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2 years ago

Why do densities of haloalkanes decrease when the size of alkyl group increases?

Chemistry

1 answer:

Verizon [17]2 years ago

8 0

Answer:

Here's what I get.

Explanation:

Your argument makes sense. For example, the density of C₅H₁₂ is about 0.63 g/mL, and it increases gradually to a limit of about 0.83 g/mL.

With the lower alkyl halides, the halogen atom is the major contributor to the mass of the molecule. The density of chloromethane is 2.2 g/mL.

As you increase the number of C atoms, the Cl atom becomes a smaller part of the molecule. The alkyl halide becomes more like an alkane.

By the time you have reached four C atoms, the density has reached a limiting value of about 0.89 g/mL.

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The 1H-NMR of a compound with molecular formula C6H12O2 consists of four signals: 1.1 (triplet, integrating to 3 Hydrogens), 1.2

Tamiku [17]

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.

4 0

3 years ago

Find ∆G for a cell whose potential is +0.24 V with 4 moles of electrons exchanged.

Setler [38]

We need to use the following formula
Δ $G= -nF E_{cell}$

$E_{cell}= 0.24V$
n= 4 moles
F= constant= 96500C/mol

let's plug in the values.

ΔG= -(4)(96500)(0.24)= -92640 J or -92.6 kJ

6 0

2 years ago

Read 2 more answers

The products of a neutralization reaction are ________ and ________ a salt.

Alla [95]

Explanation:

Neutralization reaction is a type of chemical reaction in which an acid and a base react together to form a salt and water as products.

7 0

2 years ago

Read 2 more answers

Why is it necessary to use a new salt bridge for each salt bridge for each electrochemical cell?

Irina-Kira [14]

Answer:

salt bridge balances the charge when electrons move from one half cell to another half cell.

Explanation:

Explanation: A salt bridge balances the charge when electrons move from one half cell to another half cell. During this process the salt bridge uses its electrolyte solution which further helps in balancing charges in both the half cells. ... Therefore, for each electrochemical cell a new salt bridge is used.

6 0

2 years ago

Read 2 more answers

Cu + 2AgNO3 →

netineya [11]

Answer: 292.54g of Ag

Explanation:

Cu + 2AgNO3 →Cu(NO3)2 + 2 Ag

mass conc. Of Ag = n x molar Mass

Mass conc. Of Ag = 2 x 108 = 216g

From the equation,

63.5g of Cu produced 216g of Ag

Therefore, 86g of Cu will produce Xg of Ag. i.e

Xg of Ag = (86 x 216)/63.5 = 292.54g

5 0

2 years ago