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Bingel [31]
3 years ago
12

Rewrite each of the following equations for phase changes, to include the heat required for the phase change. Use values from la

tent heat and specific heat constant tables when necessary. Indicate whether each phase change is endothermic or exothermic.
(a) H2O(l) -> H2O(s)
(b) H2O(l) -> H2O(g)
Chemistry
1 answer:
nika2105 [10]3 years ago
5 0

Answer:

The complete equation (a) is:

H_2O_{(l)}->H_2O_{(l)}+333.55 J/g, and is an exothermica phase change

The complete equation (b) is:

H_2O_{(l)}+2257J/g->H_2O_{(g)}, and is an endothermic phase change.

Explanation:

<em>Equation (a):</em>

H_2O_{(l)}->H_2O_{(l)}+333.55 J/g, is showing a phase change from a more-energetic state (liquid) to a relative-less-energetic state (solid), that means, that in order to have the phase change, it is requiered to remove heat (energy) from the water, this is known an exothermic (releases heat).

<em>Equation (b).</em>

H_2O_{(l)}+2257J/g->H_2O_{(g)}, shows that to change from a relative-less-energetic phase (liquid) to a more-energetic one (gas), it would be needed to supply energy in order to acomplish this. And it is called endothermic (absorves heat)

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7 0
3 years ago
1) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data
earnstyle [38]

Answer:

1) The absolute zero temperature is -272.74 °C

2) The absolute zero temperature is -269.91°C

Explanation:

1) The specific volume of air at the given temperatures are;

At -85°C, v =  1/(1.877) = 0.533 dm³/g, the pressure =

At 0°C, v = 1/1.294 ≈ 0.773 dm³/g

At 100°C, v = 1/0.946 ≈ 1.057 dm³/g

We therefore have;

v₁ = 0.533 T₁ = 188.15  K

v₂ = 0.773 T₂ = 273.15  K

v₃ = 1.057 T₃ = 373.15  K

The slope of the graph formed by the above data is therefore given as follows;

m = (1.057 - 0.533)/(373.15 - 188.5) = 0.00284 dm³/K

The equation is therefor;

v - 0.533 = 0.00284×(T - 188.15)

v = 0.00284×T - 0.534 + 0.533  = 0.00284×T - 0.001167

v =  0.00284×T - 0.001167

Therefore, when the temperature, at absolute 0, we have;

v = 0

Which gives;

0 =  0.00284×T - 0.001167

0.00284×T = 0.001167

T = 0.001167/0.00284 ≈ 0.411 K

Which is 0.411  -273.15 ≈ -272.74 °C

The absolute zero temperature is -272.74 °C

2) The given volume of the gas = 20.00 dm³ at 0°C and 1.000 atm

The slope of the volume temperature graph at constant pressure = 0.0741 dm³/(°C)

The temperature is converted to Kelvin temperature, in order to apply Charles law as follows;

0°C = 0 + 273.15 K = 273.15 K

Therefore, the equation of the graph can be presented as follows;

v - 20.00 = 0.0741 × (T - 273.15)

Which gives;

v = 0.0741·T - 0.0741 ×(273.15) + 20

v = 0.0741·T - 20.2404125 + 20

v = 0.0741·T - 0.240415

Therefore at absolute 0, v = 0, we have;

0 = 0.0741·T - 0.240415

0.0741·T = 0.240415

T = 0.240415/0.0741 = 3.2445 K

The temperature in degrees is therefore;

3.2445 K - 273.15 ≈ -269.91°C

The absolute zero temperature is therefore, -269.91°C

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