Tag: chemistry of non-metals

Questions Related to chemistry of non-metals

Two oxides of a metal contain 27.6% and 30.0% of oxygen, respectively. If the formula of the first oxide be ${ { M } _{ 3 }O } _{ 4 }$, find that of the second oxide.

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. $MO _3$

  2. $M _2O$

  3. $M _2O _3$

  4. $MO$

Show answer
Correct Option: C
Explanation:
Given that, formula of first oxide$ = M _3O _4$
Let mass of the metal = x
% of metal in $M _3O _4$ $=\dfrac{3x}{3x+64}\times100$
But  as given % age $= (100-27.6) = 72.4 $%
so, $\dfrac{3x}{3x+64}\times100$ = 72.4
or $x = 56$.
In 2nd oxide,
Given, oxygen = 30%, So metal = 70%
So, the ratio is
$M : O$=$\dfrac{70}{56 }:\dfrac{ 30}{16}$
$\implies 1.25 : 1.875$
$\implies 2 : 3$
So, 2nd oxide is $M _2O _3$

The binary compounds of metallic or non-metallic elements with oxygen are called ________.

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. oxoacid

  2. oxide

  3. oxo

  4. none of these

Show answer
Correct Option: B
Explanation:

The binary compound of elements with oxygen is called oxides. Here oxygen is anionic part. So suffix is 'ide'.

Example: Carbon dioxide $CO _2$

Oxides which show both acidic and basic nature are called as ______ oxides.

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. dual

  2. amphoteric

  3. aprotic

  4. none of these

Show answer
Correct Option: B
Explanation:

Amphoteric oxides show acidic and basic nature. They react with both acids and bases. 

$Al _2O _3$ is an example of amphoteric oxide.

Among the second period which group element form most acidic oxide.

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. $17^{th}$ group

  2. $14^{th}$ group

  3. $15^{th}$ group

  4. $16^{th}$ group

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Correct Option: A
Explanation:

Group element having high electronegativity will be most acidic.

And we know group 17 elements are highly electronegative.

A metal $X$ when burnt in air, $X$-forms oxide and nitride both. $X$ can be:

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. $Rb$

  2. $Mg$

  3. $Na$

  4. $K$

Show answer
Correct Option: B
Explanation:

$Mg$ forms oxide as well as nitride.

$Mg+O _2+N _2\longrightarrow MgO+{Mg} _3N _2$

A metal forms two oxides. The higher oxide contains $80\%$ metal. $0.72 g$ of the lower oxide gave $0.8 g$ of higher oxide when oxidized. Then the ratio of the weight of oxygen that combines with the fixed weight of the metal in the two oxides will be:

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. $2 :3$

  2. $2: 1$

  3. $4 :5$

  4. $3 : 2$

Show answer
Correct Option: B
Explanation:
Given that ,in higher oxide 80% metal is true.
wt. of metal$=\dfrac { 80 }{ 100 } \times 0.8\\$       
                    $ =0.64gm\\ $
wt. of oxygen$=0.16gm$
in lower oxide,
wt. of metal will be same as that of higher oxidation state(+2)

wt. of oxygen$=(72-0.64)=0.08g$

so,ratio of weight of $O _2$ & meals in two oxide is $0.16:0.08=2:1$.
1000ml of 0.75 M HCl = 0.75 mol

So, 25 ml of HCl will contain HCl = $0.75 \times 25 /1000$  = 0.01875

2 mol of HCl reacts with 1 mol of $CaCO _3$

So, 0.01875 mol of HCl will react with $\frac{1}{2}$ $\times$ 0.01875  = $0.009375$ mol

Molar mass of $CaCO _3$ = 100 g

Hence, the mass of 0.009375 mol of $CaCO _3$  = no. of moles $\times $ molar mass$
                                                                          
 $0.009375$ $\times100$  
                                                                   
Hence, the correct option is $C$.

Metal X form 2 oxide. Formula of ${ 1 }^{ st }$ is ${ XO } _{ 2 }$ ${ 1 }^{ st }$ oxide contain $50\% { O } _{ 2 }$. If ${ 2 }^{ nd }$ oxide contain $60\% { O } _{ 2 }$ .Formula of ${ 2 }^{ nd }$ oxide?

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. ${ X } _{ 2 }O$

  2. ${ XO } _{ 3 }$

  3. ${ X } _{ 2 }{ O } _{ 3 }$

  4. ${ X } _{ 3 }{ O } _{ 2 }$

Show answer
Correct Option: C
Explanation:

Given that in${ XO } _{ 2 },50%$% of $O _2$ is present

${ 2 }O _{ 2 }=32\ \frac { 32 }{ M+32 } =0.5$
now ,given that is second oxide 60% of   is present let no. of metal X contained in this oxide.is n then,
$\cfrac { 32 }{ n\times 32+32 } =0.6\ \cfrac { 1 }{ n+1 } =0.6\ 1=0.6n+0.6\ n=\cfrac { 0.4 }{ 0.6 } =\cfrac { 2 }{ 3 } \ { X } _{ 2 }{ O } _{ 3 }$  
is formula of compound

$K0 {2}$(potassium super oxide) is used in oxygen cylinder in space and submarines. This is because it ____________.

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. absords $CO _{2}$ and increase $O _{2}$ content

  2. eliminates moisture

  3. absords $CO _{2}$

  4. produces ozone

Show answer
Correct Option: A
Explanation:

Because it absorbs $CO _2$ and increases $O _2$ concentration according to the following reaction

$4K{ O } _{ 2 }+2{ CO } _{ 2 }\rightarrow 2{ K } _{ 2 }{ CO } _{ 3 }+3{ O } _{ 2 }$

$Pb{O} _{2}$ oxide is:

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. basic

  2. acidic

  3. neutral

  4. amphoteric

Show answer
Correct Option: D
Explanation:

The $Pb{ O } _{ 2 }$ behaves as acidic and basic property so it is called as amphoteric.

When $Pb{ O } _{ 2 }$ dissolved in a strong base $Pb{ O }{ H } _{ 6 }$ is obtained so it is an acidic property.
when $Pb{ O } _{ 2 }$ dissolved in HCl,$PbCl _{ 6 }$ is formed so it is a basic property.
Thus $PbO _{ 2 }$ behaves as amphoteric in nature.

Two oxides of a metal contain $50$% and $40$% metal (M) respectively. If the formula of first oxide is $MO _2$, the formula of second oxide will be:

chemistry chemistry of non-metals oxides of elements simple oxides oxygen

  1. $MO _2$

  2. $MO _3$

  3. $M _2O$

  4. $M _2O _5$

Show answer
Correct Option: B
Explanation:
$\text{Compound } 1\quad\quad\quad\quad 50\% M+50\% O$
$\text{formula}\quad\quad\quad\quad MO _2$
Let atomic weight of metal be $X,$
$\Longrightarrow MO _2:$    $\cfrac{16}{X}=\cfrac{1}{2}$
Hence atomic weight of metal $=32$
$\text{Compound }2:$ $40\% M=\cfrac{40}{32}=1.25=1\\60\%O=\cfrac{60}{16}=3.75=3$
$\therefore$ Formula is $MO _3$.