C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) ∆H1 = −758 kJ mol−1 ....1)
2C(s) + 2H2(g) → C2H4(g) ∆H2 = +52 kJ mol−1 ....2)
H2(g) + O2(g) → H2O(g) ∆H3 = −242 kJ mol−1 ....3)
Now, enthalpy of formation of carbon monoxide is given by :
∆H = ∆H1 + ∆H2 - ∆H3
∆H = ( -758 + 52 - ( -242 ) ) kJ mol−1
∆H = −464 kJ mol−1
Therefore, the enthalpy of formation of carbon monoxide is -464 kJ mol−1.
Hence, this is the required solution.
The density of a substance is equal to its mass divided by its volume. In this case, it is equal to 693.8 g divided by 92.5 mL. The density of the substance is then equal to 7.50 g/mL. It has 3 significant figures because in division, the answer will have the least number of significant figures from the given.
V1/T1 =V2/T2 (using charles law)
V1=6.00
V2=?
T1=273
T2=273
Making V2 the subject of formula the equation then becomes
V2= V1xT2/T1
6.00x263/273=6.0L
The fusion reaction in the sun is a combination of hydrogen atoms fusing to create helium. The fusion reaction in larger stars involve much heavier elements like oxygen and iron. In supernovas, often elements like gold are produced