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AlladinOne [14]
4 years ago
9

the concentration of solutes is the same both inside and outside of the cell the concentration of solutes is less outside the ce

ll than inside the cell the concentration of solutes is greater outside the cell than inside it the diffusion of water across a selectively permeable membrane the movement of molecules from a region of high concentration to a region of low concentration by means of random molecular motion a membrane that allows only specific molecules to move into or out of the cell A. diffusion B. isotonic C. hypotonic D. hypertonic E. osmosis F. selectively permeable
Chemistry
1 answer:
Volgvan4 years ago
6 0

Answer: The concentration of solutes is the same both inside and outside of the cell : isotonic

The concentration of solutes is less outside the cell than inside the cell : hypotonic

the concentration of solutes is greater outside the cell than inside it: hypertonic

the diffusion of water across a selectively permeable membrane: osmosis

the movement of molecules from a region of high concentration to a region of low concentration by means of random molecular motion :   diffusion

a membrane that allows only specific molecules to move into or out of the cell : selectively permeable

Explanation:

The solutions having same osmotic pressure and hence same molar concentration are termed as isotonic.

If a solution is of lower osmotic pressure it is called as hypotonic with respect to the concentrate solution.

The more concentrated solution is said to be hypertonic with respect to the dilute solution.

Osmosis is defined as the process in which molecules of a solvent tend to move through a semipermeable membrane from a region of low concentration solution to a region of more concentrated solution.

Diffusion can be defined as a phenomenon that occurs in liquid or gases which is the process of movement of solute particles from area of high concentration to low concentration.

Selectively permeable membrane is a membrane which allows selective movement of molecules.

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Explanation:

1.138 has 4 significant figures, which are 1, 1, 3 and 8. The numbers after the decimal point are decimals and are significant figures.

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Nitrogen (N2) enters a well-insulated diffuser operating at steady state at 0.656 bar, 300 K with a velocity of 282 m/s. The inl
Tasya [4]

Answer:

  1. The exit temperature of the Nitrogen would be 331.4 K.
  2. The area at the exit of the diffuser would be 7*10^{-3} m^2.
  3. The rate of entropy production would be 0.

Explanation:

  1. First it is assumed that the diffuser works as a isentropic device. A isentropic device is such that the entropy at the inlet is equal that the entropy T the exit.
  2. It will be used the subscript <em>1 for the</em> <em>inlet conditions of the nitrogen</em>, and the subscript <em>2 for the exit conditions of the nitrogen</em>.
  3. It will be called: <em>v</em> the velocity of the nitrogen stream, <em>T</em> the nitrogen temperature, <em>V</em> the volumetric flow of the specific stream, <em>A</em> the area at the inlet or exit of the diffuser and, <em>P</em> the pressure of the nitrogen flow.
  4. It is known that <em>for a fluid flowing, its volumetric flow is obtain as:</em> V=v*A,
  5. Then for the inlet of the diffuser: V_1=v_1*A_1=282\frac{m}{s}*4.8*10^{-3}m^2=1.35\frac{m^3}{s}
  6. For an ideal gas working in an isentropic process, it follows that: \frac{T_{2} }{T_1}=(\frac{P_2}{P_1})^k where each variable is defined according with what was presented in step 2 and 3, and <em>k </em>is the heat values relationship, 1.4 for nitrogen.
  7. Then <em>solving</em> for T_2, the temperature of the nitrogen at the exit conditions: T_2=T_1(\frac{P_2}{P_1})^k then, T_2=300 K (\frac{0.9 bar}{0.656 bar})^{(\frac{1.4-1}{1.4})}=331.4 K
  8. Also, for an ideal gas working in an isentropic process, it follows that:  \frac{P_2}{P_1}= (\frac{V_1}{V_2})^k, where each variable is defined according with what was presented in step 2 and 3, and <em>k</em> is the heat values relationship, 1.4 for nitrogen.
  9. Then <em>solving</em> for V_2 the volumetric flow at the exit of the diffuser: V_2=V_1*\frac{1}{\sqrt[k]{\frac{P_2}{P_1}}}=\frac{1.35\frac{m^3}{s}}{\sqrt[1.4]{\frac{0.9bar}{0.656bar} }}=1.080\frac{m^3}{s}.
  10. Knowing that V_2=1.080\frac{m^3}{s}, it is possible to calculate the area at the exit of the diffuser, using the relationship presented in step 4, and solving for the required parameter: A_2=\frac{V_2}{v_2}=\frac{1.08\frac{m^3}{s} }{140\frac{m}{s}}=7.71*10^{-3}m^2.
  11. <em>To determine the rate of entropy production in the diffuser,</em> it is required to do a second law balance (entropy balance) in the control volume of the device. This balance is: S_1+S_{gen}-S_2=\Delta S_{system}, where: S_1 and S_2 are the entropy of the stream entering and leaving the control volume respectively, S_{gen} is the rate of entropy production and, \Delta S_{system} is the change of entropy of the system.
  12. If the diffuser is operating at stable state is assumed then \Delta S_{system}=0. Applying the entropy balance and solving the rate of entropy generation: S_{gen}=S_2-S_1.
  13. Finally, it was assume that the process is isentropic, it is: S_1=S_2, then S_{gen}=0.
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3 years ago
Which spheres of earth are represented in tropical rainforests?
tresset_1 [31]

Answer:

Biosphere

Explanation:

because of the rain it gets from huge water sources.

I hope that was useful.

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