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2 years ago

What is the name of this hydrocarbon?

Chemistry

2 answers:

slavikrds [6]2 years ago

5 0

<u>Answer:</u> The name of the given hydrocarbon is hexane.

<u>Explanation:</u>

The given hydrocarbon has all the single bond between carbon and carbon atoms. Thus, this compound is considered as an alkane.

The IUPAC nomenclature of alkanes are given as follows:

Select the longest possible carbon chain.
For the number of carbon atom, we add prefix as 'meth' for 1, 'eth' for 2, 'prop' for 3, 'but' for 4, 'pent' for 5, 'hex' for 6, 'sept' for 7, 'oct' for 8, 'nona' for 9 and 'deca' for 10.
A suffix '-ane' is added at the end of the name.
If two of more similar alkyl groups are present, then the words 'di', 'tri' 'tetra' and so on are used to specify the number of times these alkyl groups appear in the chain.

Hence, the name for given hydrocarbon will be hexane.

Step2247 [10]2 years ago

3 0

There are 6 Carbon atoms and 14 Hydrogen which makes it a Hexane

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Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).

nalin [4]

The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE for C≡O = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

To learn more about enthalpy here

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8 months ago

1: At which temperature would a reaction withΔH = -102 kJ/mol, ΔS = -0.188 kJ/(mol×K) be spontaneous? 2: At which temperature wo

Naddik [55]

Answer:

1: At temperatures below 542.55 K

2: At temperatures above 660 K

Explanation:

Hello there!

In this case, according to the thermodynamic definition of the Gibbs free energy, it is possible to write the following expression:

$\Delta G=\Delta H-T\Delta S$

Whereas ΔG=0 for the spontaneous transition. In such a way, we proceed as follows:

$0=\Delta H-T\Delta S\\\\T=\frac{-102kJ/mol}{-0.188kJ/mol-K} \\\\T=542.55K$

It means that at temperatures lower than 542.55 K the reaction will be spontaneous.

$0=\Delta H-T\Delta S\\\\T=\frac{132kJ/mol}{0.200kJ/mol-K} \\\\T=660K$

It means that at temperatures higher than 660 K the reaction will be spontaneous.

Best regards!

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2 years ago

The vapor pressure of water at 25.0°C is 23.8 torr. Determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.

mrs_skeptik [129]

According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.

the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

$\frac{p^{0-}p}{p^{0}}=x_{B}$