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klasskru [66]
3 years ago
5

This compound, an eight-carbon alkane, serves as the 100 octane standard for gasoline. It has five methyl groups, one quaternary

carbon, tests negative for C-C double bonds and gives four monochloro substitution products, C8H17Cl. Propose a structure for this compound.

Chemistry
1 answer:
Reika [66]3 years ago
5 0

Answer:

2,2,4-Trimethyl-pentane

Explanation:

Structural characteristics of the compound is as follows:

  • Has five methyl group
  • Has one quaternary carbon
  • No. double bond
  • Gives four monochloro substitution products

Compound must have straight chain of 5 carbons.

Three methyl substituents are attached to 2 and 4 carbons.

Therefore, IUPAC name of the compound will be 2,2,4-Trimethyl-pentane.

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How many grams of CO2 are used when 6.0 g of O2 are produced? Express your answer with the appropriate units.
Ipatiy [6.2K]

Answer:

that is why co2 is in the power of 2ik

8 0
3 years ago
An elimination reaction can best be described as a reaction in which an elimination reaction can best be described as a reaction
tresset_1 [31]
So I’m not 100% sure what you’re asking but I’m going to give it a go. The elimination reaction is a term used in organic chemistry that describes a type of reactions. The name kinda tells you what’s going to happen. Something is going to be removed/eliminated from initial reactant/substrate and as a result, an alkene (double bond containing compound) will form.

In elimination reactions a hydrogen atom is first removed (as a H+) from the beta carbon. As a result, the left behind electrons create a pi bond between the beta carbon and the neighboring alpha carbon. This promotes the electronegative atom, on the alpha carbon, to leaves the substrate taking both electrons from the shared sigma bond with the alpha carbon.
4 0
3 years ago
1.5 moles of Al is how many<br> atoms?
aev [14]

Answer:

<h2>9.03 × 10²³ atoms </h2>

Explanation:

The number of atoms of Al can be found by using the formula

<h3>N = n × L</h3>

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 1.5 × 6.02 × 10²³

We have the final answer as

<h3>9.03 × 10²³ atoms</h3>

Hope this helps you

4 0
2 years ago
How can a chemical reaction be sped up?Select from the drop-down menu to correctly complete the statement.
lara31 [8.8K]

Answer: Catalyst

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8 0
3 years ago
Calculate E ° for the half‑reaction, AgCl ( s ) + e − − ⇀ ↽ − Ag ( s ) + Cl − ( aq ) given that the solubility product constant
antoniya [11.8K]

Answer: The value of E^{o} for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)

Its solubility product will be as follows.

       K_{sp} = [Ag^{+}][Cl^{-}]

Cell reaction for this equation is as follows.

     Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)

Reduction half-reaction: Ag^{+} + 1e^{-} \rightarrow Ag(s),  E^{o}_{Ag^{+}/Ag} = 0.799 V

Oxidation half-reaction: Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-},   E^{o}_{AgCl/Ag} = ?

Cell reaction: Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, K_{sp} = [Ag^{+}][Cl^{-}]

                                               = 1.77 \times 10^{-10}

Therefore, according to the Nernst equation

           E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

At equilibrium, E_{cell} = 0.00 V

Putting the given values into the above formula as follows.

         E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}

        0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}    

       E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}

                  = 0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}

       0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}

       0.577 V = 0.799 V - E^{o}_{AgCl/Ag}

       E^{o}_{AgCl/Ag} = 0.222 V

Thus, we can conclude that value of E^{o} for the half-cell reaction is 0.222 V.

3 0
3 years ago
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