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erastova [34]
2 years ago
11

Each of the following names is wrong. Draw structures based on them, and correct the names:(b) 1,1,1-trimethylheptane

Chemistry
1 answer:
devlian [24]2 years ago
5 0

The correct IUPAC name of 1,1,1-trimethylheptane is 2,2-dimethyloctane.

<h3>IUPAC NAME:</h3>

It is systematic way of nomenclature of organic compounds.

It is based on the position of functional groups, preference of the functional group, long or short chain of carbon, preference of double bond, single and triple bond, branching of carbon chain, etc.

The given compound is 1,1,1-trimethylheptane.

The given name is wrong according to IUPAC name because the numbering of carbon atom should be done in that way in which the carbon atom chain is largest.

Here, in this case the numbering is done from right side. Thus the largest carbon chain have 8 carbon atom.

If the numbering is done according to the question, number of carbon atom in straight chain is 7.

Thus, we concluded that the Correct IUPAC name of 1,1,1-trimethylheptane is 2,2-dimethyloctane.

learn more about Nomenclature:

brainly.com/question/14094007

#SPJ4

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How molecules of N2 gas can be present in a 2.5 L flask at 50°C and 650 mmHg?
ratelena [41]

Answer:

0.482 ×10²³ molecules

Explanation:

Given data:

Volume of gas = 2.5 L

Temperature of gas = 50°C (50+273 = 323 k)

Pressure of gas = 650 mmHg (650/760 =0.86 atm)

Molecules of N₂= ?

Solution:

PV= nRT

n = PV/RT

n = 0.86 atm × 2.5 L /0.0821 atm. mol⁻¹. k⁻¹. L × 323 k

n = 2.15 atm. L /26.52 atm. mol⁻¹.L

n = 0.08 mol

Number of moles of N₂ are 0.08 mol.

Number of molecules:

one mole = 6.022 ×10²³ molecules

0.08×6.022 ×10²³ = 0.482 ×10²³ molecules

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3 years ago
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Fat

Alkali

Explanation:

Fat and alkali are the two primary raw materials needed to manufacture soap.

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Using the reaction below: 2 CO2(g) + 2 H2O(l) → C2H4(g) + 3 O2(g) ΔHrxn= +1411.1 kJ What would be the heat of reaction for this
maw [93]

Answer:  d) -705.55 kJ

Explanation:

Heat of reaction is the change of enthalpy during a chemical reaction with all substances in their standard states.

2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g) \Delta H=+1411.1kJ

Reversing the reaction, changes the sign of \Delta H

C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)

\Delta H=-1411.1kJ

On multiplying the reaction by \frac{1}{2} , enthalpy gets half:

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