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 is: molarity is 0,155 M.
V(solution) = 90,0 mL = 0,09 L.
ω(NaCl) = 0,92% ÷ 100% = 0,0092.
d(solution) = 1 g/mL.
m(solution) = V(solution) · d(solution).
m(solution) = 90 mL · 1 g/mL = 90 g.
m(NaCl) = 90 g · 0,0092 = 0,828 g.
n(NaCl) = 0,828 g ÷ 58,4 g/mol.
n(NaCl) = 0,014 mol.
c(solution) = 0,014 mol ÷ 0,09 L.
c(solution) = 0,155 mol/L.
Molar mass of N = 14 g/molMolar mass of O2 = 32 g/molAdding both masses = 46 g/molActual molar mass/ Empirical molar mass = 138.02 / 46 = 3Now multiplying this co effecient with empirical fomula NO2 = 3(NO2) = N3O6So according to above explanation,D) N3O6, is the correct answer.
Answer:
electron gain
Explanation:
e- is for electron gain while
e+ is for electron loss
