<span>Ionic bonding is the entire exchange of valence electron(s) between particles. It is a kind of substance security that creates two oppositely charged particles. In ionic bonds, the metal loses electrons to wind up plainly an emphatically charged cation, though the nonmetal acknowledges those electrons to end up noticeably a contrarily charged anion.</span>
Answer:
See below.
Explanation:
It is equal to the energy transferred to (or work done on) an object when a force of one newton acts on that object in the direction of the force's.
Best Regards!
Answer is: c. 1.204 × 10²⁴ atoms
of carbon.
n(C) = 2 mol; amount of substance of carbon.
Na = 6.02·10²³ 1/mol; Avogadro constant (the number of constituent particles, in this example atoms, that are contained in the amount of substance given by one mole).
N(C) = n(C) · Na.
N(C) = 2 mol · 6.02·10²³ 1/mol.
N(C) = 12.04·10²³ = 1.204·10²⁴; number of carbon atoms in a sample.
Answer:
Option d.
1 mole AlCl3in 500 g water
Explanation:
ΔT = Kf . m . i
Freezing T° of solution = - (Kf . m . i)
In order to have the lowest freezing T° of solution, we need to know which solution has the highest value for the product (Kf . m . i)
Kf is a constant, so stays the same and m stays also the same because we have the same moles, in the same amount of solvent. In conclussion, same molality to all.
i defines everything. The i refers to the Van't Hoff factor which are the number of ions dissolved in solution. We assume 100 & of ionization so:
a. Glucose → i = 1
Glucose is non electrolytic, no ions formed
b. MgF₂ → Mg²⁺ + 2F⁻
i = 3. 1 mol of magnessium cation and 2 fluorides.
c. KBr → K⁺ + Br⁻
i = 2. 1 mol potassium cation and 1 mol of bromide anion
d. AlCl₃ → Al³⁺ + 3Cl⁻
i = 4. 1 mol of aluminum cation and 3 mol of chlorides.
Kf . m . 4 → option d will has the highest product, therefore will be the lowest freezing point.
