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kumpel [21]
3 years ago
11

Categorize each hydrocarbon as being saturated or unsaturated.

Chemistry
2 answers:
Bad White [126]3 years ago
8 0
<h3>Answer:</h3>

             Saturated: i) 3-Methyloctane ii) Heptane iii) Propane

             Unsaturated:  i) Heptyne  ii) Cyclopentene

<h3>Explanation:</h3>

                     Saturated Hydrocarbons are those organic compounds which are saturated <em>i.e. </em>all the bonds between carbon atoms are single. In other words the bonds between carbon atoms are sigma bonds (formed due to head to head overlap of orbitals)

Examples of Saturated Hydrocarbons are highlighted in Red color in attached figure.

                     Unsaturated Hydrocarbons are those organic compounds which are unsaturated <em>i.e. </em>they contain either one or more than one double or triple bonds along with or without single bonds.. In other words one or more bonds between carbon atoms are pi bonds (formed due to side wise overlap of p orbitals)

Examples of unsaturated Hydrocarbons are highlighted in Blue color in attached figure.

MA_775_DIABLO [31]3 years ago
6 0

Answer:

For PLATO: they left out one

         Saturated: 3-Methyloctane,  Heptane, Propane

           Unsaturated: Propyne, Heptyne, Cyclopentene

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Without consulting Appendix B, arrange each group in order of decreasing standard molar entropy (S°). Explain.(b) NO₂(g), NO(g),
denpristay [2]

The decreasing order of standard molar entropy (S°) is as follow:

NO₂(g) > NO(g) > N₂(g)

<h3>What is Entropy? </h3>

Entropy is defined as the randomness of the particle. It depends on temperature and pressure or number of particle per unit volume.

It is directly proportional to the temperature and pressure of the gas.

<h3>What is Standard Molar Entropy? </h3>

The standard molar entropy is defined as the entropy content of the one mole of pure substance at the standard state of temperature and pressure of interest.

The standard molar entropy is also defined as the total amount of entropy which 1 mole of the substance acquire, as it is brought from 0K to standard conditions of temperature and pressure.

The standard molar entropy depends on the molas mass of atom, molecules or compound.

N₂ has lower standard molar entropy. This can be explained as this molecule consist of same atom.

While, Complexity increases from NO to NO₂(g). Therefore, the standard molar entropy of NO₂(g) is greater than NO.

Thus, we concluded that the decreasing order of standard molar entropy (S°) is as follow:

NO₂(g) > NO(g) > N₂(g)

learn more about standard molar entropy:

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8 0
1 year ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

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