Answer:
7.32 g of F₂
Solution:
The equation is as follow,
2 LiI + F₂ → 2 LiF + I₂
According to equation,
51.88 g (2 mole) of LiF is produced from = 37.99 g (1 mole) F₂
So,
10 g of LiF will be produced by = X g of F₂
Solving for X,
X = (10 g × 37.99 g) ÷ 51.88 g
X = 7.32 g of F₂
Answer:
chlorine is therefore an anion
Answer:
84.24 g
Explanation:
Given data:
Mass of oxygen = 75 g
Mass of Al required to react = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 75 g/ 32 g/mol
Number of moles = 2.34 mol
Now we will compare the moles of oxygen with Al.
O₂ : Al
3 : 4
2.34 : 4/3×2.34 = 3.12 mol
Mass of Al required:
Mass = number of moles × molar mass
Mass = 3.12 mol × 27 g/mol
Mass = 84.24 g
Answer: I would say its either exit door or the last one.
Explanation: Hope this helps plz mark brainliest.