer:
5.72 × 10^23 Na+ ions

Moles of Ag = 89.5 g / 107.87 g/mol = 0.83 mol
From stoichiometry: 2 Ag : 1 Cu moles Cu = 0.83 / 2 = 0.415 mol
Mass of Cu = 0.415 × 63.55 = 26.4 g
Answer: Mass of Cu = 26.4 g

Write the balanced equation
4 NH3(g) + 5 O2(g) : 4 NO(g) + 6 H2O(l)
Determine the mole ratio
From the equation: 4 moles NH3 produce 4 moles NO + 6 moles H2O
Total moles of products per 4 moles NH3 = 4 + 6 = 10 moles
Calculate total moles for 1.20 moles of NH3
Total moles of products = (10 / 4) × 1.20
Total moles of products = 3.0
Answer: 3.0 moles of products are formed