Rubidium is an element that belongs to Group 1. As such it will have physical properties similar to the other Group 1 elements. Rubidium is below
Potassium in the periodic table but above
Cesium. As such it would be most like one of those two elements.
<u>Given:</u>
The initial energy of the electron Einitial = 16.32 * 10⁻¹⁹ J
The energy released i.e the change in energy ΔE = 5.4 * 10⁻¹⁹ J
<u>To determine:</u>
The final energy state Efinal of the electron
<u>Explanation:</u>
Since energy is being released, this suggests that Efinal < Einitial
i.e. ΔE = Einitial - Efinal
Efinal = Einitial - ΔE = (16.32 - 5.4)*10⁻¹⁹ = 10.92 * 10⁻¹⁹ J
Ans: A)
The electron moved down to an energy level and has an energy of 10.92 * 10⁻¹⁹ J
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:
Option B. Cation that is smaller than the original atom.
Explanation:
Magnesium is a divalent element. This implies that magnesium can give up 2 electrons to become an ion (cation) as shown below:
Mg —> Mg²⁺ + 2e¯
Next, we shall write the electronic configuration of magnesium atom (Mg) and magnesium ion (Mg²⁺). This can be written as follow:
Mg (12) = 2, 8, 2
Mg²⁺ (10) = 2, 8
From the above illustration, we can see that the magnesium atom (Mg) has 3 shells while the magnesium ion (Mg²⁺) has 2 shells.
This simply means that the magnesium ion (Mg²⁺) i.e cation is smaller that the original magnesium atom (Mg).
