Answer:
127.3° C, (This is not a choice)
Explanation:
This is about the colligative property of boiling point.
ΔT = Kb . m . i
Where:
ΔT = T° boling of solution - T° boiling of pure solvent
Kb = Boiling constant
m = molal (mol/kg)
i = Van't Hoff factor (number of particles dissolved in solution)
Water is not a ionic compound, but we assume that i = 2
H₂O → H⁺ + OH⁻
T° boling of solution - 118.1°C = 0.52°C . m . 2
Mass of solvent = Solvent volume / Solvent density
Mass of solvent = 500 mL / 1.049g/mL → 476.6 g
Mol of water are mass / molar mass
76 g / 18g/m = 4.22 moles
These moles are in 476.6 g
Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m
T° boling of solution = 0.52°C . 8.85 m . 2 + 118.1°C = 127.3°C
Answer:
<u>[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2</u>
Explanation:
2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)
The equilibrium constant expression in terms of concentrations is:
Kc=<u>[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2</u><u>.</u>
d = √((x1 - x2)2 + (y1 - y2)2)
( -2 , 5 ) ( 12 , -1 )
↑ ↑ ↑ ↑
x1 y1 x2 y2
d = √((-2 - 12)2 + (5 - (-1))2) = √((-14)2 + 62) = √(196 + 36) = √232 = 2√58 ≈ 15.23
Ethane
see the picture for explanation