I’m pretty sure the answer would be B.
C. 10 amu i am 100 sure that it is 10 amu
Answer:
Percent yield = 79.79 %
Explanation:
Given data:
Mass of Al₂O₃ = 821 g
Mass of Al = 349 g
Percent yield = ?
Solution:
Chemical equation:
2Al₂O₃ + 3C → 4Al + 3CO₂
Number of moles of Al₂O₃:
Number of moles = mass/molar mass
Number of moles = 821 g/ 101.96 g/mol
Number of moles = 8.1 mol
Now we will compare the moles of Al with Al₂O₃.
Al₂O₃ : Al
2 : 4
8.1 : 4/2×8.1 = 16.2 mol
Mass of Al:
Mass = number of moles × molar mass
Mass = 16.2 mol × 27 g/mol
Mass = 437.4 g
Percent yield:
Percent yield = (actual yield / theoretical yield)× 100
Percent yield = (349 g/ 437.4 g) × 100
Percent yield = 79.79 %
Answer:
True?
Im sorry I have no clue...
Answer:
D. 1s2,2s2,2p6,3s2,3p6,3d10,4s2,4p6,4d10,5s2,5p6
Explanation:
Hello!
In this case, since the standard iodine atom has 53 electrons, when it forms the iodide ion it is known it gains one spare electron so now it has 54; it means we need to write the new electron configuration up to 54 as shown below:
Thus, the answer should be:
D. 1s2,2s2,2p6,3s2,3p6,3d10,4s2,4p6,4d10,5s2,5p6
Even when the order is not the adequate one.
Regards!