Oxygen: 16.00 g/mol
Hydrogen: 1.01 g/mol
The mass would be 17.01 g/mol
Let <span>A= ebc </span>
<span> 0.765 = 86,300*1.3*c </span>
<span>
Solve for c = approximately 7E-6 Molar
= mols/L soln. </span>
<span>g = 7E-6*893.49
= about 0.006 g chlorophyll/L soln. </span>
<span>
1000 x 0.785 = 785 g ethanol. </span>
<span>Conc. = about 0.006g chlorophyll/785 g soln. </span>
<span>
Change that to ppm. by using formula:
(0.006/785)*1E6</span>
Answer:
Butan-2-one
Explanation:
1. 1700 cm⁻¹
A strong peak near 1700 cm⁻¹ is almost certainly a carbonyl (C=O) group.
2. Triplet-quartet
A triplet-quartet pattern indicates an ethyl group.
The 2H quartet is a CH₂ adjacent to a CH₃. The peak normally occurs at δ 1.3, but it is shifted 1.2 ppm downfield to δ 2.47 by an adjacent C=O group.
The 3H triplet at δ 1.05 is the methyl group. It, too, is shifted downfield from its normal position at δ 0.9. The effect is smaller, because the methyl group is further from the carbonyl.
3. 3H(s) at δ 2.13
This indicates a CH₃ group with no adjacent hydrogen atoms.
It is shifted 0.8 ppm downfield to δ 2.13 by the adjacent C=O group.
4. Identification
The identified pieces are CH₃CH₂-, -(CO)-, and -CH₃. There is only one way to put them together: CH₃CH₂-(C=O)-CH₃.
The compound is butan-2-one.
Answer:
Strong attraction between ions and the second one High melting points
Explanation: