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Burka [1]
3 years ago
11

A. draw the structure of the tetrahedral intermediate initially formed in the reaction shown. you do not have to consider stereo

chemistry. do not include counter-ions,
e.g., na+, i-, in your answer. in cases where there is more than one answer, just draw one. move clear erase undo redo increase scale decrease scale copy paste set label more labels attributes more attributes single bond recessed bond protruding bond double bond other bond more bonds cyclohexane ring benzene ring other ring more rings bracket
b. draw the structures of the final acyl transfer products obtained.

Chemistry
1 answer:
julsineya [31]3 years ago
4 0
The reaction is shown below, Acid protonates the carbonyl oxygen and makes the carbonyl carbon more electrophilic. Water attacks on activated carbonyl group and forms a tetrahedral intermediate.

Intermediate:
                    Structure of Intermediate is shown both in 2-D and 3-D (below attached). 

Carbonyl group is regenerated with the elimination of ethanol.

Reaction is as below, The final product is carboxylic acid.

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Which number is different for isotopes of the same element?
tankabanditka [31]

Answer:

D number of protons or atomic mass

4 0
2 years ago
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When iridium-192 is used in cancer treatment, a small cylindrical piece of 192ir, 0.6 mm in diameter and 3.5 mm long, is surgica
Ivan

Iridium-192 is used in cancer treatment, a small cylindrical piece of 192 Ir, 0.6 mm in diameter (0.3mm radius) and 3.5 mm long, is surgically inserted into the tumor. if the density of iridium is 22.42 g/cm3, how many iridium atoms are present in the sample?

Let us start by computing for the volume of the cylinder. V = π(r^2)*h where r and h are the radius and height of the cylinder, respectively. Let's convert all given dimensions to cm first. Radius = 0.03 cm, height is 0.35cm long.

V = π * (0.03cm)^2 * 0.35 cm = 9.896*10^-4 cm^3

Now we have the volume of 192-Ir, let's use the density provided to get it's mass, and once we have the mass let's use the molar mass to get the amount of moles. After getting the amount of moles, we use Avogadro's number to convert moles into number of atoms. See the calculation below and see if all units "cancel":

9.896*10^-4 cm^3 * (22.42 g/cm3) * (1 mole / 191.963 g) * (6.022x10^23 atoms /mole)

= 6.96 x 10^19 atoms of Ir-122 are present.

4 0
3 years ago
20 points!
Paladinen [302]

Answer:

c and d

Explanation:

mitochondria and vacoule

8 0
3 years ago
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Describe an organic molecule.
GarryVolchara [31]

Answer: molecule of the kind normally found in living systems. Organic molecules are usually composed of carbon atoms in rings or long chains, to which are attached other atoms of such elements as hydrogen, oxygen, and nitrogen.

Explanation:

6 0
3 years ago
A 25.0 ml sample of a 0.2900 m solution of aqueous trimethylamine is titrated with a 0.3625 m solution of hcl. calculate the ph
Flauer [41]
a) After adding 10 mL of HCl

first, we need to get moles of (CH3)3N = molarity * volume

                                                                 = 0.29 m * 0.025 L
                                                                 = 0.00725M moles
then, we need to get moles of HCl = molarity * volume

                                                          = 0.3625 m * 0.01L
                                                          = 0.003625 moles

so moles of (CH3)3N remaining    = moles of (CH3)3N - moles of HCl
                                                         = 0.00725 - 0.003625

                                                      = 0.003625 moles

and when the total volume = 0.01 L + 0.025L = 0.035 L

∴ [(CH3)3N] = moles remaining / total volume

                    = 0.003625 moles / 0.035L
                    = 0.104 M

when we have Pkb so we can get Kb :

pKb = - ㏒Kb
4.19 = -㏒Kb

∴Kb = 6.5 x 10^-5

when Kb = [(CH3)3NH+] [OH-] / [(CH3)3N]

and by using ICE table we assume we have:

[(CH3)3NH+] = X & [OH] = X 

and [(CH3)3N] = 0.104 -X

by substitution:

∴ 6.5 x 10^-5 = X^2 / (0.104-X)  by solving for X

∴X = 0.00257 M
∴[OH-] = X = 0.00257 M

∴POH = -㏒[OH]

           = -㏒0.00257
           = 2.5

∴ PH = 14 - POH
         = 14 - 2.5
         = 11.5
b) after adding 20ML of HCL:

moles of HCl = molarity * volume
                       = 0.3625 m * 0.02 L

                       = 0.00725 moles

  

the complete neutralizes of (CH3)3N we make 0.003625 moles of (CH3)3NH+ So, now we need the Ka value of (CH3)3NH+:

and when the total volume = 0.02L + 0.025 = 0.045L

∴ [ (CH3)3NH+] = moles / total volume

                          = 0.003625 / 0.045L
                          = 0.08 M
 
when Ka = Kw / Kb 

and we have Kb = 6.5 x 10^-5 & we know that Kw = 1 x 10^-14 

so, by substitution:

Ka = (1 x 10^-14) / (6.5 x 10^-5)

    = 1.5 x 10^-10


when Ka expression = [(CH3)3N][H+] / [(CH3)3NH+]

by substitution:

∴ 1.5 x 10^-10 = X^2 / (0.08 - X)  by solving for X

∴X = 3.5 x 10^-6  M

∴ [H+]= X = 3.5 x 10^-6 M

∴PH = -㏒[H+]

        = -㏒(3.5 x 10^-6)

       = 5.5


C) after adding 30ML of HCl:

moles of HCl = molarity * volume 

                       = 0.3625m * 0.03L

                       = 0.011 moles

and when moles of (CH3)3N neutralized = 0.003625 moles 

∴ moles of HCl remaining    = moles HCl - moles (CH3)3N neutralized

                                                = 0.011moles - 0.003625moles

                                                = 0.007375 moles
when total volume = 0.025L + 0.03L

                                = 0.055L

∴[H+] = moles / total volume

           = 0.007375 mol / 0.055L

            = 0.134 M

∴PH = -㏒[H+]

        = -㏒ 0.134

        = 0.87
8 0
3 years ago
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