Explanation:
Copper(II) sulfide reacts with oxygen gas to give solid copper(II) oxide and sulfur trioxide gas.
The reaction is given as:

When 1 mol copper(II) sulfide react with 2 moles of oxygen gas it gives 1 mol of solid copper(II) oxide and 1 mol of sulfur trioxide gas
The gas formed in above reaction that is sulfur trioxide reacts with water to give sulfuric acid or hydrogen sulfate.
The reaction is given as:

1 mol of sulfur trioxide gas reacts with 1 mol of liquid water to produce 1 molo of liquid hydrogen sulfate or sulfuric acid
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Since the given solubility is 350 ppm, convert it first with fraction of solubility. by dividing the solubility with 10^6
S = 350 / 10^6
s = 3.5 x 10^-4
the multiply it to the total solution to calculate the amount of substance present
m = ( 3.5 x 10^-4 ) ( 1.01 )
m = 3.535 x 10^-4 g of the substance present
Answer: 1.284M NH3
Explanation: (12.23 grams)/(17.0 gramms/mole) = 0.7191 moles
Dissolved in 560.0 ml (=0.5600L)
(0.7191 moles)/(0.560L) = 1.284M (4 sig figs)