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a reaction between liquid reactants takes place at ( 4.0^{circ}c ) in a…

Question

a reaction between liquid reactants takes place at ( 4.0^{circ}c ) in a sealed, evacuated vessel with a measured volume of ( 30.0l ). measurements show that the reaction produced ( 33g ) of sulfur tetrafluoride gas.
calculate the pressure of sulfur tetrafluoride gas in the reaction vessel after the reaction. you may ignore the volume of the liquid reactants. be sure your answer has the correct number of significant digits.
pressure: ( square atm )

Explanation:

Step1: Calculate the number of moles of \(SF_4\)

The molar mass of \(SF_4\) (\(M\)): \(M = 32.07+(4\times19.00)=32.07 + 76.00=108.07\space g/mol\)
The number of moles (\(n\)) of \(SF_4\): \(n=\frac{m}{M}\), where \(m = 33\space g\)
\(n=\frac{33\space g}{108.07\space g/mol}\approx0.305\space mol\)

Step2: Convert the temperature to Kelvin

The temperature (\(T\)): \(T=(4.0 + 273.15)\space K=277.15\space K\)
The volume (\(V\)): \(V = 30.0\space L\)
We use the ideal - gas law \(PV=nRT\), where \(R = 0.0821\space L\cdot atm/(mol\cdot K)\)

Step3: Solve for pressure (\(P\))

From \(PV=nRT\), we can express \(P\) as \(P=\frac{nRT}{V}\)
Substitute \(n = 0.305\space mol\), \(R=0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 277.15\space K\), and \(V = 30.0\space L\) into the formula:
\(P=\frac{0.305\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times277.15\space K}{30.0\space L}\)
First, calculate the numerator: \(0.305\times0.0821\times277.15=(0.305\times0.0821)\times277.15\approx0.02504\times277.15\approx6.94\)
Then, \(P=\frac{6.94}{30.0}\space atm\approx0.231\space atm\)

Answer:

\(0.23\space atm\)