Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT. Substituting a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>
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Explanation:
Molar mass Na = 23g/mol
46g = 456/2 = 2mol
1mol = 6.022*10^23 atoms
2mol = 2*6.022*10623
= 1.204*10^24 atoms
Answer:
0.486 L
Explanation:
Step 1: Write the balanced reaction
2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)
Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃
The molar mass of KCIO₃ is 122.55 g/mol.
1.52 g × 1 mol/122.55 g = 0.0124 mol
Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃
The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol
Step 4: Calculate the volume corresponding to 0.0186 moles of O₂
0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L
Cs+1
The only common oxidation state is +1.
