The right answer is the first molecule (3-methyl-butan-2-one).
There are many bands around 2900-3000 cm-1 proving the presence de C-H bound.
There an intense band in 1700 cm-1 proving the presence of C=O.
There's no band proving the presence of aromatic structure, C=C bond nor C-N bond.
An atomic mass unit is defined as a mass equal to one twelfth the mass of an atom of carbon-12. The mass of any isotope of any element is expressed in relation to the carbon-12 standard. For example, one atom of helium-4 has a mass of 4.0026 amu. An atom of sulfur-32 has a mass of 31.972 amu.
Answer:
Heres my attempt at this and hope it helps friend
.1372L * (.83M/L) = .114 mols propanoic acid
.06862L * (1.1M/L) = 0.0755 mols NaOH
using the henderson hasselbach equation
ph = pKa + log([A-]/[HA])
so
ph = 4.89 + log(.0755/.114) = 4.72
Explanation:
Identify at least three reasons the Articles of Confederations failed as a governing document. In your opinion, evaluate which defect was most debilitating, using evidence and your knowledge of American government to justify your position.
<span>Step 1 is to determine the mass of each part
Mass of Ca is 40.08 g
Mass of C is 12.01 g
Mass of O is 16.00 x 3 = 48.00 g
Step 2 is to determine the total mass of the compound
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g
Step 3 is to determine the % of each part using the following formula:
Mass of part / total mass x 100 =
40.08 / 100.09 x 100 = 40.04 % Ca
12.01 / 100.09 x 100 = 12.00 % C
48.00 / 100.09 x 100 = 47.96 % O
Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :)
40.04
+12.00
+47.96
=100.00</span><span>
</span>
