Aluminium is a self-preserving metal, because

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p-Block Elements - I

easy

Aluminium is a self-preserving metal, because

It is not tarnished by air

A thin film of basic carbonate on its surface

A non-porous layer of oxide is formed on its surface

It is not affected by salt water

Solution

Aluminum is called self-protecting metal because when it comes in contact with the atmosphere it is covered entirely with a layer of aluminum oxide which stops further corrosion.

Standard 11

Chemistry

Match List$-I$ with List$-II :$

List$-I$ List$-II$

$a$ Borax $i$ $NaBO _2$

$b$ Kernite $ii$ $Na _2 B _4 O _7 .4 H _2 O$

$c$ Orthoboric acid $iii$ $H _3 BO _3$

$d$ Borax bead $iv$ $Na _2 B _4 O _7 .10 H _2 O$

Choose the correct answer from the options given below :

medium

(NEET-2022)

$B{\left( {OH} \right)_3} + NaOH \to NaB{O_2} + Na\left[ {B{{\left( {OH} \right)}_4}} \right] + {H_2}O$

How can this reaction is made to proceed in forward direction?

normal

Which one of the following is correct statement

easy

$N{a_2}{B_4}{O_7}.10{H_2}O\xrightarrow{{Heat}}X + NaB{O_2} + {H_2}O,X + C{r_2}{O_3}\xrightarrow{{Heat}}\mathop Y\limits_{\left( {Green\,coloured} \right)} $ $X$ and $Y$ are

normal

Write balanced equations for :

$(i)\,B{F_3} + LiH \to $

$(ii)\,{B_2}{H_6} + {H_2}O \to $

$(iii)\,NaH + {B_2}{H_6} \to $

$(iv)\,{H_3}B{O_3}\xrightarrow{\Delta }$

$(v)\,Al + NaOH \to $

$(vi)\,{B_2}{H_6} + N{H_3} \to $

medium

List$-I$	List$-II$
$a$ Borax	$i$ $NaBO _2$
$b$ Kernite	$ii$ $Na _2 B _4 O _7 .4 H _2 O$
$c$ Orthoboric acid	$iii$ $H _3 BO _3$
$d$ Borax bead	$iv$ $Na _2 B _4 O _7 .10 H _2 O$

Aluminium is a self-preserving metal, because

Solution

Similar Questions

Match List$-I$ with List$-II :$ List$-I$ List$-II$ $a$ Borax $i$ $NaBO _2$ $b$ Kernite $ii$ $Na _2 B _4 O _7 .4 H _2 O$ $c$ Orthoboric acid $iii$ $H _3 BO _3$ $d$ Borax bead $iv$ $Na _2 B _4 O _7 .10 H _2 O$ Choose the correct answer from the options given below :

$B{\left( {OH} \right)_3} + NaOH \to NaB{O_2} + Na\left[ {B{{\left( {OH} \right)}_4}} \right] + {H_2}O$ How can this reaction is made to proceed in forward direction?

Which one of the following is correct statement

$N{a_2}{B_4}{O_7}.10{H_2}O\xrightarrow{{Heat}}X + NaB{O_2} + {H_2}O,X + C{r_2}{O_3}\xrightarrow{{Heat}}\mathop Y\limits_{\left( {Green\,coloured} \right)} $ $X$ and $Y$ are

Write balanced equations for : $(i)\,B{F_3} + LiH \to $ $(ii)\,{B_2}{H_6} + {H_2}O \to $ $(iii)\,NaH + {B_2}{H_6} \to $ $(iv)\,{H_3}B{O_3}\xrightarrow{\Delta }$ $(v)\,Al + NaOH \to $ $(vi)\,{B_2}{H_6} + N{H_3} \to $

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Match List$-I$ with List$-II :$

List$-I$ List$-II$

$a$ Borax $i$ $NaBO _2$

$b$ Kernite $ii$ $Na _2 B _4 O _7 .4 H _2 O$

$c$ Orthoboric acid $iii$ $H _3 BO _3$

$d$ Borax bead $iv$ $Na _2 B _4 O _7 .10 H _2 O$

Choose the correct answer from the options given below :

$B{\left( {OH} \right)_3} + NaOH \to NaB{O_2} + Na\left[ {B{{\left( {OH} \right)}_4}} \right] + {H_2}O$

How can this reaction is made to proceed in forward direction?

Write balanced equations for :

$(i)\,B{F_3} + LiH \to $

$(ii)\,{B_2}{H_6} + {H_2}O \to $

$(iii)\,NaH + {B_2}{H_6} \to $

$(iv)\,{H_3}B{O_3}\xrightarrow{\Delta }$

$(v)\,Al + NaOH \to $

$(vi)\,{B_2}{H_6} + N{H_3} \to $