Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Answer: n = 3.0 moles
V = 60.0 L
T = 400 K
From PV = nRT, you can find P
P = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L
P = 1.642 atm = 1.6 atm (to 2 significant figures)
The answer is B.
You can rule C out because divergent means moving away. Rule out A because there is an oceanic and continental plate, not 2 of the same type. Rule D out for the same reason.
Explanation:
Reversible reactions that happen in a closed system eventually reach equilibrium. At equilibrium, the concentrations of reactants and products do not change. But the forward and reverse reactions have not stopped - they are still going on, and at the same rate as each other.
