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Aloiza [94]
3 years ago
7

Pressure drop in packed column ..... a tray column

Chemistry
1 answer:
Stolb23 [73]3 years ago
5 0

Answer:

The answer is b)

Explanation:

In packed column the mass transfer area is higher and the packing can be random, that could provoque an increase in the pressure, because obstruction of the mass flux, While in a tray column the sieves can have specific porosity and the flux of mass can be facilitated, thus the pressure drop.  

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Determine the mass of MgCl2 needed to create a 100. ml solution with a concentration of 3.00 M.
kiruha [24]

Explanation:

1000ml \: contain \: 3 \: moles \\ 100 \: ml \: will \: contain \: ( \frac{100 \times 3}{1000} ) \: moles \\  = 0.3 \: moles \\ RFM = 95 \\ 1 \: mole \: weighs \: 95 \: g \\ 0.3 \: moles \: weigh \: ( \frac{(0.3 \times 95)}{1}  \: g \\  = 28.5 \: g \: of \: magnesium \: chloride

5 0
3 years ago
Light shining on a strip of metal can dislodge electrons. Do you think this is more consistent with light being made up of waves
Naya [18.7K]

Answer:

The correct answer to the following question will be "Particles".

Explanation:

  • A particle seems to be a little component of something, it's little. When you're talking about a subatomic particle, that would be a structured user likely won't see because it's quite unbelievably thin, but it has a tiny mass as well as structural integrity. Such particles seem to be tinier than that of the particles or atoms.
  • Such that the light which shines on the bit of metal could dissipate electrons, the particles seem to be more compatible with the light.
5 0
3 years ago
Read 2 more answers
(5 points)
ExtremeBDS [4]

<u>Answer 2 :</u> The given electronic configuration for a neutral atom of phosphorous in its ground state is incorrect.

Explanation :

A neutral atom of phosphorous has 15 electrons.

The given electronic configuration is incorrect.

The reason is, According to Aufbau principle, the electrons will be first filled in the sub-shell having lower orbital energy. As from the given configuration, 3p sub-shell has lower orbital energy than 4s sub-shell. So, the electrons will be filled in 3p sub-shell first. Hence, the ground state electronic configuration of neutral atom of phosphorous is,

1s^22s^22p^63s^23p^3

<u>Answer 3 :</u>

Element                     Rubidium              Magnesium                Aluminium

Symbol                             Rb                         Mg                              Al

Group number                  1                             2                               13

Number of valence          1                             2                                3

electrons

The order of general reactivity on the basis of number of valence electrons.

Rb > Mg > Al

Reason : The reactivity is determined by the number of electrons present in the outermost shell that means the element which have 1 valence electron will be more reactive because they can easily lose electrons.

5 0
3 years ago
What is the relationship between the Kuiper Belt and the status of Pluto today?
Xelga [282]

Pluto is a dwarf planet, but one of the largest known members, in the Kuiper belt.

The Kuiper Belt extends between 30 AU and 55 AU from the Sun

(1 AU = 1.5 × 10^8 km = distance from Earth to Sun).

Pluto's orbit is highly elliptical. It ranges from 30 AU to 50 AU. When Pluto is closest to the Sun, it is inside the orbit of Neptune (30 AU).

Astronomers class Pluto as a <em>resonant Kuiper belt object</em> (KBO). Because it gets so close to Neptune, its orbit is in <em>resonance</em> with that of Neptune. Pluto makes two orbits for every three of Neptune.

4 0
3 years ago
Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.
Marianna [84]

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
  • Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
  • Initial temperature (T₁) = –25 °C
  • Final temperature (T₂) = 0 °
  • Change in temperature (ΔT) = 0 – (–38) = 38 °C
  • Specific heat capacity (C) = 2050 J/(kg·°C)
  • Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
  • Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 0 °C
  • Final temperature (T₂) = 100 °C
  • Change in temperature (ΔT) = 100 – 0 = 100 °C
  • Specific heat capacity (C) = 4180 J/(kg·°C)
  • Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
  • Mass (m) = 0.4 Kg
  • Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
  • Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
  • Mass (M) = 0.4 Kg
  • Initial temperature (T₁) = 100 °C
  • Final temperature (T₂) = 160 °C
  • Change in temperature (ΔT) = 160 – 100 = 60 °C
  • Specific heat capacity (C) = 1996 J/(kg·°C)
  • Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
  • Heat for –38 °C to 0°C (Q₁) = 31160 J
  • Heat for melting (Q₂) = 133600 J
  • Heat for 0 °C to 100 °C (Q₃) = 167200 J
  • Heat for vaporization (Q₄) = 904000 J
  • Heat for 100 °C to 160 °C (Q₅) = 47904 J
  • Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

7 0
2 years ago
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