Question

how many moles of oxygen at stp in problem 7 above ? the volume at stp was 7.329 l. this is a missing variable problem, we need to define the variables: v = 7.329 l t = 273.15 k p = 1.000 atm n = ? r = 0.08205 l atm/mol k solving for n n = \frac{p}{r} putting in the numbers n = \frac{\text{atm} \text{l}}{\text{k}} = 0.3270 mol we can also calculate the grams of oxygen by converting grams into moles (\text{mol} o_2)(\frac{\text{g}}{1 mol o_2}) = 10.46 g o_2 a. n_2 b. 20.0 c. 473.15 d. 323.15 e. 1.000 f. p g. p_2 h. v i. v_2 j. n_1 k. t l. t_2 m. 1.464 n. r o. - 0.7543

Explanation:

Step1: Ideal gas law formula

The ideal gas law is \(PV = nRT\). Solving for \(n\) gives \(n=\frac{PV}{RT}\).

Step2: Substitute values

Substitute \(P = 1.000\) atm, \(V=7.329\) L, \(R = 0.08205\) L atm/mol K, and \(T = 273.15\) K into the formula. So \(n=\frac{1.000\times7.329}{0.08205\times273.15}\).

Step3: Calculate moles

\(n=\frac{7.329}{22.40}=0.3270\) mol.

Step4: Calculate grams

To find grams of \(O_2\), use \(m=n\times M\). The molar mass \(M\) of \(O_2\) is \(32.0\) g/mol. So \(m = 0.3270\times32.0=10.46\) g.

Answer:

1. \(H\) (for \(V\) in the formula \(n=\frac{PV}{RT}\))
2. \(K\) (for \(T\) in the formula \(n=\frac{PV}{RT}\))
3. \(E\) (value of \(P = 1.000\) atm)
4. \(H\) (value of \(V = 7.329\) L)
5. \(N\) (value of \(R = 0.08205\) L atm/mol K)
6. \(K\) (value of \(T = 273.15\) K)
7. \(32.0\) (molar mass of \(O_2\) in g/mol)
8. \(0.3270\) (moles of \(O_2\) calculated)