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arlik [135]
1 year ago
5

The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &S carvone is measured a

s (-) 23°. Which enantiomer is in excess and what is its enantiomeric excess? What are the percentages of (R) - & (S) - carvone in the sample​
Chemistry
2 answers:
jasenka [17]1 year ago
8 0

Answer:

Explanation:

ee= -23/-61 × 100= 38%

S + R = 100%

S - R = 38%

Solve simultaneously;

S= 69% ( excess)

R= 100-69= 31%

shusha [124]1 year ago
7 0

Answer:

See explanation

Explanation:

% optical purity = specific rotation of mixture/specific rotation of pure enantiomer  * 100/1

specific rotation of mixture = 23°

specific rotation of pure enantiomer = 61°

Hence;

% optical purity = 23/61 * 100 = 38 %

More abundant enantiomer = 100% - 38 % = 62%

Hence the pure  (S) carvone is (-) 62° is the more abundant enantiomer.

Enantiomeric excess = 62 - 50/50 * 100 = 24%

Hence

(R) - carvone  =  38 %

(S) - carvone = 62%

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As per the Charles law, here the volume is decreased from 50 ml to 25 ml so the boiling point also decreases.

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Dust and smoke.

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If a sample containing 2.50 ml of nitroglycerin (density=1.592g/ml) is detonated, how many total moles of gas are produced?
masya89 [10]
<span>0.127 moles The formula for nitroglycerin is C3H5N3O9 so let's first calculate the molar mass of it. Carbon = 12.0107 Nitrogen = 14.0067 Hydrogen = 1.00794 Oxygen = 15.999 C3H5N3O9 = 3 * 12.0107 + 5 * 1.00794 + 3 * 14.0067 + 9 * 15.999 = 227.0829 Now calculate the number of moles of nitroglycerin you have by dividing the mass by the molar mass 2.50 ml * 1.592 g/ml / 227.0829 g/mol = 0.017527 mol The balanced formula for when nitroglycerin explodes is 4 C3H5N3O9 => 12 CO2 + 10 H2O + O2 + 6 N2 Since all of the products are gasses at the time of the explosion, there is a total of 29 moles of gas produced for every 4 moles of nitroglycerin Now multiply the number of moles of nitroglycerin by 29/4 0.017527 mol * 29/4 = 0.12707075 moles Round to 3 significant figures, giving 0.127 moles</span>
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2 years ago
Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)
marissa [1.9K]

Answer:

The\Delta H_{rxn} of the given reaction is -129.6 kJ

Explanation:

The given chemical reaction is as follows.

H_{2}C=CH_{2}(g)+H_{2}(g)\rightarrow H_{3}C-CH_{3}(g)

Enthalpy of each reactant and products are as follows.

\Delta H_{C=C}\,=615.0\,kJ\,mol^{-1}

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\Delta H _{C-H}\,=416.2\,kJ\,mol^{-1}

In the given chemical reaction involved two C-H bonds in the reactant side and one C-C bond in the product side therefore, the enthalpy of formation will be the negative.

\Delta H_{rxn}=-\Delta H_{C-C}-2\Delta H_{C-H}+\Delta H_{C=C}+\Delta H_{H-H}

=-347.4-2\times416.2+615.0+435.1

=-129.6 \,kJ

Therefore, The\Delta H_{rxn} of the given reaction is -129.6 kJ

4 0
1 year ago
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