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seropon [69]
2 years ago
7

If an ice cube melts at 0ºC but water also freezes at 0ºC, what is the difference between melting and freezing in terms of…….

Chemistry
1 answer:
devlian [24]2 years ago
4 0

Answer:

<em>8. When melting the energy of the ice cubes increases and when frezzing the energy of the water decreases.</em>

<em />

<em>9. When melting the otion of the particles speed up, when freezing the motion of the particles slow down.</em>

Explanation:

<em><u>8. The energy involved in the phase changes</u></em>

When an ice cube melts at 0ºC it must absorb or gain energy from the surroundings. This is an endothermic process, because the energy of the liquid water shall be higher than the energy of the starting ice cubes.

In conclusion, when melting the energy of the water increases.

On the other hand, when the liquid water freezes, also at 0ºC, the flow of heat energy is from the liquid water to the surroundings. This is an exothermic process because the water releases heat. At the end, the energy of the frozen water will be lower than the energy of the liquid water.

In conclusion, when freezing the enery of the water decreases.

<u><em>9. The motion of the particles of the substances during the phase changes</em></u>

<u><em /></u>

The particles of liquid substances are in constant motion, which is what allows them to flow. This is because the particles (molecules or atoms) are relatively distant from each other, with relatively low intermolecular forces.  Thus, they can slide on each other.

On the other hand, the solid substances have the particles in fixed position in structures called crystals. The particles vibrate but do not translation motion.

In conclusion, the difference is that the motion of the particles during melting increases and the motion of th particles during freezing decreases.

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MariettaO [177]

Answer:

P_2 =0.51  atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}

0.008389261745 \times 273 = 4.5P_2

P_2 =\dfrac{0.008389261745 \times 273 }{4.5}

P_2 =0.51 \ atm

4 0
2 years ago
An aqueous solution containing 17.5 g of an unknown molecular (nonelectrolyte) compound in 100.0 g of water has a freezing point
Sonbull [250]

Answer:

molar mass = 180.833 g/mol

Explanation:

  • mass sln = mass solute + mass solvent

∴  solute: unknown molecular (nonelectrolyte)

∴ solvent: water

∴ mass solute = 17.5 g

∴ mass solvent =  100.0 g = 0.1 Kg

⇒ mass sln = 117.5 g

freezing point:

  • ΔTc = - Kc×m

∴ ΔTc = -1.8 °C

∴ Kc H2O = 1.86 °C.Kg/mol

∴ m: molality (mol solute/Kg solvent)

⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)

⇒ m = 0.9677 mol solute/Kg solvent

  • molar mass (Mw) [=] g/mol

∴ mol solute = ( m )×(Kg solvent)

⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )

⇒ mol solute = 0.09677 mol

⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )

⇒ Mw solute = 180.833 g/mol

6 0
3 years ago
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According to Gayle Lusac's law, pressure is proportional to absolute temperature of a gas. Thus:
P/T = constant

So if the temperature becomes 3T, the pressure would increase to 3P
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