Vinegar is an aqueous solution of acetic acid, ch3cooh. a 5.00 ml sample of a particular vinegar requires 26.90 ml of 0.175 m na
1 answer:
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In this kind of exercises, you should use the "ideal gas" rules: PV = nRT
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa
V should be in cubic meter:
16L = 0.016 m3
R =
= constant
= 
==> P1 * V1 = P2 * V2
V2 =
= 
V2 = 0.00581 m3 = 5.81 L
P should be in Pascal:
445mmHg = 59328Pa
1225mmHg = 163319Pa
V should be in cubic meter:
16L = 0.016 m3
R =
==> P1 * V1 = P2 * V2
V2 =
V2 = 0.00581 m3 = 5.81 L
Answer:
Here's what I get.
Explanation:
The frequency of a vibration depends on the strength of the bond (the force constant).
The stronger the bond, the more energy is needed for the vibration, so the frequency (f) and the wavenumber increase.
Acetophenone
Resonance interactions with the aromatic ring give the C=O bond in acetophenone a mix of single- and double-bond character, and the bond frequency = 1685 cm⁻¹.
p-Aminoacetophenone
The +R effect of the amino group increases the single-bond character of the C=O bond. The bond lengthens, so it becomes weaker.
The vibrational energy decreases, so wavenumber decreases to 1652 cm⁻¹.
p-Nitroacetophenone
The nitro group puts a partial positive charge on C-1. The -I effect withdraws electrons from the acetyl group.
As electron density moves toward C-1, the double bond character of the C=O group increases.
The bond length decreases, so the bond becomes stronger, and wavenumber increases to 1693 cm¹.
