This problem handles<em> boiling-point elevation</em>, which means we will use the formula:
ΔT = Kb * m
Where ΔT is the difference of Temperature between boiling points of the solution and the pure solvent (Tsolution - Tsolvent). Kb is the ebullioscopic constant of the solvent (2.64 for benzene), and m is the molality of the solution.
Knowing that benzene's boiling point is 80.1°C, we <u>solve for m</u>:
Tsolution - Tsolvent = Kb * m
80.23 - 80.1 = 2.64 * m
m = 0.049 m
We use the definition of molality to <u>calculate the moles of azulene</u>:
0.049 m = Xmoles azulene / 0.099 kgBenzene
Xmoles azulene = 4.87 x10⁻³ moles azulene
We use the mass and the moles of azulene to<u> calculate its molecular weight</u>:
0.640 g / 4.875 x10⁻³ mol = 130.28 g/mol
<em>A molecular formula that would fulfill that molecular weight</em> is C₁₀H₁₀. So that's the result of solving this problem.
The actual molecular formula of azulene is C₁₀H₈.
Answer:
false thought ia ion of neon = clarity active
Explanation:
The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]
In NH3 , let oxidation number of N be x
x + (+1)3 = 0
x = -3
In HNO3 , let oxidation number of N be x
1 + x + (-2)3 = 0
x = +5
In NO2 , let oxidation number of N be x
x + (-2)2 = 0
x = +4
