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2 years ago

Sublimation is used to describe

2 answers:

6 0

Sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state; therefore:

c. the process of ice (solid) changing into water vapor (gas)

hope this helps:)

NNADVOKAT [17]2 years ago

5 0

The answer is c as this phase change refers to any change of a solid to gas.

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What is the current in a 120V circuit if the resistance is 20Ω?

Kaylis [27]

We have: $I=\frac{U}{R}=\frac{120}{20}=6A$

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2 years ago

In a football game, running back is at the 10 yard line and running up the field towards the 50 yard line, and runs for 3 second

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If he keeps that pace he will be at the 34 yard line

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3 years ago

a glass beaker has a mass of 50g. a liquid of density 1.8g/cm3 is poured into the beaker until it reaches the 200cm3 mark. calcu

xz_007 [3.2K]

<u>Answer:</u>

total mass = 410 g

<u>Explanation:</u>

density = 1.8 g/cm³

volume = 200 cm³

density = mass / volume

mass (of liquid) = density x volume

= 1.8 x 200

= 360 g

total mass (beaker + liquid) = 50 + 360 = 410 g [Ans]

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3 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

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3 years ago

If you help me with the correct answer I will give you the brainlist

gulaghasi [49]

Answer:

1. A force is a push or pull upon an object resulting from the object's interaction with another object.

Explanation:

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2 years ago