Answer:
Explanation:
We can calculate the volume of the oxygen molecule as the radius of oxygen molecule is given as 2×10⁻¹⁰m.
We know that volume=4/3×πr³
volume =4/3×π(2.0×10⁻¹⁰m)³
volume=33.40×10⁻³⁰m³
Volume of oxygen molecule=33.40×10⁻³⁰m³
we know the ideal gas equation as:
PV=nRT
k=R/Na
R=k×Na
PV=n×k×Na×T
n×Na=N
PV=Nkt
p is pressure of gas
v is volume of gas
T is temperature of gas
N is numbetr of molecules
Na is avagadros number
k is boltzmann constant =1.38×10⁻²³J/K
R is real gas constant
So to calculate pressure using the formula;
PV=NkT
P=NkT/V
Since there is only one molecule of oxygen so N=1
P=[1×1.38×10⁻²³J/K×300]/[33.40×10⁻³⁰m³
p=12.39×10⁷Pascal
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
First gap is BY USING MAGNETS
2nd is ELECTROLYSIS
3rd is EVAPORATION
