Answer:
Specififc rotation [∝] = 0.5° mL/g.dm
Explanation:
Given that:
mass = 400 mg
volume = 10 mL
For a solution,
The Concentration = mass/volume
Concentration = 400/10
Concentration = 40 g/mL
The path length l = 20 cm = 2 dm
Observed rotation [∝] = + 40°
Specififc rotation [∝] = ∝/l × c
where;
l = path length
c = concentration
Specififc rotation [∝] = (40 / 2 × 40)
Specififc rotation [∝] = 0.5° mL/g.dm
The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)
The statement which describes how NO2- reacts in this equilibrium:
<span>H2SO3(aq) + NO2-(aq) HSO3-(aq) + HNO2(aq
is the second option - </span><span>B. as a Brønsted-Lowry base by accepting a proton.
</span>This is because bases take proton H+ in order to become HNO2.
