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Feliz [49]
3 years ago
5

When the optically active carboxylic acid below is decarboxylated using the conditions typical in the acetoacetate synthesis, wi

ll the ketone product also be optically active?

Chemistry
1 answer:
maria [59]3 years ago
5 0

Answer:

ye, it will be optically active

Explanation:

a compound is said to be optically active if it can optically rotate.

the removal of carboxyl group and release of cabon dioxide from carboxylic acid in acetoacetate synthesis which will result in production of ketone as given the attachment below.

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3 years ago
Someone help me answer this please! I’ll mark you brainliest
kondor19780726 [428]

b. increase in surface area

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8 0
3 years ago
calculate the ph for each case in the titration of 50.0 ml of 0.130 m hclo(aq) with 0.130 m koh(aq). use the ionization constant
-BARSIC- [3]

The ph before the addition of any Koh is<u> 10.105.</u>

Concentration is the abundance of a constituent divided by way of the overall volume of an aggregate. several sorts of mathematical descriptions may be outstanding: mass concentration, molar concentration, variety concentration, and extent awareness.

After the addition of 50 ml KOH,

moles of KOH = 50 * 0.13 =<u> 6.5 mmol </u>

<u>moles </u><u>of HClO = 50 * 0.13 = 6.5 mmol </u>

occurred hydrolysis solution,

pH = 0.5(14 + pKa + log [base conjugate])

pH = 0.5(14 + (- log (4 * 10^-8)) + log (6.5/(50 + 50)))

pH = <u>10.105</u>

The concentration of a substance is the quantity of solute found in a given amount of solution. Concentrations are normally expressed in terms of molarity, defined because of the variety of moles of solute in 1 L of answer.

The Concentration of an answer is a measure of the quantity of solute that has been dissolved in a given amount of solvent or answer. A concentrated answer is one that has a rather huge quantity of dissolved solute.

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4 0
1 year ago
How are conclusions and evidence related
serg [7]
You need evidence to support a conclustion 
3 0
3 years ago
Read 2 more answers
The maximum amount of nickel(II) cyanide that will dissolve in a 0.220 M nickel(II) nitrate solution is...?
sweet [91]

Answer : The maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

Explanation :

The solubility equilibrium reaction will be:

                       Ni(CN)_2\rightleftharpoons Ni^{2+}+2CN^-

Initial conc.                        0.220       0

At eqm.                             (0.220+s)   2s

The expression for solubility constant for this reaction will be,

K_{sp}=[Ni^{2+}][CN^-]^2

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

3.0\times 10^{-23}=(0.220+s)\times (2s)^2

s=5.84\times 10^{-12}M

Therefore, the maximum amount of nickel(II) cyanide is 5.84\times 10^{-12}M

7 0
3 years ago
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