A) \[{{H}_{2}}C=CH-C\equiv CH\]
B) \[HC\equiv C-C{{H}_{2}}-C\equiv CH\]
C) \[NaOH\]
D) \[{{104}^{o}}40\]
Correct Answer: B
Solution :
By \[2A+\frac{C}{2}-B,\] we get \[N{{a}_{2}}O+S{{O}_{3}}\xrightarrow{{}}N{{a}_{2}}S{{O}_{4}};\] \[\Delta H=-2\times 146+\frac{259}{2}-418\] or \[\Delta H=-580.5\approx -581kJ\]You need to login to perform this action.
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