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valina [46]
3 years ago
15

Question 2: Naming Hydrocarbons

Chemistry
1 answer:
oksian1 [2.3K]3 years ago
4 0

Answer:

1) 1-Pentyne

2) But-2-yne

3) Ethyne

4) 1-Propyne

5) 1-Butyne

6) Pent-2-yne

Explanation:

1) The number of carbon on the longest chain = 5, (two carbon atoms between the triple bond) therefore the prefix is <em>pen-</em>

The functional group is the triple bond on the first carbon with the suffix <em>-yne</em>

The name of the compound is 1-Pentyne

2) There are 4 carbon atoms on the longest chain, therefore, the prefix is <em>but-</em>

The location of the triple bond functional group (suffix; <em>-yne</em>) is on the second carbon atom

Therefore, the name of the compound is But-2-yne

3) The triple bond functional group is between the only two carbon atoms, therefore, the name is Ethyne

4) The number of carbon atoms = 3, with a prefix <em>prop-</em>, and the functional group on the first carbon atom on the longest chain is the triple bond, with suffix <em>-yne</em>

Therefore, the name of the compound is 1-Propyne

5) The number of carbon atoms on the longest chain = 4, the prefix is <em>but-</em>

The location of the triple bond functional group (suffix<em> -yne</em>) is on the first carbon atom

Therefore, the name is 1-Butyne

6) The number of carbon atoms on the longest chain = 5, the prefix of the compound is <em>pen-</em>

The location of the triple bond functional group (suffix; <em>-yne</em>) is on the second carbon atom, therefore;

The name of the compound is Pent-2-yne

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3 years ago
Analysis of a compound of sulfur, oxygen and fluorine showed that it is 31.42% S and 31.35% O, with F accounting for the remaind
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Answer:

The molecular formula is  SO2F2

Explanation:

Step 1: Data given

Suppose the mass of compound = 100 grams

The compound contains:

31.42 % S = 31.42 grams S

31.35 % O = 31.35 grams O

100 - 31.42 - 31.35 = 37.23 F

Molar mass of S = 32.065 g/mol

Molar mass F = 19.00 g/mol

Molar mass O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles S = 31.42 grams / 32.065 g/mol

Moles S = 0.9799 moles

Moles 0 = 31.35 grams / 16.00 g/mol

Moles 0 = 1.959 moles

Moles F = 37.23 grams / 19.00 g/mol

Moles F = 1.959 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.9799 / 0.9799 = 1

F: 1.959/ 0.9799 = 2

O : 1.959 / 0.9799 = 2

The empirical formula is SO2F2

This formula has a molecular mass of 102.06 g/mol

This means the empirical formula is also the molecular formula : SO2F2

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In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in
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Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

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From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

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(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

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