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2 years ago

Identify the functional groups present in each molecule.

Chemistry

1 answer:

Colt1911 [192]2 years ago

5 0

Answer:

Carbon 3 is double bonded to an oxygen and attached to carbon 2 and carbon 4. :

Answer: Carbonyl group ( Ketone or aldehyde)

Carbon 17 is attached to an oxygen, which is attached to a hydrogen. :

Answer: Carboxyl group (Carboxylic acid)

A central carbon is attached to an amine, two hydrogens, and a carbon that is double bonded to an oxygen and single bonded to an oxygen attached to a hydrogen. :

Answer: Amide group

An amide group contains both amine and carboxyl

$.$

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Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T

Ostrovityanka [42]

Explanation:

The given data is as follows.

$\Delta H$ = 286 kJ = $286 kJ \times \frac{1000 J}{1 kJ}$

= 286000 J

$S_{H_{2}O} = 70 J/^{o}K$ , $S_{H_{2}} = 131 J/^{o}K$

$S_{O_{2}} = 205 J/^{o}K$

Hence, formula to calculate entropy change of the reaction is as follows.

$\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)$

= $[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]$

= $[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]$

= 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

$\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}$

= $286000 J - (163.5 J/K \times 298 K)$

= 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

6 0

3 years ago

What type of bonding is occuring in the compound below?

rewona [7]

Answer:

(B). it's metallic bonding

8 0

2 years ago

→

marin [14]

Answer:

[A]²

Explanation:

Since the formation is independent of D, D is 0 order.

Since a quadruples when it is doubled it can be written as

2A^X= 4

To find the unknown power we can assume A= 1 to make the math simple. So When a = 2 (Because you doubled it) raised to X power it will equal 4

so the unknown power is 2

Making the rate law

[a]²[b]⁰

or simply just

[A]²

8 0

2 years ago

The first method of determining the chemical composition of substances in space was to

ehidna [41]

<span>The first method to determine the chemical composition of a substance in space was using light. By determining red shift in the observed spectrum of light they could determine the elements they were observing. Different elements change the way light behaves and from this scientists can determine the makeup of things such as stars and nebulas.</span>

5 0

2 years ago

Read 2 more answers

Suppose you have 75 gas-phase molecules of methanol (CH3OH) at T = 470 K. These molecules are contained in a spherical container

Katarina [22]

Answer:

The average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

Explanation:

Here Pressure in a container is given as

$P=\frac{1}{3} \rho$

Here

P is the pressure which is to be calculated

ρ is the density of the gas which is to be calculated as below

$\rho =\frac{mass}{Volume of container}$

Here

mass is to be calculated for 75 gas phase molecules as

$m=n_{molecules} \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=75 \times \frac{1 mol}{6.022 \times 10^{23} molecules} \times \frac{32 g/mol}{1 mol}\\m=3.98 \times 10^{-21} g$

Volume of container is 0.5 lts

So density is given as

$\rho =\frac{mass}{Volume of container}\\\rho =\frac{3.98 \times 10^{-21} \times 10^{-3} kg}{0.5 \times 10^{-3} m^3}\\\rho =7.97 \times 10^{-21}\, kg/m^3$

is the mean squared velocity which is given as

$=RMS^2$

Here RMS is the Root Mean Square speed given as 605 m/s so

$=RMS^2\\=(605)^2\\=366025$

Substituting the values in the equation and solving

$P=\frac{1}{3} \rho \\P=\frac{1}{3} \times 7.97 \times 10^{-21} \times 366025\\P=9.72 \times 10^{-16} Pa$

So the average pressure in the container due to these 75 gas molecules is $P=9.72 \times 10^{-16} Pa$

6 0

3 years ago