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Sedaia [141]
3 years ago
6

The process by which water diffuses from a region of greater concentration to a region of lesser concentration is called

Physics
2 answers:
DiKsa [7]3 years ago
8 0

Answer:

osmosis is the correct answer.

Explanation:

The process by which water diffuses from a region of greater concentration to a region of lesser concentration is called osmosis.

Osmosis is a process in which solvent moves through a selective semipermeable membrane into a more concentrated region.

There are three types of Osmotic solutions

  • isotonic solution where the osmotic pressure is the same across the membrane.
  • hypotonic solution where the osmotic pressure is lower compared to another solution.
  • hypertonic solution where the solute concentration is higher as compared to the other solution

Natasha_Volkova [10]3 years ago
5 0
Osmosis is the process by which water diffuses from a region of greater concentration to a region of lesser concentration
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6 0
3 years ago
What is the rarefaction of a longitudinal
Softa [21]
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8 0
3 years ago
An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum
Tju [1.3M]

Answer:

The pressure is  P_2  =  4.25 \ a.t.m

Explanation:

From the question we are told that

   The initial pressure is P_1  =  11.2\ a.t.m

   The  temperature is  T_1 =   299 \ K

   

Let the first volume be  V_1 Then the final volume will be  2 V_1

 Generally for a diatomic  gas

           P_1 V_1 ^r  =  P_2  V_2  ^r

Here r is the radius of the molecules which is  mathematically represented as

    r =  \frac{C_p}{C_v}

Where C_p \  and\   C_v are the molar specific heat of a gas at constant pressure and  the molar specific heat of a gas at constant volume with values

     C_p=7 \  and\   C_v=5

=>   r =  \frac{7}{5}

=>  11.2*( V_1 ^{\frac{7}{5} } ) =  P_2  *  (2 V_1 ^{\frac{7}{5} } )

=>   P_2  =  [\frac{1}{2} ]^{\frac{7}{5} } * 11.2

=>  P_2  =  4.25 \ a.t.m

8 0
3 years ago
A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt
Yuri [45]
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P=A\epsilon \sigma T^4
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A is the surface of the object (in our problem, A=1.25 m^2
\epsilon is the emissivity of the object (in our problem, \epsilon=1)
\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4) is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is T=100^{\circ} C=373 K

Substituting these values, we find the power emitted by radiation:
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6 0
3 years ago
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Answer:

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