As you move around there is a change in: electronegativies, ionisation energies, atomic radius etc. different amounts of these properties are going to effect how the element acts
Answer:
T₂ = 721 k
Explanation:
Given data:
Initial volume = 285 mL
Initial pressure = 1.88 atm
Initial temperature = 355 K
Final temperature = ?
Final volume = 435 mL
Final pressure = 2.50 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁ / P₁V₁
T₂ = 2.50 atm × 435 mL × 355 K / 1.88 atm × 285 mL
T₂ = 386062.5 atm. mL. K /535.8 atm. mL
T₂ = 721 k
Answer: 72.4 kJ/mol
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
![\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCO_2%7D%5Ctimes%20%5CDelta%20H_%7BCO_2%7D%29%2B%28n_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_%7BH_2O%7D%29%5D-%5B%28n_%7BO_2%7D%5Ctimes%20%5CDelta%20H_%7BO_2%7D%29%2B%28n_%7BC_3H_8%7D%5Ctimes%20%5CDelta%20H_%7BC_3H_8%7D%29%5D)
where,
n = number of moles
(as heat of formation of substances in their standard state is zero
Now put all the given values in this expression, we get
![-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]](https://tex.z-dn.net/?f=-2220.1%3D%5B%283%5Ctimes%20-393.5%29%2B%284%5Ctimes%20-241.8%29%5D-%5B%285%5Ctimes%200%29%2B%281%5Ctimes%20%5CDelta%20H_%7BC_3H_8%7D%29%5D)
Therefore, the heat of formation of propane is 72.4 kJ/mol
Number
The atomic number is the number of protons in one atom of an element.
Hope this helps!
Answer:
a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly.
Explanation:
