<span>First:
46.7 g of N with 53.3 g of O,
=> mass ratio O to N = 53.3 / 46.7 = 1.1413
Second
17.9 g of N and 82.0 g of O.
mass ratio of O to N = 82.0 / 17.9 = 4.5810
Third
Ratio of the mass ratio of O to N in the second compound
to the mass ratio of O to N in the first compound =
= 4.5810 / 1.1413 = 4.013 ≈ 4
Answer: 4
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Answer:
The concentration of chloride ion is 
Explanation:
We know that 1 ppm is equal to 1 mg/L.
So, the 
content 100 ppm suggests the presence of 100 mg of in 1 L of solution.
The molar mass of is equal to the molar mass of Cl atom as the mass of the excess electron in is negligible as compared to the mass of Cl atom.
So, the molar mass of is 35.453 g/mol.
Number of moles = (Mass)/(Molar mass)
Hence, the number of moles (N) of present in 100 mg (0.100 g) of is calculated as shown below:
So, there is 
of present in 1 L of solution.
Answer:
23.2 g of Al will be left over when the reaction is complete
Explanation:
2Al + 3S → Al₂S₃
1 mol of Al = 26.98 g
1 mol of S = 32.06 g
Mole = Mass / Molar mass
63.8 g/ 26.98 g/m = 2.36 mole of Al
72.3 g / 32.06 g/m = 2.25 mole of S
2 mole of Aluminun react with 3 mole of sulfur
2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S
As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.
3 mole of S react with 2 mole of Al
2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole
I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.
2.36 mole of Al - 1.50 mole of Al = 0.86 mole
This is the quantity of Al without reaction.
Molar mass . mole = Mass → 26.98 g/m . 0.86 m = 23.2 g
The answer is package c because each cd would cost $1.87, which is the lowest price
