no it is not possible, because they both have the same number of valence electrons in each element. in a compound you are supposed to have two or more elements that have different numbers of valence electrons so when put together they for a compound.
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Answer:
24.32
Explanation:
From the question given above, the following data were obtained:
Isotope A:
Mass of A = 24
Abundance (A%) = 78.70%
Isotope B
Mass of B = 25
Abundance (B%) = 10.13%
Isotope C:
Mass of C = 26
Abundance (C%) = 11.17%
Average atomic mass of Mg =..?
The average atomic mass of Mg can be obtained as illustrated below:
Average atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]
Average atomic mass = [(24 × 78.70)/100] + [(25 × 10.13)/100] + [(26 × 11.17)/100]
= 18.888 + 2.5325 + 2.9042
= 24.3247 ≈ 24.32
Therefore, the average atomic mass of magnesium (Mg) is 24.32
Explanation:
The balanced equation of the reaction is given as;
Mg(OH)2 (s) + 2 HBr (aq) → MgBr2 (aq) + 2 H2O (l)
1. How many grams of MgBr2 will be produced from 18.3 grams of HBr?
From the reaction;
2 mol of HBr produces 1 mol of MgBr2
Converting to masses using;
Mass = Number of moles * Molar mass
Molar mass of HBr = 80.91 g/mol
Molar mass of MgBr2 = 184.113 g/mol
This means;
(2 * 80.91 = 161.82g) of HBr produces (1 * 184.113 = 184.113g) MgBr2
18.3g would produce x
161.82 = 184.113
18.3 = x
x = (184.113 * 18.3 ) / 161.82 = 20.8 g
2. How many moles of H2O will be produced from 18.3 grams of HBr?
Converting the mass to mol;
Number of moles = Mass / Molar mass = 18.3 / 80.91 = 0.226 mol
From the reaction;
2 mol of HBr produces 2 mol of H2O
0.226 mol would produce x
2 =2
0.226 = x
x = 0.226 * 2 / 2 = 0.226 mol
3. How many grams of Mg(OH)2 are needed to completely react with 18.3 grams of HBr?
From the reaction;
2 mol of HBr reacts with 1 mol of Mg(OH)2
18.3g of HBr = 0.226 mol
2 = 1
0.226 = x
x = 0.226 * 1 /2
x = 0.113 mol
