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yawa3891 [41]
4 years ago
15

Lons that are present before and after a neutralization reaction are

Chemistry
1 answer:
My name is Ann [436]4 years ago
5 0

Answer:A

Explanation:A

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If 10.5-g of Na2SO4 are actually recovered experimentally, that is the percent yield?
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<span>"If they recover 10.5 it means you shall add with the limiting 10.8, giving you 21.3g out of 142 times 100, giving 15%, I believe."
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3 years ago
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.
6 0
3 years ago
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