Tag: chemistry

Questions Related to chemistry

Which of the following statements are correct? 

chemistry amines introduction to diazonium salts structure, preparation and physical properties of diazonium salts diazonium salts

  1. Aryldiazonium ions are more stable than alkyldiazonium ions.

  2. Electron release from the ortho- and para-positions of the ring stabilises the aryldiazonium ion.

  3. The increased stability of aryldiazonium is due to the great difficulty of forming $\displaystyle { Ar }^{ \oplus }$ as compared to $\displaystyle { R }^{ \oplus }$.

  4. Alkyldiazonium is more stable than aryldiazonium ion.

Show answer
Correct Option: A,B,C

Arrange the following resonating structure according to their contribution towards resonance hybrid?
(a) $CH _2=\overset {\oplus}{N}=\overset {\ominus}{N}$ (b) $\overset {\ominus}{C}H _2-N=\overset {\oplus}{N}:$ (c) $\overset {\oplus}{C}H _2-\underset {. .}{N}=\overset {\ominus}{N}$ (d) $\overset {\ominus}{C}H _2-\overset {\oplus}{N}\equiv \overset {. .}{N}$

chemistry amines introduction to diazonium salts structure, preparation and physical properties of diazonium salts diazonium salts

  1. a > d > c > b

  2. b > a > c > d

  3. a > c > b > d

  4. d > a > b > c

Show answer
Correct Option: A
Explanation:
The correct order of the contribution of resonating structure towards resonance hybrid is
$\displaystyle a (CH _2=\overset {\oplus}{N}=\overset {\ominus}{N}) > d (\overset {\ominus}{C}H _2-\overset {\oplus}{N}\equiv \overset {. .}{N}) > c (
\overset {\oplus}{C}H _2-\underset {. .}{N}=\overset {\ominus}{N}
) > b (\overset {\ominus}{C}H _2-N=\overset {\oplus}{N}:)$
a and d have higher contribution as they have less charge separation.
b and c have lower contribution as they have more charge separation.

Magnesium metal is heated to redness in the presence of nitrogen and on cooling water is added. The gas evolved is:

some important compounds of magnesium and calcium the s-block elements chemistry

  1. $N _{2}$

  2. $H _{2}$

  3. $NH _{3}$

  4. $O _{2}$

Show answer
Correct Option: C
Explanation:

The gas evolved is $NH _{3}$.

When magensium is heated to redness in the presence of nitrogen, it forms magnesium nitride.
$ 3Mg + N _{2} \rightarrow Mg _{3}N _{2}$

This magnesium nitride when reacts with water forms ammonia gas.
$Mg _{3}N _{2} + 6H _{2}O \rightarrow3Mg(OH) _{2} + 2NH _{3}$

The formula of anhydrite is:

some important compounds of magnesium and calcium the s-block elements chemistry

  1. $CaSO _{4}.2H _{2}O$

  2. $2CaSO _{4}.H _{2}O$

  3. $CaSO _{4}$

  4. $Mg(ClO _{4}) _{2}$

Show answer
Correct Option: C
Explanation:

In nature, calcium sulphate occurs in the form of  anhydrite $(CaSO _4)$ and gypsum $(CaSO _4 \cdot 2H _2O)$.


Hence, option $C$ is correct.

Magnesium may be called a refractory because it?

some important compounds of magnesium and calcium the s-block elements chemistry

  1. has high melting point

  2. is a good conductor of heat

  3. is chemically inert as well as an electrical insulator

  4. forms Grignard's reagent

Show answer
Correct Option: A
Explanation:
  • A refractory material or refractory is a heat-resistant material
  • Magnesium oxide is used to make linings for some furnaces. It is known as a refractory material - which just means that it is resistant to heat. You may have come across this on the CIE page about ceramics. Magnesium oxide has a high melting point (2852°C) and so resists the high temperatures in a furnace.

Which of the following is useful in diagnosing stomach in X-ray picture?

some important compounds of magnesium and calcium the s-block elements chemistry

  1. $Mg(OH) _2$

  2. $BaSO _4$

  3. $Na _2CO _3$

  4. $Ba(OH) _2$

Show answer
Correct Option: B
Explanation:

A suspension of Barium sulphate $(BaSO _4)$, a chalky powder, is ingested in to patient at the time of X-ray. Barium coated areas of the digestive tract then appear on an X-ray as white.

Gypsum, $CaSO _4.2H _2O$, on heating to about $120^0C$ forms a compound which has the chemical composition represented by 

some important compounds of magnesium and calcium the s-block elements chemistry

  1. $CaSO _4.1/2H _2O$

  2. $CaSO _4$

  3. $CaSO _4.H _2O$

  4. $2CaSO _4.3H _2O$

Show answer
Correct Option: A

In aqueous solution, the most stable sulphate is :

some important compounds of magnesium and calcium the s-block elements chemistry

  1. $Be{SO} _{4}$

  2. $Mg{SO} _{4}$

  3. $Ca{SO} _{4}$

  4. $Ba{SO} _{4}$

Show answer
Correct Option: B

Which of the following is INCORRECT statement:

some important compounds of magnesium and calcium the s-block elements chemistry

  1. Above 373 k,$Na _2CO _3,10H _2O$ converted into anhydrous form

  2. Hemihydrate of $CaSO _4$ is known as Gypsum

  3. $NaHCO _3$ act as weak antiseptic for skin infections

  4. All of these

Show answer
Correct Option: B
Explanation:

Gypsum is a hydrated calcium sulphate (CaSO4. 2H2O). Plaster of Paris is calcium sulphate hemihydrate. 

Solubility product of $\displaystyle { Mg\left( OH \right)  } _{ 2 }$ at ordinary temperature is $\displaystyle 1.96\times { 10 }^{ -11 }$. $\displaystyle pH$ of a saturated solution of $\displaystyle { Mg\left( OH \right)  } _{ 2 }$ will be

some important compounds of magnesium and calcium the s-block elements chemistry

  1. 10.53

  2. 8.47

  3. 6.94

  4. 3.47

Show answer
Correct Option: A
Explanation:

$\displaystyle { Mg\left( OH \right)  } _{ 2 }\rightleftharpoons M{ \underset { S }{ g }  }^{ 2+ }+2{ \underset { 2S }{ OH }  }^{ - }$
$\displaystyle { K } _{ sp }{ Mg\left( OH \right)  } _{ 2 }=\left[ { Mg }^{ 2+ } \right] { \left[ { OH }^{ - } \right]  }^{ 2 }$
$\displaystyle \Rightarrow \quad { K } _{ sp }{ Mg\left( OH \right)  } _{ 2 }=4{ S }^{ 3 }$
$\displaystyle 1.96\times { 10 }^{ -11 }=4{ S }^{ 3 }$
or $\displaystyle S={ \left[ \frac { 1.96\times { 10 }^{ -11 } }{ 4 }  \right]  }^{ 1/3 }$
or $\displaystyle S={ \left( 4.9\times { 10 }^{ -12 } \right)  }^{ 1/3 }$
$\displaystyle \therefore \quad S=1.69\times { 10 }^{ -4 }$
So, concentration of $\displaystyle \left[ { OH }^{ - } \right]=2S $
$\displaystyle \therefore \quad \left[ { OH }^{ - } \right] =3.38\times { 10 }^{ -4 }$
$\displaystyle \Rightarrow \quad pOH=-\log { \left[ { OH }^{ - } \right]  } $
$\displaystyle =-\log { \left[ 3.38\times { 10 }^{ -4 } \right]  } $
$\displaystyle \therefore \quad pOH=3.471$
$\displaystyle pH=14-pOH$
$\displaystyle =14-3.471$
$\displaystyle \therefore \quad pH=10.529$