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2 years ago

Would you expect the water or sugar solution to have the most supercooling? WHY? PLEASE HELP

Chemistry

2 answers:

g100num [7]2 years ago

6 0

Answer:

The water solution

Explanation:

Super freezing occurs when energy is quickly withdrawn from a liquid and it cools below the solidification temperature. This phenomenon occurs due to the fact that the cooling is very fast and does not occur the formation of an initial "point" or crystal, which is generally formed to start the freezing. A small crystal formed will attract the molecules nearby, forming a large crystal, that is, the frozen liquid.

The sugar solution should be distilled so that Super Cooling occurs more efficiently than the water solution because the presence of the sugar would create nucleation points and allow the formation of ice crystals. The super cooled water solution turns very quickly into ice, much faster than the sugar solution. It is also possible to supercooled the water after the homogeneous nucleation point, in which case it eventually solidifies into a type of glass.

il63 [147K]2 years ago

3 0

Supercooled water can be prompted to trun into ice by tends to make it more difficult to achieve low nucleation Trehalose and other sugar solutions at low.

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lana66690 [7]

Answer:

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

Explanation:

Step 1: Data given

Mass of glucose = 0.305 grams

Combustion of 0.305 grams causes a raise of 6.30 °C

Calorimeter has a heat capacity of 755 J/°C

Molar mass of glucose = 180.2 g/mol

Step 2: The balanced equation

C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g)

Step 3:

ΔH = (m * C * ΔT + c(calorimeter) * ΔT)

with m = mass of the solutin = 0.305 grams

with C = heat capacity of water = 4.184 J/g°C

with ΔT = the change in temperature = 6.30 °C

with c(calorimeter) = 755 J/°C

ΔH = 0.305 * 4.184 *6.30 + 755 * 6.30 = 4764.5 J ( negative because it's exothermic)

Step 4: Calculate moles of glucose

Moles glucose = mass glucose / Molar mass glucose

Moles glucose = 0.305 grams / 180.2 g/mol

Moles glucose = 0.00169 moles

Step 5: Calculate molar enthalpy

Molar enthalpy = -4764.5 J / 0.00169 moles

Molar enthalpy = - 2819254.2 J/moles = -2819.3 kJ/moles

The molar enthalpy of combustion of glucose is -2819.3 kJ/mol

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