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charle [14.2K]
3 years ago
6

The melting point of phenol is 40.5∘C and that of toluene is −95∘C. What is the best explanation for this difference? Select the

correct answer below: a. The (−OH) group on phenol can form hydrogen bonds, and the −CH3 group on toluene cannot. b. Phenol has only one hydrogen on the −OH group available to form hydrogen bonds, so the hydrogen bond is stronger. c. In toluene, the hydrogen bond is spread over all three hydrogens on the methyl group, so the interaction is weaker overall. d. Phenol has a higher molecular mass than toluene. e. None of the above.
Chemistry
2 answers:
Simora [160]3 years ago
7 0

Answer:

a. The (−OH) group on phenol can form hydrogen bonds, and the −CH3 group on toluene cannot.

Explanation:

Hello,

We firs must consider that the hydroxyl functional group is present in phenol as a highly polar section into its structure. Thus, phenol molecules are strongly associated by the presence of hydrogen bonds which toluene does not have due to its apolarity.

Consequently, since associating interactions are present in the phenol but absent in the toluene, more energy must be supplied to the phenol to melt it down, that is why phenol's melting point is higher than toluene's that one.

Best energy

romanna [79]3 years ago
3 0

Answer:

None of the above

Explanation:

The (−OH) group on phenol can form hydrogen bonds, and the −CH3 group on toluene cannot.

Phenol has only one hydrogen on the −OH group available to form hydrogen bonds, so the hydrogen bond is stronger. In toluene, the hydrogen bond is spread over all three hydrogens on the methyl group, so the interaction is weaker overall.

Phenol has a higher molecular mass than toluene.

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3 years ago
As ocean depth increases, _____. temperature increases density decreases salinity increases pressure decreases
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Answer:

The correct option is: salinity increases  

Explanation:

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4 0
3 years ago
Decane (C10H22) is used in diesel. The combustion for decane follows the equation: 2 C10H22 + 31 O2 à 20 CO2 + 22 H2O. Calculate
creativ13 [48]

The mass of water produced is 792 grams by the combustion of 568 grams of decane.

Given:

Combustion of 568 grams of decane with 2979 grams of oxygen.

2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O

To find:

The mass of water produced by combustion of 568 grams of decane.

Solution:

Mass of decane = 568 g

Moles of decane :

= \frac{568 g}{142 g/mol}=4 mol

Mass of oxygen gas = 2976 g

Moles of oxygen gas:

= \frac{2976 g}{32 g/mol}=93 mol

2 C_{10}H_{22 }+ 31 O_2 \rightarrow 20 CO_2 + 22 H_2O

According to reaction, 2 moles of decane reacts with 31 moles of oxygen, then 4 moles of decane will react with:

=\frac{31}{2}\times 4mol=62\text{ mol of}O_2

But according to the question, we have 93.0 moles of oxygen gas which is more than 62 moles of oxygen gas.

So, this means that oxygen gas is present in an excessive amount. Which simply means:

  • Oxygen gas is an excessive reagent.
  • Decane is a limiting reagent.
  • Decane being limiting reagent will be responsible for the amount of water produced after the reaction.

According to reaction, 22 moles of water is produced from 2 moles of decane, then 4 moles of decane will produce:

=\frac{22}{2}\times 4mol=44\text{mol of }H_2O

Mass of 44 moles of water ;

=44mol\times 18g/mol=792g

792 grams of water is produced by the combustion of 568 grams of decane.

Learn more about limiting reagent and excessive reagent here:

brainly.com/question/14225536?referrer=searchResults

brainly.com/question/7144022?referrer=searchResults

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Consider the general reaction: 2 A + b B → c C and the following average rate data over a specific time period \Delta t: - \frac
lawyer [7]

Answer:

c = 4

Explanation:

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       a A     +    b B     ⇒ c C   +    d D

the rate is given by:

rate = - 1/a ΔA/Δt = - 1/b ΔB/Δt = + 1/c ΔC/Δt = + 1/d ΔD/Δt

this is done so as to express the rate in a standarized way which is the same to all the reactants and products irrespective of their stoichiometric coefficients.

For this question in particular we know the coefficient of A and need to determine the coefficient c.

- 1/2 ΔA/Δt = + 1/c ΔC/Δt

- 1/2 (-0.0080 ) = + 1/c ( 0.0160 mol L⁻¹s⁻¹ )

0.0040 mol L⁻¹s⁻¹  c = 0.0160 mol L⁻¹s⁻¹

∴ c = 0.0160 / 0.0040 = 4

7 0
3 years ago
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