Anyone really good with Organic chemistry

Consider the aldol-dehydration reaction. Draw the two possible products of the reaction between benzaldehyde and methylethylketo

Dafna1 [17]

Solution :

An $\text{aldol condensation}$ reaction is a type of $\text{condensation reaction}$ in organic chemistry where the enol or an enolate ion reacts with the carbonyl compound and forms a $\beta$ -hydroxyaldehyde or a $\beta$ -hydroxyketone, and then followed by a dehydration to give conjugated enone.

Benzaldehyde reacts with methylketone and forms two products:

5 0

3 years ago

If you burn 10 kilograms of wood in a fire (combustion) what is the weight of the products after the fire has finished burning t

sleet_krkn [62]

The answer is: the weight of products is is equal the weight of the wood plus the weight of oxygen that was used to burn that wood, so weigh of the product is greater than 10 kilograms.

Conservation of mass (mass is never lost or gained in chemical reactions), during chemical reaction no particles are created or destroyed, the atoms are rearranged from the reactants to the products.

In this example wood (mostly carbon) and oxygen are reactants and carbon dioxide (mostly) is product of reaction.

3 0

3 years ago

The fire department needs information on friction losses occurring between a water main and an open fire hydrant. At maximum mai

lisov135 [29]

Explanation:

The given data is as follows.

$P_{1}$ = 85 psig, $P_{2}$ = $P_{atm}$ = 15 psia

Q = 1620 gpm, d = 2.5 inch, l = 8 ft = 2.4384 m

According to Darey-Weisbach equation,

$h_{l} = \frac{4fl \nu^{2}}{2gD}$ ......... (1)

Value of 'f' will be decided on the basis of Reynold number.

As, it is known that $R_{l} = \frac{\rho \nu d}{\mu}$

where, $\mu_{water}$ = $10^{-3}$ kg/ms

As it is known that 1 gpm = $\frac{1}{3.67} m^{3}/hr$

So, $1 m^{3}/hr$ = 3.67 gpm

Therefore, Q = $1620 \times \frac{1}{3.67}$

= $441.4168 m^{3}/hr$

= 0.1226 $m^{3}/s$

In, 1 inch = 2.54 cm = 0.0254 m

Therefore, d = 2.5 \times 0.0254 = 0.0635 m

V = $\frac{Q}{\frac{\pi}{4}d^{2}}$

= $\frac{0.1226}{0.785 \times (0.0635)^{2}}$

= 38.73 m/s

Hence, we will calculate Reynold number as follows.

$R_{l}$ = $\frac{1000 \times 38.73 \times 0.0635}{10^{-3}}$

= 2459355

As $R_{l}$ > 2000 then, it means that flow is turbulent.

As, f = 0.079 $R^{-0.25}_{l}$

= 0.001994

Putting all the values into equation (1) formula as follows.

$h_{l} = \frac{4fl \nu^{2}}{2gD}$

= $\frac{4 \times 0.001994 \times 2.4384 \times (38.73)^{2}}{2 \times 9.81 \times 0.0635}$

= $1.04069 \times 10^{5} m$

Thus, we can conclude that friction loss from the main to the discharge point is $1.04069 \times 10^{5} m$ .

4 0

3 years ago

The half life of a certain radioactive element is 1,250 years what percent of the atoms remain after 7500 years

andrezito [222]

Each half-life results in ~50% (1/2) of the original element remaining.

7500/1250 = 6 half-lives, so 100(1/2)^6
= 100(0.015625)
= 1.5625% of the original element would remain

7 0

3 years ago

Read 2 more answers

Please help!!!!!!!!!!!!!!!!!! Now that the lab is complete, it’s time to write your lab report. The purpose of this guide is to

Inessa [10]

Answer:

what was the lab?

Explanation:

8 0

4 years ago