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sergeinik [125]
3 years ago
11

Why is a solution of 4% acetic acid in 95% ethanol used to wash the crude aldol-dehydration product?

Physics
1 answer:
raketka [301]3 years ago
5 0

A cold acetic acid solution is used to wash the residue of the reagent in preparation of an aldol condensation product after vacuum filtration.  The main reason in washing with the acetic acid rinse is to neutralize any sodium hydroxide.

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An aircraft flying in a straight
Lelechka [254]

Answer:

t = 10.1 s

d = 2020 m

Explanation:

Time to drop from vertical rest

h = ½gt²

t = √(2h/g) = √(2(500)/9.8) = 10.1 s

d = vt = 200(10.1) = 2020 m  

3 0
3 years ago
6) Why do electromagnetic waves not require a medium for travel?
sergij07 [2.7K]
Electromagnetic waves are a type of
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implies a wave, these waves indicate
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8 0
3 years ago
A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the
erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, T_{b} = 75^{\circ}C

Air temperature, T_{infty} = 20^{\circ}

k = 388 W/mK

h = 20\ W/m^{2}K

Now,

Perimeter of the fin, p = \pi D = 0.005\pi \ m

Cross-sectional area of the fin, A = \frac{\pi}{4}D^{2}

A = \frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}

To calculate the heat transfer rate:

Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})

where

m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237

Now,

Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W

5 0
3 years ago
An X-Ray machine delivers a radiation dose of 5mRem/hr. at 3ft from the machine. How far will the X-Ray technician have to move
kupik [55]

Answer:

4.7ft

Explanation:

Pls see attached file

8 0
3 years ago
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

7 0
3 years ago
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