Answer: In CaF2, the oxidation number of Ca is +2
, and that of F is -1
. In H2SO4, the oxidation number of H is +1
, that of S is +6
, and that of O is -2
. In CaSO4, the oxidation number of Ca is +2
, that of S is +6
, and that of O is -2
. In HF, the oxidation number of H is +1
, and that of F is -1
The answer is: lose electrons and form positive ions.
Most metals have strong metallic bond, because of strong electrostatic attractive force between valence electrons (metals usually have low ionization energy and lose electrons easy) and positively charged metal ions.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
For example, magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.
Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.
Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
Answer:
Subtract the charge from the atomic number. When an ion has a positive charge, the atom has lost electrons. To calculate the remaining number of electrons, you subtract the amount of extra charge from the atomic number. In the case of a positive ion, there are more protons than electrons.
Explanation:
