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Mrac [35]
3 years ago
12

An 8.0 L sample of neon gas at 300. K exerts a pressure of 900. kPa. If the gas is compressed to 2.0 L and the temperature is ra

ised to 600. K, what will be the new pressure?
Chemistry
1 answer:
svetoff [14.1K]3 years ago
6 0

Answer:

7200 kPa

Explanation:

Applying,

PV/T = P'V'/T'................ Equation 1

Where P = Initial pressure of neon gas, V = Initial volume of neon gas, T = Initial temperature of neon gas, P' = Final pressure of neon gas, V' = Final volume of neon gas, T' = Final Temperature of neon gas

Make P' the subject of the equation

P' = PVT'/V'T.............. Equation 2

Given: P = 900 kPa, V = 8.0 L, T = 300 K, V' = 2.0 L, T' = 600 K

Substitute these values into equation 2

P' = (900×8×600)/(2×300)

P' = 7200 kPa

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<u>Net ionic equation</u>

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Explanation:

Let's consider two kind of equations:

  • Full ionic equation: includes all ions and species that do not dissociate in water.
  • Net ionic equation: includes only ions that participate in the reaction (<em>not spectator ions</em>) and species that do not dissociate in water.
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