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Evgen [1.6K]
4 years ago
7

A hammer taps on the end of a 4.0-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pu

lses of sound, one that travels through the metal and one that travels through the air. The pulses are separated in time by 11.0 ms.
1. What is the speed of sound in this metal? v = 6060 m/s
2.If the bar was twice as long, what would be the time difference?
Physics
1 answer:
Mashcka [7]4 years ago
6 0

Answer: c*m on now

Explanation:

u dumb b***h

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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

4 0
4 years ago
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