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melisa1 [442]
3 years ago
12

After the reaction between sodium borohydride and the ketone is complete, the reaction mixture is treated with water and H2SO4 t

o produce the desired alcohol. Explain the reaction by clearly indicating the source of the hydrogen atom that ends up on the oxygen
Chemistry
1 answer:
In-s [12.5K]3 years ago
7 0

Answer:

The hydrogen can be gotten from the added Acid or water during "workup".

Explanation:

Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.

For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.

So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.

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Average Molarity for HCl is .391
Ira Lisetskai [31]

Answer:

#1: 0.00144 mmolHCl/mg Sample

#2: 0.00155 mmolHCl/mg Sample

#3: 0.00153 mmolHCl/mg Sample

Explanation:

A antiacid (weak base) will react with the HCl thus:

Antiacid + HCl → Water + Salt.

In the titration of antiacid, the strong acid (HCl)  is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.

Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:

Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl

Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl

Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl

The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.

Thus, mmoles HCl /mg OF SAMPLE<em> </em>for each trial is:

#1: 2.178mmol / 1515mg

#2: 2.253mmol / 1452mg

#3: 2.219mmol / 1443mg

<h3>#1: 0.00144 mmolHCl/mg Sample</h3><h3>#2: 0.00155 mmolHCl/mg Sample</h3><h3>#3: 0.00153 mmolHCl/mg Sample</h3>
8 0
3 years ago
Suppose that in the citric acid cycle, oxalacetate is labeled with 14C in the carboxyl carbon farthest from the keto group. Wher
Nadusha1986 [10]

Answer:

All of alpha-ketoglutarate formed in the citric acid cycle would contain the radioactive ^{14}C while none of succinate would contain ^{14}C, and all of carbon dioxide released would contain ^{14}C.

Explanation:

When oxaloacetate in the citric acid cycle is labeled with ^{14}C in carboxyl carbon atom which is farthest from keto group, alpha ketoglutarate formed from this oxaloactetate has the full radioactive label. In the next step, the carboxylic group (that contains the ^{14}C) is eliminated in the form of the release of the carbon dioxide and succinate is formed. Succinate thus will not have radioactivity. CO_2 released had all the radioactivity.

8 0
4 years ago
How many atoms are in 3.89 moles of Cu?
AleksandrR [38]

Answer:

2.34×10^24

Explanation:

N=n×Na

N=3.89×(6.02×10^23)

N=2.34×+0^24 atoms of Cu

3 0
4 years ago
Scientific method for medal
earnstyle [38]
The method is cations
6 0
3 years ago
Which combination can be used to make a buffer with a pH of 4.2? The Ka of benzoic acid, C6H5COOH, is 6.5×10^–5 and the Ka of hy
ankoles [38]

50.0 mL of 0.500 M HCN and 25.0 mL of 0.500 M sodium hydroxide combination can be used to make a buffer with a pH of 4.2. The KK

an of benzoic acid, C6H5COOH, is 6.5×10-5 and the an of hydrocyanic acid, HCN, is 4.9×10-10.

Benzoic acid

A white (or colourless) solid with the chemical formula C6H5CO2H, benzoic acid is pronounced: "bnzo.k." The simplest aromatic carboxylic acid is this one. The term comes from gum benzoin, which was the substance's exclusive source for a very long period. Many plants naturally contain benzoic acid, which is used as an intermediary in the formation of numerous secondary metabolites[9]. Benzoic acid salts are employed as food preservatives. An essential starting point for the industrial synthesis of many other chemical compounds is benzoic acid. The term "benzoates" refers to the salts and esters of benzoic acid.

To learn more about benzoic acid refer here:

brainly.com/question/24052816

#SPJ 4

8 0
1 year ago
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