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# $5H_2O+NaB_4O_7+2HCl\rightarrow 2NaCl+x$. In this equation the $x$ is

$\begin{array}{1 1}(a)\;Orthoboric\;acid\\(b)\;Hydroboric\;acid\\(c)\;Metaboric\;acid\\(d)\;Tetra\;boric\;acid\end{array}$

$4H_3BO_3$