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Answer:
Explanation:
Hello there!
In this case, since the radioactive reaction for the alpha emission of astatine-218 to bismith-214 involve the release of a helium atom as shown below:
Whereas the atomic number decreases by 2 and the mass number by 4 in agreement to the release of the Helium atom.
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Answer:
a. Vₐ = 111.5282 + 1.29396m
b. For m = 0.100m; Vₐ = 111.6576
Explanation:
The partial molar volume of compound A in a mixture of A and B is defined as :
Where V is volume and n are moles of a.
a. As molality is proportional to moles of substance, partial molar volume of glucose can be defined as:
Vₐ = dV / dm = d(1001.93 + 111.5282m + 0.64698m²) / dm
<em>Vₐ = 111.5282 + 1.29396m</em>
b. Replacing for m = 0.100m:
Vₐ = 111.5282 + 1.29396×0.100
<em>Vₐ = 111.6576</em>
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Answer:
Isopropyl propionate
Explanation:
1. Information from formula
The formula is C₆H₁₂O₂. A six-carbon alkane would have the formula C₆H₁₄. The deficiency of two H atoms indicates the presence of either a ring or a double bond.
2. Information from the spectrum
(a) Triplet-quartet
A 3H triplet and a 2H quartet is the classic pattern for a CH₃CH₂- (ethyl) group
(b) Septet-doublet
A 1H septet and a 6H doublet is the classic pattern for a -CH(CH₃)₂ (isopropyl) group
(c) The rest of the molecule
The ethyl and isopropyl groups together add up to C₇H₁₂.
The rest of the molecule must have the formula CO₂ and one unit of unsaturation. That must be a C=O group.
The compound is either
CH₃CH₂-COO-CH(CH₃)₂ or (CH₃)₂CH-COO-CH₂CH₃.
(d) Well, which is it?
The O atom of the ester function should have the greatest effect on the H atom on the adjacent carbon atom.
The CH of an isopropyl is normally at 1.7. The adjacent O atom should pull it down perhaps 3.2 units to 4.9.
The CH₂ of an ethyl group is normally at 1.2. The adjacent O atom should pull it down to about 4.4.
We see a signal at 5.0 but none near 4.4. The compound is isopropyl propionate.
3. Summary
My peak assignments are shown in the diagram below.
