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melomori [17]
3 years ago
10

Draw the structure of a compound of molecular formula C4H8O that has a signal in its 13C NMR spectrum at ≥ 160 ppm. Then draw th

e structure of an isomer with no rings of molecular formula C4H8O that has all of its 13C NMR signals at < 160 ppm.

Chemistry
1 answer:
-Dominant- [34]3 years ago
4 0

Answer:

1. Butyraldehyde; 2. but-3-en-1-ol

Explanation:

1. Peak ≥ 160 ppm

The formula C₄H₈O shows that the Index of Hydrogen Deficiency = 1.

It could be caused by either a ring or a double bond.

A peak at ≥ 160 ppm strongly indicates a C=O group, so the rest of the molecule can contain no rings or double bonds.

There are no other heteroatoms, so the compound most be an aldehyde or a ketone.

One compound that meets these criteria is butyraldehyde, CH₃CH₂CH₂CH=O (see Fig. 1.)

2. No rings; all peaks < 160 ppm

If all peaks are < 160 ppm, there can no C=O group.

There is no ring, so there must be a C=C double bond.

There is no other unsaturation, so the O atom must be present as an alcohol or an ether.

One compound that meets these criteria is but-3-en-1-ol, CH₂=CHCH₂CH₂OH (see Fig. 2).

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the molecular mass of hydrogen sulphide, which contains two atoms of hydrogen and one atom of sulphur is = 2 — 1 + 1 — 32 = 34 a.m.u.

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From a laboratory process designed to separate water into hydrogen and oxygen gas, a student collected 20.0g of Hydrogen and 158
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A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5
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Answer:

0.263M of CH₃COOH is the concentration of the solution.

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

<em>1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.</em>

<em />

In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:

0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.

As the sample of acetic acid had a volume of 25.0mL = 0.025L:

6.56x10⁻³ moles of CH₃COOH / 0.0250L =

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3 years ago
A sample of gas contains 0.1700 mol of NH3(g) and 0.2125 mol of O2(g) and occupies a volume of 17.8 L. The following reaction ta
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Answer:

The volume of the sample after the reaction takes place is 19.78 L.

Explanation:

The given variables are;

Number of moles of NH₃(g) = 0.1700 mol

Number of moles of O₂(g) = 0.2125 mol

Volume occupied by the mixture = 17.8 L

The reaction

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Then takes place

That is 4 moles of NH₃(g) reacts with 5 moles of O₂(g) to produce 4 moles of NO(g) and 6 moles of H₂O(g).

Since there are less number of moles of NH₃(g) (= 0.1700 mol) in the mixture, we factor the above equation by the number of moles of NH₃(g)  present.

That is,

1 moles of NH₃(g) reacts with 5/4 moles of O₂(g) to produce 1 moles of NO(g) and 3/2 moles of H₂O(g).

Therefore,

0.1700 mol of NH₃(g) reacts with 5/4×0.1700  moles of O₂(g) to produce 0.1700  moles of NO(g) and 3/2×0.1700  moles of H₂O(g).

Which gives

0.1700 mol of NH₃(g) reacts with 0.2125  moles of O₂(g) to produce 0.1700  moles of NO(g) and 0.255  moles of H₂O(g).

Therefore, all of the NH₃(g) and O₂(g)  are consumed in the reaction and the present gases in sample then becomes

0.1700  moles of NO(g) and 0.255  moles of H₂O(g).

Total number of moles of reactant = 0.17 + 0.2125 = 0.3825

Total number of moles of product formed = 0.17 + 0.255 = 0.425

However, Avogadro's law states that equal volume of all gases at the same temperature and pressure contains equal number of molecules.

That is volume occupied by  0.3825 moles of gas = 17.8 L

Therefore the volume occupied by  0.425 moles of gas = 17.8×0.425/0.3825 L = 19.78 L

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