A steel container of volume 0.35 L can withstand pressures up to 88 atm before exploding. Part A Find the mass of helium that can be stored in this container at 313 K. Express your answer using two significant figures. m = g
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Given: Pressure (P) = 88 atm Volume (V) = 0.35 L Temperature (T) = 313 K Gas constant (R) = 0.0821 L.atm/mol.K \[ n = \frac{PV}{RT} \] \[ n = \frac{(88 atm)(0.35 L)}{(0.0821 L.atm/mol.K)(313 K)} \] \[ n = \frac{30.8}{25.6573} \] \[ n \approx 1.2 mol \] Show more…
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