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Olegator [25]
3 years ago
11

Air flows from a reservoir in which the pressure is kept at 124 kPa through a convergent–divergent nozzle and exhausts to the

atmosphere where the pressure is 101.3 kPa. Under these conditions, the nozzle is choked and the flow is subsonic on both sides of the throat. To what value must the pressure in the reservoir be increased so that there is a normal shock on the nozzle exit plane?

Physics
1 answer:
ch4aika [34]3 years ago
4 0

Answer:

ΔP = 986 Kpa

Explanation:

The solution is given in the pictures below

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72) What is the freezing point (°C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115
Rina8888 [55]

<u>Answer:</u> The freezing point of solution is -3.34°C

<u>Explanation:</u>

Vant hoff factor for ionic solute is the number of ions that are present in a solution. The equation for the ionization of calcium nitrate follows:

Ca(NO_3)_2(aq.)\rightarrow Ca^{2+}(aq.)+2NO_3^-(aq.)

The total number of ions present in the solution are 3.

  • To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (Ca(NO_3)_2) = 11.3 g

M_{solute} = Molar mass of solute (Ca(NO_3)_2) = 164  g/mol

W_{solvent} = Mass of solvent (water) = 115 g

Putting values in above equation, we get:

\text{Molality of }Ca(NO_3)_2=\frac{11.3\times 1000}{164\times 115}\\\\\text{Molality of }Ca(NO_3)_2=0.599m

  • To calculate the depression in freezing point, we use the equation:

\Delta T=iK_fm

where,

i = Vant hoff factor = 3

K_f = molal freezing point depression constant = 1.86°C/m.g

m = molality of solution = 0.599 m

Putting values in above equation, we get:

\Delta T=3\times 1.86^oC/m.g\times 0.599m\\\\\Delta T=3.34^oC

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T=\text{freezing point of water}-\text{freezing point of solution}

\Delta T = 3.34 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

3.34^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-3.34^oC

Hence, the freezing point of solution is -3.34°C

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