Answer:
6
Explanation:
FCC is face centered cubic lattice. In FCC structure, there are eight atoms at the eight corner of the cubic unit cell and one atom centered in each of the faces. FCC unit cells consist of four atoms, (8/8) at the corners and (6/2) in the faces.
Given that, Cu has FCC structure and it contains a vacancy at origin (0, 0, 0). And there is no other vacancy directly adjacent to the vacancy at the origin. So, all the adjacent positions contain Cu atoms. Hence, the total number of adjacent atoms of the vacancy at origin can jump into this vacancy.
the above FCC unit cell clearly indicates that there are six adjacent atoms adjacent to the vacancy at origin
So, the total number of adjacent atoms of the vacancy at origin can jump into this vacancy is 6.
Answer:
c is not a true statement
<u>Given:</u>
Moles of Al = 0.4
Moles of O2 = 0.4
<u>To determine:</u>
Moles of Al2O3 produced
<u>Explanation:</u>
4Al + 3O2 → 2Al2O3
Based on the reaction stoichiometry:
4 moles of Al produces 2 moles of Al2O3
Therefore, 0.4 moles of Al will produce:
0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3
Similarly;
3 moles O2 produces 2 moles Al2O3
0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles
Thus Al will be the limiting reactant.
Ans: Maximum moles of Al2O3 = 0.2 moles
Answer:
a) No. of moles of hydrogen needed = 5.4 mol
b) Grams of ammonia produced = 27.2 g
Explanation:
a)
No. of moles of nitrogen = 1.80 mol
1 mole of nitrogen reacts with 3 moles of hydrogen
1.80 moles of nitrogen will react with
= 1.80 × 3 = 5.4 moles of hydrogen
b)
No. of moles of hydrogen = 2.4 mol
It is given that nitrogen is present in sufficient amount.
3 moles of hydrogen produce 2 moles of 
2.4 moles of hydrogen will produce
= 
Molar mass of ammonia = 17 g/mol
Mass in gram = No. of moles × Molar mass
Mass of ammonia in g = 1.6 × 17
= 27.2 g
