A) -1, 6
B) 6, 0
C) 6, 3
D) None of these
Correct Answer: B
Solution :
We have, \[=2.35\times {{10}^{-2}}m\] \[r=2.35m\] \[L=\frac{{{\mu }_{r}}.{{\mu }_{0}}{{N}^{2}}A}{l}\] \[=\frac{600\times 4\pi \times {{10}^{-7}}\times {{\left( 2000 \right)}^{2}}\times \left( 1.5 \right)\times {{10}^{-4}}}{0.3}\] \[2C+3{{H}_{2}}\to {{C}_{2}}{{H}_{6}};\Delta H=-21.1\] \[C+OL2\to C{{O}_{2}};\Delta H=-94.1\] \[{{H}_{2}}+\frac{1}{2}{{O}_{2}}\to {{H}_{2}}O;\Delta H=-68.3\] Hence, the greatest and least value are 6 and 0.You need to login to perform this action.
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