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ivanzaharov [21]
3 years ago
11

Acetone (C3H6O) is a liquid that is used to dissolve substances that do not mix with water. The boiling point of acetone is 56°C

. If you have some gaseous acetone at 100°C and you cool it down, at what temperature in degrees C, will it start to condense? JUST give a number, no units)
Chemistry
1 answer:
serg [7]3 years ago
8 0

Answer : Acetone start to condense below 56 °C

Explanation :

Boiling point : It is defined as the temperature at which vapor pressure of the liquid is equal to the atmospheric pressure of the liquid.

That means, the temperature at which the liquid changes into gas.

Condensation point : It is defined as the temperature at which the phase changes from a gaseous to liquid.

As per question, the boiling point of acetone is 56 °C. As, the gas is at 100 °C then the condensation of acetone will take place at 56 °C but below this temperature acetone exists in a liquid phase.

Hence, below 56 °C acetone start to condense.

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What is the energy of a light wave with a frequency of 9.8 x 10^20 Hz?
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Answer:

Regions of the Electromagnetic Spectrum

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7 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

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Weight of the balloon = 2.0 g

Six weights each of mass 30.0 g is added to the balloon.

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Assuming the balloon to be a sphere,

Volume of the sphere = \frac{4}{3}πr^{3}

178 cm^{3} = \frac{4}{3}(\frac{22}{7})r^{3}

r = 3.49 cm

Radius of the balloon = 3.49 cm

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Answer:

  B

Explanation:

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