To determine the standard heat of reaction, ΔHrxn°, let's apply the Hess' Law.
ΔHrxn° = ∑(ν×ΔHf° of products) - ∑(ν×ΔHf° of reactants)
where
ν si the stoichiometric coefficient of the substances in the reaction
ΔHf° is the standard heat of formation
The ΔHf° for the substances are the following:
CH₃OH(l) = -238.4 kJ/mol
CH₄(g) = -74.7 kJ/mol
O₂(g) = 0 kJ/mol
ΔHrxn° = (1 mol×-74.7 kJ/mol) - ∑(1 mol×-238.4 kJ/mol)
ΔHrxn° = +163.7 kJ
Answer:
Atomic number of this isotope = 77
Explanation:
Given that,
Mass number = 193
No of neutrons = 116
We need to find the atomic no of this isotope.
We know that,
Atomic mass = No of protons + No. of neutrons
Also, atomic no = no of protons
So,
Atomic mass = atomic no + No. of neutrons
⇒ Atomic no = Atomic mass - no of neutrons
Atomic no = 193 - 116
Atomic no = 77
Hence, 77 is the atomic no of the isotope.
