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Scorpion4ik [409]
3 years ago
7

Butanoic acid + 2 propanol

Chemistry
1 answer:
Anuta_ua [19.1K]3 years ago
4 0

Answer:

Butanoic acid and 2-propanol reacts to form isopropyl butyrate.

Explanation:

brainliest plz

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A sample of the chiral molecule limonene is 79% enantiopure. what percentage of each enantiomer is present? what is the percent
Degger [83]

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%



7 0
3 years ago
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Is ink a compound or mixture?
Vladimir79 [104]
A mixture, because it contains multiple dyes and compounds

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8 0
3 years ago
Manganese nodules are considered a valuable resource from the ocean floor because they are abundant and contain not only mangane
MAVERICK [17]

Answer: iron (oxides of iron)

Explanation: manganese nodules usually contain layers of iron and manganese oxides in a concentric arrangement and are called polymetallic nodules.

4 0
3 years ago
The pKa of chloroacetic acid is 2.9. If an organic chemist has a 0.10 M aqueous solution of chloroacetic acid at pH 2.4, what pe
Trava [24]

Answer:

The answer is given below.

Explanation:

We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.

AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M

                      HA →              H+         +      A-

Initial concentration:  0.1    →  3.98 ^10-3       +      0

equilibrium concentration:  0.1(1-α) →   3.98 * 10-3 + 0.1α              0.1α

pKa of chloroacetic acid is 2.9

-log(Ka) = 2.9

Ka = 1.26 * 10-3

From the equation, Ka = [H+] * [A-] / [HA]

1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)      

Since α<<1, we assume 1-α = 1

Solving the equation, we have: α = 0.094

Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%

6 0
3 years ago
Determine the molecular formula of a compound with an empirical formula of NH2 and a formula mass of 32.06 amu
Lilit [14]
N2H4

<span>Each nitrogen weighs 14.01 and each H weighs 1.01. !4.01+14.01+1.01+1.01 = 32.06 (roughly) </span>

4 0
3 years ago
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