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Kaylis [27]
3 years ago
5

Air has

Chemistry
2 answers:
katovenus [111]3 years ago
6 0

Answer:

Neither

Explanation:

mrs_skeptik [129]3 years ago
4 0

Answer would be neither!

The reason being that it can take the shape and volume of anything.

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Cierta cantidad de hidrogeno esta encerrada en una camara de platino a volumen constante. Cuando la camara se sumerge en un baño
salantis [7]

Answer:

-245.7°C es la temperatura del gas bajo 100 torr

Explanation:

Para resolver esta pregunta debemos hacer uso de la ley de Boyle que establece que la presión de un gas es directamente proporcional a la temperatura de este cuando el volumen permanece constante. La ecuación es:

P1T2 = P2T1

<em>Donde P es presión y T temperatura absoluta del estado inicial, 1, y final, 2.</em>

<em />

Reemplazando:

P1 = 1000torr

T2 = ? -Incógnita-

P2 = 100torr

T1 = 273K -Temperatura del hielo fundido = 0°C = 273K

1000torrT2 = 100torr273

T2 = 27.3K

27.3K - 273 =

<h3>-245.7°C es la temperatura del gas bajo 100 torr</h3>

<em />

3 0
3 years ago
We can also use the equation for enthalpy change for physical phase changes. Consider the phase change H2O(l) → H2O(g). Calculat
BlackZzzverrR [31]
<span>44.0 kJ is the answer</span>
4 0
3 years ago
Read 2 more answers
3.
sineoko [7]

Answer:

3. V = 0.2673 L

4. V = 2.4314 L

5. V = 0.262 L

6. V = 2.224 L

Explanation:

3. assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 225 L

∴ T1 = 175 K

∴ P1 = 150 KPa = 1.48038 atm

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))

⇒ n = 0.043 mol

∴ T2 = 112 K

∴ P2 = P1 = 150 KPa = 1.48038 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)

⇒ V2 = 0.2673 L

4. gas is heated at a constant pressure

∴ T1 = 180 K

∴ P = 1 atm

∴ V1 = 44.8 L

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))

⇒ n = 0.3295 mol

∴ T2 = 90 K

⇒ V2 = RT2n/P

⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)

⇒ V2 = 2.4314 L

5.  V1 = 200 L

∴ P1 = 50 KPa = 0.4935 atm

∴ T1 = 271 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))

⇒ n = 0.2251 mol

∴ P2 = 100 Kpa = 0.9869 atm

∴ T2 = 14 K

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)

⇒ V2 = 0.262 L

6.a)  ∴ V1 = 24.6 L

∴ P1 = 10 atm

∴ T1 = 25°C = 298 K

⇒ n = RT/PV

⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))

⇒ n = 0.0993 mol

∴ T2 = 273 K

∴ P2 = 101.3 KPa = 0.9997 atm

⇒ V2 = RT2n/P2

⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)

⇒ V2 = 2.224 L

3 0
3 years ago
Calculating Density Warm Up
Artist 52 [7]

Answer:

2

Explanation:

33-25=8

48/8=6

6 0
4 years ago
What is the name of the gas produced when nitric acid is added to copper metal? please help! much appreciated!
Lemur [1.5K]
The answer to your question is nitrogen dioxide

8 0
3 years ago
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