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lilavasa [31]
3 years ago
8

Which classes of isomers are easier to separate from each other in the laboratory: enantiomers or diastereomers?

Chemistry
1 answer:
Varvara68 [4.7K]3 years ago
8 0

Answer:

diastereomers

Explanation:

Enantiomers are optical isomers that are mirror images of each other. Enantiomers can not be separated by physical methods because the both compounds have almost the same physical properties such as melting points, boiling points, etc. This makes it difficult to separate enanatiomers by conventional methods.

A common method of separating enantiomers is by converting them into diasteromers which are easily separated since diasteromers often have different chemical and physical properties.

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Compare and contrast the compositions of binary ionic and binary molecular compounds
ICE Princess25 [194]
<span>Both are composed of two elements (that means binary).
Ionic is composed of two ions, metallic cation-positive ions, and nonmetallic anion - negative ions. For example NaCl (sodium-chloride).
Molecular compounds are composed of two nonmetallic elements. For example HCl (hydrochloric acid).</span>
7 0
3 years ago
what kind of change occurs when sodium and chlorine combine ? Describe the matter before and after the change
DaniilM [7]
Sodium is a silver-colored metal which is soft enough to cut with a knife. It is an extremely reactive metal, and is always found naturally in ionic compounds, not in its pure metallic form. Pure sodium metal reacts violently (and sometimes explosively) with water, producing sodium hydroxide, hydrogen gas, and heat:


Chlorine is a poisonous, yellow-green gas, with a very sharp odor, a

Chemical change happen cuz these 2 elements will combine to form a new compound which’s NaCl ( table salt )
5 0
3 years ago
Which of the following will have a larger radius than zinc ? Gallium (Ga) Magnesium (Mg) Aluminum (Al) Strontium (St)
oksian1 [2.3K]

Answer:

Strontium (St)

Explanation:

The atom that will have a larger radius than zinc is strontium, Sr.

Atomic radius is defined as the half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.

  • Across the period atomic radii decreases progressively due to the increase in nuclear charge.
  • Down a group atomic radii increase progressively due to the successive shells of electrons being added.
  • Since strontium satisfies the criteria, it has the larger atomic radius.
7 0
3 years ago
Air is compressed from an inlet condition of 100 kPa, 300 K to an exit pressure of 1000 kPa by an internally reversible compress
ElenaW [278]

Answer:

(a) W_{isoentropic}=8.125\frac{kJ}{mol}

(b) W_{polytropic}=7.579\frac{kJ}{mol}

(c) W_{isothermal}=5.743\frac{kJ}{mol}

Explanation:

Hello,

(a) In this case, since entropy remains unchanged, the constant k should be computed for air as an ideal gas by:

\frac{R}{Cp_{air}}=1-\frac{1}{k}  \\\\\frac{8.314}{29.11} =1-\frac{1}{k}\\

0.2856=1-\frac{1}{k}\\\\k=1.4

Next, we compute the final temperature:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/k}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.4}=579.21K

Thus, the work is computed by:

W_{isoentropic}=\frac{kR(T_2-T_1)}{k-1} =\frac{1.4*8.314\frac{J}{mol*K}(579.21K-300K)}{1.4-1}\\\\W_{isoentropic}=8.125\frac{kJ}{mol}

(b) In this case, since n is given, we compute the final temperature as well:

T_2=T_1(\frac{p_2}{p_1} )^{1-1/n}=300K(\frac{1000kPa}{100kPa} )^{1-1/1.3}=510.38K

And the isentropic work:

W_{polytropic}=\frac{nR(T_2-T_1)}{n-1} =\frac{1.3*8.314\frac{J}{mol*K}(510.38-300K)}{1.3-1}\\\\W_{polytropic}=7.579\frac{kJ}{mol}

(c) Finally, for isothermal, final temperature is not required as it could be computed as:

W_{isothermal}=RTln(\frac{p_2}{p_1} )=8.314\frac{J}{mol*K}*300K*ln(\frac{1000kPa}{100kPa} ) \\\\W_{isothermal}=5.743\frac{kJ}{mol}

Regards.

8 0
3 years ago
A solution of KCIO3 is prepared using 75 grams of the solute in enough water to make 0.250 liters of solution. The gram-formula
charle [14.2K]
Using the mass/volume percentage method for percentages of the solution, you simply divide the grams of solute by the volume of the solution and multiply by 100 to get your percentage.
(75.0g/250mL)•100 = 30.0% solute
3 0
3 years ago
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