An ice cube tray contains enough water at 22.0 c to make 18 ice cubes that each have a mass of 30.0g. the tray is placed in a freezer that uses cf2cl2 as a refridgerant. the heat of vap of cf2cl2 is 158 j/g. what mass of [email  protected] must be vaporized in the refridgeration cycle to convert all the water at 22.0 c to ice at -5.o c.
this is all i have so far. ? h capacity h20 (s) = 2.08 j/g c h capacity h20 (l) = 4.18 j/g c ethanply of fusion of ice =6.02 kj/mol.

Heat capacity can be illustrated as the amount of heat absorbed or released to modify the temperature of the compound. It can be represented as, q = mcΔT

Here, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

On the basis of the given values,

Mass of one ice cube = 30 g

Mass of 10 ice cubes = 30 × 18 = 540 g

Temperature of water = 22 °C

Temperature of ice = -5 °C

Specific heat capacity of water (s) = 2.08 J/g °C

Specific heat capacity of water (l) = 4.18 J/g °C

ΔHfusion (ice) = 6.02 kJ/g

ΔHvap (CF₂Cl₂) = 158 J/g

The heat change is calculated when temperature of water changes from 22°C to 0°C,

q1 = mwater.cwater.ΔT

= 540 g × 4.18 J/g °C * (0°C -22°C)

= -49.74 kJ

Heat change when 540 g of water freezes at O°C is termed as enthalpy of fusion that is given as,

q2 = ΔHfusion × mass of water

= -6.02 kJ/mol × 540 g / 18 g/mol

= -180.6 kJ

The heat change has to be calculated when temperature of 540 g ice changes from 0°C to -5°C (Tf) as,

q3 = mice.cice.ΔT

= 540 g × 2.08 J/g°C * (-5°C - 0°C)

= -5.62 kJ

Total heat lost by the system = q1 +q2 + q3

= (49.74 + 180.6 + 5.62)

= 235.96 kJ

Now, the heat gained by CF₂Cl₂ = Heat lost by the system

ΔHvap × mass of CF₂Cl₂ vaporized = 235.96 kJ

Mass of CF₂Cl₂ vaporized = 235.96 kJ / ΔHvap

= 235.96 kJ / 158 × 10⁻³ kJ/g

= 1.5 × 10³ g

Hence, the amount of CF₂Cl₂ vaporizes in the overall conversion is 1.5 × 10³ g

it should be all of them because it asks if you included all of those in your answer

explanation:

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3