A) Meta borate
B) Tetra borate
C) Double oxide
D) Ortho borate
Correct Answer: A
Solution :
\[N{{a}_{2}}{{B}_{4}}{{O}_{7}}.10{{H}_{2}}O\xrightarrow[-10\,\,{{H}_{2}}O]{}N{{a}_{2}}{{B}_{4}}{{O}_{7}}\]\[\xrightarrow{\Delta }2NaB{{O}_{2}}+{{B}_{2}}{{O}_{3}}\]\[CuO+{{B}_{2}}{{O}_{3}}\to \underset{(\text{Copper meta borate blue)}}{\mathop{Cu{{(B{{O}_{2}})}_{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\,\]
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