Answer:
2H2S (g) + 3O2 (g) → 2H2O (l) + 2SO2 (g)
Calculate ΔH° from the given data. Is the reaction exothermic or endothermic?
ΔH°f (H2S) = -20.15 kJ/mol; ΔH°f (O2) = 0 kJ/, mol; ΔH°f (H2O) = -285.8 kJ/mol; ΔH°f (SO2) = -296.4 kJ/mol
Sugar would not increase the boiling point of the water as much as the others.To show this, we recall the equation for boiling point elevation ∆t:
∆t = i Kb m
The difference in these four solutions will be in the product of the van't Hoff Factor i and molality m: i * m.
If we assume that we have 0.1m of each solute, the products i*m for these solutions are:
Solution Identity of particles i i * m
0.1 m NaCl Na+, Cl- 2 2 * 0.1 = 0.2
0.1 m MgSO4 Mg^2+, SO4^2- 2 2 * 0.1 = 0.2
0.1 m K2SO4 two K+, SO4^2- 3 3 * 0.1 = 0.3
0.1 m C12H22O11 covalently bonded molecules 1 1 * 0.1 = 0.1
Therefore, having the lowest i * m, sucrose will have the lowest boiling point elevation.
Answer: C
Explanation:
A. Shows 3-Hexyne (NOT 2-HEXYNE)
B. Shows 7 carbons (too many) (NOT 2-HEXYNE)
C. Shows a triple bond (yne) and 6 carbons and it's on the second carbon (2-HEXYNE)
D. Shows two substitent on the second carbon but the triple bond is on the 3rd carbon so it's 2,2-dimethyl-3-heptyne (NOT 2-HEXYNE)
Answer:
helium family or neon family
Explanation:
you can use p
