Na_2B_4O_7 cdot 10H_2O xrightarrow{Delta} 2Na | vedclass

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(C) The thermal decomposition of borax $(Na_2B_4O_7 \cdot 10H_2O)$ results in the formation of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O

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$B_2O_3, Mn(BO_2)_2$

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(C) The thermal decomposition of borax $(Na_2B_4O_7 \cdot 10H_2O)$ results in the formation of sodium metaborate $(NaBO_2)$ and boric anhydride $(B_2O_3)$.Thus,$(A) = B_2O_3$.When boric anhydride $(B_2O_3)$ reacts with metal oxides like $MnO$ upon heating,it forms metal metaborates,which are often colored beads.$B_2O_3 + MnO \xrightarrow{\Delta} Mn(BO_2)_2$.Thus,$(B) = Mn(BO_2)_2$.Therefore,$(A)$ is $B_2O_3$ and $(B)$ is $Mn(BO_2)_2$.

$Na_2B_4O_7 \cdot 10H_2O \xrightarrow{\Delta} 2NaBO_2 + B_2O_3 (A) + 10H_2O$. $(A) + MnO \xrightarrow{\Delta} Mn(BO_2)_2 (B)$. $(A)$ and $(B)$ are

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