All categories

2 years ago

A compound with molecular formula C6H9No that has an amide functional group and does not have an alkene group

Chemistry

1 answer:

Flauer [41]2 years ago

3 0

Answer:

Following are the responses to the given question:

Explanation:

$\to C_6H_9NO$

calculating the HOI:

$= \frac{2C+2+N-H-X}{2} \\\\ =\frac{2\times 6+2+1-9-0}{2}\\\\ =\frac{12-6}{2}\\\\=\frac{6}{2}\\\\=3$

So, in the above compund there are three bounds or we can say that one is double and two bound is single.

Please find the attachment file of the Function group:

In the above given conditon compound doesn't include the alkene functional group that is absense of $-C=C-$ , and compound include $-C\equiv C-$ So, please find the attached file of the structure of the compound:

You might be interested in

What are some HYDROPHILIC materials?

alisha [4.7K]

Including the alkanes, oils , fats and greasy substances in general

8 0

3 years ago

Use the provided reduction potentials to calculate ArGº for the following balanced redox reaction: Pb2+(aq) + Cu(s) → Pb(s) + Cu

nydimaria [60]

Answer : The correct option is, +91 kJ/mole

Solution :

The balanced cell reaction will be,

$Cu(s)+Pb^{2+}(aq)\rightarrow Cu^{2+}(aq)+Pb(s)$

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^0_{[Pb^{2+}/Pb]}=-0.13V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

$E^0_{cell}=E^0_{cathode}-E^0_{anode}$

$E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}$

$E^0_{cell}=-0.13V-(0.34V)=-0.47V$

Now we have to calculate the standard Gibbs free energy.

Formula used :

$\Delta G^o=-nFE^o_{cell}$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

n = number of electrons = 2

F = Faraday constant = 96500 C/mole

$E^o$ = standard e.m.f of cell = -0.47 V

Now put all the given values in this formula, we get the Gibbs free energy.

$\Delta G^o=-(2\times 96500\times (-0.47))=+90710J/mole=+90.71kJ/mole\approx +91kJ/mole$

Therefore, the standard Gibbs free energy is +91 kJ/mole

6 0

2 years ago

The hammer, anvil, and stirrup are all bones found in the ear.

AfilCa [17]

Answer: Hammer, anvil and stirrup are small bones in the ear behind ear drum and before cochlea. These bones helps in transfering the vibrations from ear drum to the cochlea which is further passed to auditory nerve and then to brain

Explanation:

4 0

3 years ago

The colour of RBC is Red .why?

RoseWind [281]

Answer: fat show the first half and a lot of people in a row in this rr

7 0

3 years ago

The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and

Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

Moles KSCN = Moles Ag⁺ in the filtrate:

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

Total moles Ag⁺:

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

Moles of Ag⁺ in the precipitate:

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

Moles AsO₄ = Moles As:

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

Moles As₂O₃:

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

Mass As₂O₃:

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

7 0

3 years ago