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Murljashka [212]

3 years ago

14

Fill in the blank, please help. (55 points)

1 answer:

Degger [83]3 years ago

7 0

The answer would be evaporation takes place at the surface of an ocean, lake, stream, or other body of water

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The vapor pressure of ethanol at

Karo-lina-s [1.5K]

482VP I think is the correct answer.

5 0

3 years ago

If the p-side has a higher doping concentration, explain how to keep tuning to the same radio channel?

zzz [600]

Answer:

increase in temperature of the intrinsic semiconductor

Explanation:

If the p-side has a higher doping concentration, it implies that number of holes (positive ion) increased which is greater than number of electron (negative ion) in the n-side
in order to balance the intrinsic concentration, that is to balance the number of holes and electrons which depends on temperature.
an increase in the temperature of the intrinsic semiconductor (p-side), increases the number of electron but number of holes remains constant.

A balance in the intrinsic concentration helps in tuning to the same radio channel.

6 0

3 years ago

(6) Compare a CSTR with a PFR below. a. A flow of 0.3 m3/s enters a CSTR (volume of 200 m3) with an initial concentration of spe

Dmitry [639]

Answer:

Explanation:

Given that:

The flow rate Q = 0.3 m³/s

Volume (V) = 200 m³

Initial concentration $C_o$ = 2.00 ms/l

reaction rate K = 5.09 hr⁻¹

Recall that:

$time (t) = \dfrac{V}{Q}$

$time (t) = \dfrac{200}{0.3}$

$time (t) = 666.66 \ sec$

$time (t) = \dfrac{666.66 }{3600} hrs$

$time (t) = 0.185 hrs$

$\text{Using First Order Reaction:}$

$\dfrac{dc}{dt}=kc$

where;

$t = \dfrac{1}{k} \Big( \dfrac{C_o}{C_e}-1 \Big)$

$0.185 = \dfrac{1}{5.09} \Big ( \dfrac{200}{C_e}- 1 \Big)$

$0.942 = \Big ( \dfrac{200}{C_e}- 1 \Big)$

$1+ 0.942 = \Big ( \dfrac{200}{C_e} \Big)$

$\dfrac{200}{C_e} = 1.942$

$C_e = \dfrac{200}{1.942}$

$\mathbf{C_e = 102.98 \ mg/l}$

Thus; the concentration of species in the reactant = 102.98 mg/l

b). If the plug flow reactor has the same efficiency as CSTR, Then:

$t _{PFR} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{k} \Big [ In ( \dfrac{C_o}{C_e}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} = \dfrac{1}{5.09} \Big [ In ( \dfrac{200}{102.96}) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196 \Big [ In ( 1.942) \Big ]$

$\dfrac{V_{PFR}}{Q_{PFR}} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3 hrs} =0.196(0.663)$

$\dfrac{V_{PFR}}{0.3*3600 sec} =0.196(0.663)$

$V_{PFR} =0.196(0.663)*0.3*3600$

$V_{PFR} = 140.34 \ m^3$

The volume of the PFR is ≅ 140 m³

3 0

3 years ago

How many oxygen atoms are present in 30g of glucose?

Alex17521 [72]

Jdjejsbdjwwoanwjediekemwmkssk

5 0

3 years ago

During an experiment, Sandy recorded incorrect observations and measurements. What would be the most

ruslelena [56]

Sandy would reach an incorrect outcome.that's because she will repeat this mistake in the future too.

3 0

3 years ago