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Dovator [93]
3 years ago
15

Does a large ball of aluminum foil have the same properties as a small ball?

Chemistry
1 answer:
dsp733 years ago
5 0

Answer:

yes

Explanation:

same properties different size

can i have brainliest pls

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2 years ago
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a 2.7 L of N2 is collected at 121kpa and 288 K . if the pressure increases to 202 kpa and the temperature rises to 303 K , what
jok3333 [9.3K]

Answer:

The gas will occupy a volume of 1.702 liters.

Explanation:

Let suppose that the gas behaves ideally. The equation of state for ideal gas is:

P\cdot V = n\cdot R_{u}\cdot T (1)

Where:

P - Pressure, measured in kilopascals.

V - Volume, measured in liters.

n - Molar quantity, measured in moles.

T - Temperature, measured in Kelvin.

R_{u} - Ideal gas constant, measured in kilopascal-liters per mole-Kelvin.

We can simplify the equation by constructing the following relationship:

\frac{P_{1}\cdot V_{1}}{T_{1}} = \frac{P_{2}\cdot V_{2}}{T_{2}} (2)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kilopascals.

V_{1}, V_{2} - Initial and final volume, measured in liters.

T_{1}, T_{2} - Initial and final temperature, measured in Kelvin.

If we know that P_{1} = 121\,kPa, P_{2} = 202\,kPa, V_{1} = 2.7\,L, T_{1} = 288\,K and T_{2} = 303\,K, the final volume of the gas is:

V_{2} = \left(\frac{T_{2}}{T_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}} \right)\cdot V_{1}

V_{2} = 1.702\,L

The gas will occupy a volume of 1.702 liters.

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