The atomic number of Be is 4, and so it has 2 shells. There are valence electrons in the second, which is the outermost, shell of Be. To get the element with one more shell, there would be 3 shells on the new element, and 1 less valence electron, so the new element should have 1 valence electron. Sodium is the element with 3 shells, and one valence electron which fits perfectly into the description.
Answer;
The total pressure is 1.107 atm.
Explanation;
The total pressure is the sum of the pressures of the three gases in the flask
Pressure (total) = 0.215 atm + 0.066 atm + 0.826 atm = 1.107 atm
= 1.107 atm.
The number of atoms present in 0.58 mole of magnesium, Mg is 3.49×10²³ atoms
<h3>Avogadro's hypothesis </h3>
1 mole of Mg = 6.02×10²³ atoms
<h3>How to determine the atoms in 0.58 mole of Mg </h3>
1 mole of Mg = 6.02×10²³ atoms
Therefore,
0.58 mole of Mg = 0.58 × 6.02×10²³
0.58 mole of Mg = 3.49×10²³ atoms
Thus, 3.49×10²³ atoms are present in 0.58 mole of Mg
Learn more about Avogadro's number:
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Answer:
(A) is 0.0773 mol B2H6
(C) is 2.79 x 10^23 H atoms
Explanation:
Questions (A) and (B) are the same.
2.14 g B2H6 x (1 mol B2H6/27.668g B2H6) = 0.0773 mol B2H6 (A)
<u>27.668 is the molar mass of B2H6 calculated from the period table: </u>
(2 x 10.81) + (6 x 1.008) = 27.668
1.008 is the mass of H and 10.81 is the mass of B
(C)
0.0773 mol B2H6 x (6 mol H/ 1 mol B2H6) x (6.022 x 10^23 H atoms/1 mol H)
= 2.79 x 10^23 hydrogen atoms
Further Explanation:
- For every 1 mol of B2H6, there are 6 moles of H (indicated by the subscript)
- 6.022 x 10^23 is Avogrado's number and it equals to 1 mol of anything
- Avogrado's number can be in units of atoms, molecules, or particles
Answer:
Answers are in the explanation
Explanation:
Ksp of CdF₂ is:
CdF₂(s) ⇄ Cd²⁺(aq) + 2F⁻(aq)
Ksp = 6.44x10⁻³ = [Cd²⁺] [F⁻]²
When an excess of solid is present, the solution is saturated, the molarity of Cd²⁺ is X and F⁻ 2X:
6.44x10⁻³ = [X] [2X]²
6.44x10⁻³ = 4X³
X = 0.1172M
<h3>[F⁻] = 0.2344M</h3><h3 />
Ksp of LiF is:
LiF(s) ⇄ Li⁺(aq) + F⁻(aq)
Ksp = 1.84x10⁻³ = [Li⁺] [F⁻]
When an excess of solid is present, the solution is saturated, the molarity of Li⁺ and F⁻ is XX:
1.84x10⁻³ = [X] [X]
1.84x10⁻³ = X²
X = 0.0429
<h3>[F⁻] = 0.0429M</h3><h3 /><h3>The solution of CdF₂ has the higher fluoride ion concentration</h3>
