Answer:
8M IM NOT SURE SO. IF YOU CAN GET A CLEARED PICTURE OF IT WOULD REALLY HELP
Answer:
104.969 amu.
Explanation:
From the question given above, the following data were obtained:
Isotope A:
Mass of A = 107.977 amu
Abundance (A%) = 0.1620%
Isotope B:
Mass of B = 106.976 amu
Abundance (B%) = 1.568%
Isotope C:
Mass of C = 105.974 amu
Abundance (C%) = 47.14%
Isotope D:
Mass of D = 103.973 amu
Abundance (D%) = 51.13%
Average atomic mass =?
The average atomic mass of the element can be obtained as follow:
Average atomic mass = [(Mass of A × A%) /100] + [(Mass of B × B%) /100] + [(Mass of C × C%) /100] + [(Mass of D × D%) /100]
Average atomic mass = [(107.977 × 0.1620)/100] + [(106.976 × 1.568)/100] + [(105.974 × 47.14)/100] + [(103.973 × 51.13)/100]
= 0.175 + 1.677 + 49.956 + 53.161
= 104.969 amu
Therefore, the average atomic mass of the element is 104.969 amu.
Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
What country and grade are you?
Please tell me! I’m going to help you
Answer:
D
Explanation:
Silver is a metal with high conductivity ....
Hope its right !