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motikmotik
2 years ago
11

Each of the following names is wrong. Draw structures based on them, and correct the names:(a) 4-methylhexane

Chemistry
1 answer:
Keith_Richards [23]2 years ago
5 0

The correct name is 3-methylhexane.

<h3>Structures : </h3>

The structure of 4-methylhexane is

                          CH3

                           |

CH3-CH2-CH2-CH-CH2-CH3

In the systematic nomenclature, the longest carbon chain is numbered so as to give the lowest number to the substituent.

                          CH3

                           |

CH3-CH2-CH2-CH-CH2-CH3

6       5       4      3     2      1

Hence the correct name is 3-methylhexane.

What is 3-methylhexane?

A branching hydrocarbon having two enantiomers is 3-methylhexane. It is one of heptane's isomers. The chirality of the molecule makes it one of only two structural isomers of heptane with this feature, the other being 2,3-dimethylpentane. 3-methylhexane and 3-methylhexane are the enantiomers.

<h3>Rules for naming the compound ? </h3>
  • Typically, a molecular molecule consists of two or more nonmetal components.
  • Molecular compounds are termed by using the stem of the first element's name plus the suffix -ide, followed by the second element. The number of atoms in a molecule is specified using numerical prefixes.

To know more about 3-methylhexane please click here : brainly.com/question/24165638

#SPJ4

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What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)
blsea [12.9K]
<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, (C_{2}H_{5})_{3}N, the reaction will be
(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}

 and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

Now, the Kb value  = 5.3 x 10^-4

And kb =  \frac{( [( C_{2}H_{5})_{3}NH^{+} ]*  OH^{-} )}{[( C_{2}H_{5})_{3}N]}

 thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

                    =0.00516 M 

Thus, pOH = 2.30 
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6 0
4 years ago
What is the relative rate of diffusion of hydrogen and nitrogen ​
vredina [299]

Answer:

2.645

Explanation:

Rate of diffusion formula:

Sqrt(mass2/mass1)

>>sqrt(14/2)

(Note:Hydrogen must exist in dwiatomic, [H2])

3 0
2 years ago
How many moles are in 1.20 x 10^21 atoms of Phosphorus?
Ludmilka [50]

Answer:

You would get 19.969 moles

Explanation:

8 0
3 years ago
The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera
torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = \frac{0.570 g}{122.12 g/mol}=0.004667 mol

Energy released by 0.004667 moles of benzoic acid on combustion:

Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J

Heat capacity of the calorimeter = C

Change in temperature of the calorimeter = ΔT = 2.053°C

Q=C\times \Delta T

15,066.8 J=C\times 2.053^oC

C=7,338.92 J/^oC

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = \frac{2.900 g}{180.16 g/mol}=0.016097 mol

Energy released by the 0.016097 moles of calorimeter  combustion:

Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

Q'=C\times \Delta T'

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\Delta T'=6.097^oC

The temperature change from the combustion of the glucose is 6.097°C.

6 0
3 years ago
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Answer:

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As the timber market was totally undisciplined, the exploitation of the wood caused a strong deforestation, leaving the soil totally unprotected and susceptible to strong erosion.

Soil erosion has a very negative impact on the environment, requiring regulations to be made to prevent this from happening.

Based on this, we can say that one way to prevent environmental problems in the seaport of troy is by establishing laws and regulations that limit logging activities.

3 0
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