The electronic configuration of magnesium is:
1s² 2s² 2p⁶ 3s² = [Ne] 3s²
This means that for a magnesium atom, in order to have its outermost orbital full, the easiest way would be to lose the two outermost electrons. This is seen in its relatively low first two ionization energies. The third ionization energy is several times higher because the ion would move from a stable form to an highly unstable form. (Mg⁺² → Mg⁺³ + e⁻).
Sodium only has one electron in its outermost orbital, so its first ionization energy would be several times lower than the second.
The right answer is N2(g) + 3H2(g) => 2NH3(g).
Ammonia (NH3) is produced by the synthesis of nitrogen (N2) and hydrogen (H2):
N2 + 3 H2 → 2 NH3
During the first half of the 20th century, the NH3-green was produced without CO2 emission at
from hydroelectricity, water, and atmospheric air.
H2 came from the electrolysis of water, the N2 of atmospheric air.
Answer:
Mg (s) + 2HCl -----> H2(g) + MgCl2 (aq)
Explanation:
This reaction produces hydrogen gas and magnesium chloride. So, we have :
Mg (s) + HCl -----> H2 (g) + MgCl2 (aq)
In other to balance this reaction, we then add "2" to HCl on the left hand side of the equation so that the number of hydrogen and chlorine on both sides will be equal.
Thus our balanced equation would be:
Mg (s) + 2HCl (aq) -----> H2 (g) + MgCl2 (aq)
Object does not have mass
