Answer:
(c) The retention time would be higher (d) The retention time would be lower.
Explanation:
For the polar solutes which were separated using the hydrophilic interaction chromatography (HILIC) with a strongly polar bonded phase, the retention time would be higher if eluent were changed from 80 vol% to 90 vol% acetonitrile in water.
However, for the polar solutes which were separated using the normal-phase chromatography on bare silica with methyl t=butyl ether and 2-propanol solvent, the retention time would be lower if the eluent were changed from 40 vol% to 60 vol% 2-propanol.
First convert grams to moles
using molar mass of butane that is 58.1 g
3.50g C4H10 x (1 mol
C4H10)/(58.1g C4H10) = 0.06024 mol C4H10 <span>
<span>Now convert moles to molecules by using Avogadro’s number
0.06024 mol C4H10 x (6.022x10^23 molecules C4H10)/(1 mol
C4H10) = 3.627x10^22 molecules C4H10
And there are 4 carbon atoms in 1 molecule of butane, so use
the following ratio:
3.627 x 10^22 molecules C4H10 x (4 atoms C)/(1 molecule
C4H10)
<span>= 1.45 x 10^23 atoms of carbon are present</span></span></span>
Formula for hydroselenic acid: H2Se
Answer:
See explanation and image attached
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.
Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.
Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.
The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.
Answer:
7.16x10⁻⁸M = [Ag+]
Explanation:
Using the equation:
E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)
<em>Where E</em>⁰<em>= 0.4249V</em>
<em>E(Cell) = -(-0.0019V) -Measured value-</em>
<em>[Cu2+] = 1M</em>
<em />
Replacing:
0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)
-0.423V = - 0.0296 • log (1M/[Ag+]²)
14.29 = log (1M/[Ag+]²)
1.95x10¹⁴ = 1M / [Ag+]²
[Ag+]² = 5.12x10⁻¹⁵M
7.16x10⁻⁸M = [Ag+]
