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MatroZZZ [7]
3 years ago
15

When a 1:1 mixture of ethyl butanoate and ethyl acetate is treated with sodium ethoxide, four Claisen condensation products are

possible.
-Draw the structure(s) of the product(s) that have an ethyl group on the chiral center.

-Although chirality may be mentioned in the question, do not include chirality in your drawings. .

-Do not draw organic or inorganic by-products .

-Draw one structure per sketcher. Add additional sketchers using the dropdown menu in the bottom right corner. .

-Separate multiple products using the +sign from the dropdown menu

Chemistry
1 answer:
Ahat [919]3 years ago
4 0

Answer:

answer is attached:

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A baker is decorating cupcakes using 3 cherries for every 1 cupcake. If the baker has 12 cherries and 5 cupcakes, what is the th
11111nata11111 [884]
C. 4 decorated cupcakes.

Because 12 divided by 3 = 4, so you can have 4 cupcakes each with 3 cherries
7 0
3 years ago
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Give the charge and full ground-state electron configuration of the monatomic ion most likely to be formed by the element. rb
devlian [24]

Actually Rb or Rubidium in zero state has the following electron configuration:

<span>1s22s2</span><span>2p6</span><span>3s2</span><span>3p63d10</span><span>4s2</span><span>4p65s1</span>

 

However we can see that the ion has a 1 positive charge, which means that it lacks 1 electron, therefore the answer from the choices is:

<span>d. rb+: 1s22s22p63s23p64s23d104p6</span>
3 0
3 years ago
Why phenolphthalein is used in experiment?
Dominik [7]
In an acid-base titration, the neutralization reaction between the acid and base can be measured with either a color indicator or a pH meter. In this experiment, aphenolphthalein<span> color indicator will be </span>used<span>.
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4 0
3 years ago
​If 250.0 mL of a 0.96 M solution of acetic acid (C 2H 4O 2) are diluted to 800.0 mL, what will be the approximate molarity of t
adoni [48]

Answer:

0.30M HOAc

Explanation:

Given 250.0ml (0.96M HOAc) => 800ml(??M HOAc)

Use the dilution equation...

(Molarity x Volume)concentrated soln = (Molarity x Volume)dilute soln

(0.96M)(250.0ml) = (Molarity diluted soln)(800ml)

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7 0
3 years ago
Write packing efficiency of fcc ,BCC, SCC and formula also . ?​
Minchanka [31]

❖ <u>Packing Efficiency</u> ❖

➪ The percentage of total space occupied by particles is called <u>packing efficiency</u>.

\\ \qquad{\rule{200pt}{3pt}}

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

❖ Packing efficiency of simple cubic structure (SCC).❖

\rm \: {Let \: } \bf{r } \: \rm{ be  \: the \:  radius  \: of  \: a \:  sphere  }\:  \\ \rm and \: \bf{a} \:  \rm {be \:  the \:  edge \:  of  \: unit \:  cell.} \\  \\  \;  \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar

❒ Since, simple cubic unit cell contain 1 atom (sphere). So, the total volume of sphere will be :

:  : \implies \sf \: 1 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi r^3

  • Volume of unit cell = a³

  • Volume of unit cell = (2r)³

  • Volume of unit cell = 8r³

❒ <u>Now, we know that</u>,❒

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

➪ Substituting the known values in the formula, we get the following results:

:  : \implies \rm \: {\dfrac{\dfrac{4}{3}\pi r^3}{8r^3} \times 100} \\ \\ :  : \implies   \dfrac{\pi}{6} \times 100 \\\\\   :  : \implies \bf {52.4 \%}

❒ Hence, the packing efficiency of simple cubic structure is 52.4%.

\\ \qquad{\rule{200pt}{3pt}}

➪ Packing efficiency of cubic close packing (SCC)/face centred cubic structure (FCC).

\rm \: {Let \: } \bf{r } \: \rm{ be  \: the \:  radius  \: of  \: a \:  sphere  }\:  \\ \rm and \: \bf{a} \:  \rm {be \:  the \:  edge \:  of  \: unit \:  cell.} \\  \\  \;  \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

: : \implies \rm 4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3

  • Volume of unit cell = a³

Volume of unit cell = (2√2r)³

Volume of unit cell = 16√2r³

❒ <u>Now, we know that</u>,❒

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

➪Substituting the known values in the formula, we get the following results:

:   : \implies \rm {\dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100 }\\\\  \implies \rm \dfrac{\pi}{3\sqrt{2}} \times 100 \\\\\  \implies\bf52.4 \%

❖ Hence, the packing efficiency of face centred cubic structure is 74%.

\\ \qquad{\rule{200pt}{3pt}}

❖ Packing efficiency of body cubic structure (BCC).

\rm \: {Let \: } \bf{r } \: \rm{ be  \: the \:  radius  \: of  \: a \:  sphere  }\:  \\ \rm and \: \bf{a} \:  \rm {be \:  the \:  edge \:  of  \: unit \:  cell.} \\  \\  \;  \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

: : \implies  \rm \: 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3

  • Volume of unit cell = a³

Volume of unit cell = (4r/√3)³

Volume of unit cell = 64r³/3√3

❒ <u>Now, we know that</u>,❒

\bigstar \boxed{ \sf \:  \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}

➪Substituting the known values in the formula, we get the following results:

: : \implies \rm {\dfrac{\dfrac{8}{3}\pi r^3}{\dfrac{64r^3}{3\sqrt{3}}} \times 100} \\\\  \implies \dfrac{3\pi}{8} \times 100 \\\ \\  \bf  \implies \: 68 \%

❒ Hence, the packing efficiency of body centred cubic structure is 68%.

\\ \qquad{\rule{200pt}{3pt}}

6 0
2 years ago
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