Answer:
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We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.
We also know that the trend in the electronegativity is to decrease when you go up-down ia group.
The known electronegativities of the elements of this group are:
Li: 0.98
Na: 0.93
K: 0.82
Rb: 0.82
Cs: 0.79
Fr: 0.70
Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.
Answer: 2 mol
Explanation:
- According to the ideal gas law, One mole of an ideal gas at STP (standard temperature and normal pressure) occupies 22.4 liters.
- Using cross multiplication,
1 mol of (O2) → 22.4 L
? → 43.9 L
Therefore, the number of moles of oxygen in 43.9 L = (43.9 × 1)/ 22.4 = 1.96 mol≈ 2 mol..
3 mol H₂ → 2 mol NH₃
5 mol H₂ → x mol NH₃
x=2*5.0/3=3.3
n(NH₃)=3.3 mol
calculate moles of both reagents given and the moles of FeS that each of them would form if they were in excess
moles = mass / molar mass
moles Fe = 7.62 g / 55.85 g/mol
= 0.1364 moles
1 mole Fe produces 1 mole FeS
Therefore 7.62 g Fe can form 0.1364 moles FeS
moles S = 8.67 g / 32.07 g/mol
= 0.2703 moles S
1 mole S can from 1 moles FeS
So 8.67 g S can produce 0.2703 moles FeS
The limiting reagent is the one that produces the least product. So Fe is limiting.
The maximum amount of FeS possible is from complete reaction of all the limiting reagent.
We have already determined that the Fe can form up to 0.1364 moles of FeS, so this is max amount of FeS you can get.
Convert to mass
hope this helps :)
