It will form HCN Hydrogen Cyanide.
Since it talks about the number of particles, we would need the Avogadro's number which is an empirical value equal to 6.022×10²³ particles/mole. The other information that we have to know is the molar mass of Na₂SO₄ which is 142.04 g/mol.
Mass of a single particle of Na₂SO₄:
(142.04 grams Na₂SO₄/mol) * (1 mol/6.022×10²³ particles) = 2.359×10⁻²² g
Thus, each Na₂SO₄ weighs 2.359×10⁻²² g.
Number of Na₂SO₄ particles in the mixture:
(11.53 g Na₂SO₄) * (1 mol/142.04 g) * (6.022×10²³ particles/mol) = 4.89×10²² particles
Thus, the mixture contains 4.89×10²² particles is Na₂SO₄.
Answer:
there is no d electron that can be promoted via the absorption of visible light
Explanation:
One of the properties of transition elements is the possession of incompletely filled d orbitals. This property accounts for their unique colours.
The colours of transition metal compounds stem from d-d transition of electrons due to the presence of vacant d orbitals of appropriate energy to which electrons could be promoted.
For elements whose atoms have a d10 configuration, such vacant orbitals does not exist hence their compounds are not colored.
Sometimes, the colour of transition metal compounds stem from ligand to metal charge transfer(LMCT) for instance in KMnO4.
Electron configuration
an element A has 3 shells and has 3 valence electrons
so electron configuration of A: 1s², 2s², 2p⁶, 3s², 3p¹
will tend to lose 3 electron to form a cation: A³⁺
a. Nitrate of A
A(NO₃)₃
b. Chloride of A
ACl₃
c. Oxide of A
A₂O₃
Nuclear reactions involves a change in the nucleus of the atom meaning that an element can entirely be converted into another element where the atom change in atomic number, mass and type.
Based on this, the correct answer is:
<span>D. Elements are created that differ from the reactants.</span>