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a mixture of neon and nitrogen gases is maintained in a 7.94 l flask at a pressure of 1.51 atm and a temperature of 26.0 °c. if the gas mixture contains 5.35 grams of neon, the number of grams of nitrogen in the mixture is g.
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Explanation:

Step1: Calculate the total moles of gas

Use the ideal gas law \(PV = nRT\).
Given \(P = 1.51\ atm\), \(V=7.94\ L\), \(T=(26.0 + 273.15)K=299.15\ K\), and \(R = 0.0821\ L\cdot atm/(mol\cdot K)\).

$$n=\frac{PV}{RT}=\frac{1.51\times7.94}{0.0821\times299.15}$$

$$n=\frac{12.0}{24.5}=0.490\ mol$$

Step2: Calculate the moles of neon

The molar mass of neon \(M_{Ne}=20.18\ g/mol\). Given \(m_{Ne} = 5.35\ g\).

$$n_{Ne}=\frac{m_{Ne}}{M_{Ne}}=\frac{5.35}{20.18}=0.265\ mol$$

Step3: Calculate the moles of nitrogen

$$n_{N_2}=n - n_{Ne}=0.490 - 0.265 = 0.225\ mol$$

Step4: Calculate the mass of nitrogen

The molar mass of nitrogen \(M_{N_2}=28.02\ g/mol\).

$$m_{N_2}=n_{N_2}\times M_{N_2}=0.225\times28.02$$

$$m_{N_2}=6.30\ g$$

Answer:

\(6.30\)