JEE Main 2025 Chemistry question
Given : [C(graphite)] = 710 kJ mol⁻¹ = 414 kJmol⁻¹ = 436 kJmol⁻¹ = 611 kJmol⁻¹ The for is kJ mol⁻¹ (nearest integer value)
Answer: 25
Solution
kJ/mol.
Given : [C(graphite)] = 710 kJ mol⁻¹ = 414 kJmol⁻¹ = 436 kJmol⁻¹ = 611 kJmol⁻¹ The for is kJ mol⁻¹ (nearest integer value)
kJ/mol.