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3 years ago

15

A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent

enantiomeric excess?

2 answers:

Degger [83]3 years ago

7 0

Answer : The % of (+) limonene isomer = 79%

The % of (-) limonene isomer = 0%

The % of enantiomeric excess = 58%

Explanation : Enantiomeric excess (ee) is the measurement of purity used for chiral substances.

Given,

% of pure limonene enantiomer = The % of (+) limonene isomer = 79%

Therefore, The % of (-) limonene isomer = 0%

Formula used :

$\%(+)\text{ isomer}=\frac{ee}{2}+50\%$

Where, ee → enantiomeric excess

Now, put all the values in above formula, we get the value of enantiomeric excess (ee).

${ee}=\frac{\%(+)-50\%}{2}$

$=\frac{79\%-50\%}{2}$

= 58%

DaniilM [7]3 years ago

7 0

Answer:

The major enatiomer makes up 89.5% of the mixture

The minor enatiomer makes up 10.5% of the mixture

The percent enantiometric excess is 79%

Explanation:

% (+) = ee/2

79 % / 2 + 50%

= 39.5% + 50%

= 89.5%

and

% (-) = 100 % - 89.5 %

= 10.5%

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2. Which of the following is not a physical property of water? It is a colorless liquid. It is composed of hydrogen and oxygen.

S_A_V [24]

The correct answer is C. Sugar dissolves in it

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3 years ago

A chemistry graduate student is given 125.mL of a 0.20M acetic acid HCH3CO2 solution. Acetic acid is a weak acid with =Ka×1.810−

Over [174]

Answer : The mass of sodium acetate is, 1.097 grams.

Explanation : Given,

The dissociation constant for acetic acid = $K_a=1.8\times 10^{-5}$

Concentration of acetic acid (weak acid)= 0.20 M

volume of solution = 125. mL

pH = 4.47

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (1.8\times 10^{-5})$

$pK_a=5-\log (1.8)$

$pK_a=4.74$

Now we have to calculate the concentration of sodium acetate (conjugate base or salt).

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

Now put all the given values in this expression, we get:

$4.47=4.74+\log (\frac{[Salt]}{0.20})$

$[Salt]=0.107M$

Now we have to calculate the mass of sodium acetate.

$\text{Concentration}=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{\text{Molar mass of }NaCH_3CO_2\times \text{Volume of solution (in mL)}}$

$0.107M=\frac{\text{Mass of }NaCH_3CO_2\times 1000}{82g/mol\times 125mL}$

$\text{Mass of }NaCH_3CO_2=1.097g$

Therefore, the mass of sodium acetate is, 1.097 grams.

6 0

4 years ago

CHEM HELP ASAP!! <br><br> What mass of H2 would be needed to produce 208 kg of methanol?

Eddi Din [679]

So if we use the equation:

$CO+2H_{2}$ → $CH_{3}OH$

We can then determine the amount of $H_{2}$ needed to produce 208 kg of methanol.

So let's find out how many moles of methanol 208 kg is:

Methanol molar weight = 32.041g/mol

So then we can solve for moles of methanol:

$208kg*\frac{1,000g}{1kg} *\frac{1mol}{32.041g} =6,491.68mol$

So now that we have the amount of moles produced, we can use the molar ratio (from the balanced equation) of hydrogen and methanol. This ratio is 2:1 hydrogen:methanol.

Therefore, we can set up a proportion to solve for the moles of hydrogen needed:

$\frac{2}{1} =\frac{x}{6,491.68}$

$x=12,983.36mol$

So now that we have the number of moles of $H_{2}$ that are produced, we can then use the molar weight of hydrogen to solve for the mass that is needed:

$12,983.36mol*\frac{2.016g}{1mol} =26,174.45g_H_{2}$

Therefore, the amount of diatomic hydrogen ( $H_{2}$ ) that is needed to produce 208kg of methanol is $2.62x10^{4}$ g.

3 0

3 years ago

Which of the following aqueous solutions will have the lowest freezing point? a. 0.75 m (NH4)3PO4 b. 1.0 m CaSO4 c. 1.0 m LiCl d

Nataly [62]

Answer: a. 0.75m (NH4)3PO4 will have the lowest freezing point .

Explanation: Freezing point decreases as the concentration of the solute substance increases. Assuming the same solvent for all of them, for instance water.

∆T= i.Kf.b

∆T= freezing point depression

i= vant Hoff factor

Kf= molality

Assuming water to be the solvent for all Kf=1.86°C/M

VANT HOFF FACTORS :

For (NH4)3PO4

This has 3 ionic bonding and 1 covalent bonding making it 4 bond

Therefore i=4

For CaSO4

This has 1 metallic bond and 1 covalent bond making it 2 bond.

Therefore i=2

For LiCl

This has 1 metallic bond and 1 non metallic bond making it 2 bond.

Therefore i=2

For CH3OH

This has only 1 covalent bond.

Therefore i=1

MOLALITY:

(NH4)3PO4 = 0.75M

CaSO4= 1.0M

LiCl= 1.0M

CH3OH= 1.5M

FREEZING POINT DEPRESSION:

For (NH4)3PO4

∆T= 4×0.75×1.86=5.58°C

For CaSO4

∆T= 2×1.0×1.86=3.72°C

For LiCl

∆T = 2×1.0×1.86= 3.72°C

For CH3OH

∆T= 1×1.5×1.86=2.79°C

REMEMBER THE HIGHER THE FREEZING POINT DEPRESSION THE LOWER THE FREEZING POINT.

FREEZING POINT DEPRESSION IS THE CHANGE IN THE FREEZING POINT PROPORTIONAL TO THE AMOUNT OF SOLUTE ADDED THE THE SOLUTION.

THEREFORE THE ONE WITH THE LOWEST FREEZING POINT IS (NH4)3PO4

4 0

4 years ago

The compound tyrosine contains 59.66% C, 6.12% H, 7.73% N, and 26.49% O by mass. What is the empirical formula of tyrosine?

Bezzdna [24]

Answer: C9H11O3N

Explanation:

The percentage by mass of each element is divided by its relative atomic mass. The lowest ratio is now observed. Each ratio is now divided by this lowest ratio and approximated to whole numbers. All these are shown in detail in the image attached.

3 0

3 years ago