Chemistry

Chemical Compounds and Reactions

505 Questions

Chemical compounds and reactions involve understanding the properties, colors, and formation of various chemical substances. Questions focus on identifying precipitates, observing color changes during reactions, and naming common compounds. This topic is a fundamental part of the chemistry syllabus for competitive exams.

identifying precipitateschemical reaction colorscompound namingcolloidal particle propertiesacid base reactions

Chemical Compounds and Reactions Questions

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Law of constant composition is also called as the law of :

  1. Conservation of mass

  2. Conservation of energy

  3. Multiple proportion

  4. Definite proportion
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Law of constant composition is also called as the law of Definite proportion which says that the law of constant composition is a chemistry law which states samples of a pure compound always contain the same elements in the same mass proportion.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

In a compound the ratio of the atoms or element by mass remains always same irrespective of :

  1. temperature

  2. nature of compound

  3. source of compound

  4. size of compound
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Compounds are formed by the combination of two or more elements. In a compound the ratio of the atoms or element by mass remains always same irrespective of the source of compound. This means a certain compound always formed by the combination of atoms in same ratio by mass. If the ratio of mass of constituent atoms will be altered the new compound is formed.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

If the ratio of the atoms by mass is altered then :

  1. new compound formed

  2. same compound are formed

  3. no reaction takes place

  4. state of compound are changed
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a compound the ratio of the atoms or element by mass remains always same irrespective of the source of compound. This means a certain compound always formed by the combination of atoms in same ratio by mass. If the ratio of mass of constituent atoms will be altered the new compound is formed.

Multiple choice chemistry general principles of metallurgy extraction of copper and zinc extraction of copper and zinc from their oxides basics of metallurgy

Which of the following two substances react to produce blister copper?

  1. $Cu _{2}S _{(s)}$ and $Cu _{2}O _{(s)}$
  2. $Cu _{2}O _{(s)}$ and $FeS _{(s)}$
  3. $Cu _{2}S _{(s)}$ and $O _{2(g)}$
  4. $Cu _{2}S _{(s)}$ and $FeS _{2(s)}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Blast of air converts $Cu _2S$ partly into $Cu _2O$ which reacts with remaining $Cu _2S$ to give molten copper. The copper so obtained is called "Blister copper" because, as it solidifies, $SO _2$ hidden in it escapes out producing blister on its surface

option A is correct

Multiple choice enzymes and catalysts catalysis adsorption and colloids surface chemistry chemistry adsorption theory of heterogeneous catalysis

Presence of traces of arsenious oxide ($As _{2}O _{3}$) in the reacting gases $SO _{2}$ and $O _{2}$ in presence of platinised asbestos in contact process acts as:

  1. catalytic promoter

  2. catalytic poison

  3. dehydrating agent

  4. drying agent.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

  • Promoters are substances that enhance the activity of a catalyst while poisons decrease the activity of a catalyst.
  • In the manufacturing of sulfuric acid, a step involves the reaction of $SO _2$ and $O _2$ in presence of platinized asbestos to give $SO _3$
                    .i.e. $SO _2+O _2 \rightarrow SO _3$ 
In the above reaction if traces of arsenious oxide ($As _2O _3$)  is used then it acts as the catalytic poison by reducing the catalytic activity of Platinised asbestos.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

The resulting solution obtained at the end of electrolysis of concentrated aqueous solution of $NaCl:$

  1. Turns blue litmus into red

  2. Turns red litmus into blue

  3. Remains colourless with phenolphthalein

  4. The colour of red or blue litmus does not change
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

During the electrolysis of aqueous sodium chloride solution, the products are $NaOH , Cl _2$ and $H _2$.
$NaCl(aq) + H _2O(l) \rightarrow Na^+ (aq) +OH^-(aq) +\frac{1}{2} H _2(g) +\frac{1}{2}Cl _2(g)$
Due to formation of base, it turns res litmus into blue.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

Two platinum electrodes were immersed in a solution of $CuSO _4$ and electric current was passed through the solution. After some time, it was found that colour of $CuSO _4$ disappeared with evolution of gas at the electrode. The colorless solution contains:

  1. Platinum sulphate

  2. Copper hydroxide

  3. Copper sulphate

  4. Sulphuric acid
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$CuSO _4(aq)\, \xrightarrow{Electrolsis} \, Cu^{2+}(aq)\, +\, SO _4^{2-}(aq)$

At cathode: $Cu^{2+}(aq)\, +\, 2e^-\, \rightarrow\, Cu\, (reduction)$

The blue color of $CuSO _4$ disappears due to the deposition of Cu on Pt electrode.

At anode: $H _2O\, \rightarrow\, 2H^{\oplus}\, +\, 2e^-\, \frac{1}{2} O _2(g)$

Since oxidation potential of $H _2O$ > oxidation potential of $SO _4^{2-}$, so oxidation of $H _2O$ occurs and $O _2(g)$ is evolved at anode.

The colourless solution is due to the formation of $H _2SO _4$ as follows:

$2H^{\oplus}\, (from\, anode)\, +\, SO _4^{2-}\, \rightarrow\, H _2SO _4$

Multiple choice chemistry the s-block elements (alkali and alkaline earth metals) some important compounds of calcium some important compounds of magnesium and calcium compounds of s block elements

$ Y\xleftarrow[]{\Delta,205^{\circ}C} CaSO 4.2H _2O\  \xrightarrow[]{\Delta,120^{\circ}C} X $. 


$X$ and $Y$ are respectively _____.

  1. plaster of paris, dead burnt plaster

  2. dead burnt plaster, plaster of paris
  3. $CaO$, plaster of paris

  4. plaster of paris, mixture of gases
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Reaction of process:


  $CaSO _4 \xleftarrow[]{\Delta,205^{\circ}C} CaSO _4.2H _2O \xrightarrow[]{\Delta,120^{\circ}C} CaSO _4.(1/2)H _2O$                                                                                           (dead burnt plaster)                                          (plaster of paris)
Anhydrous Calcium sulphate is known as dead burnt plaster and $CaSO _4.\cfrac{1}{2} H _2O$ is called plaster of paris.
Hence, the correct option is $A$.

Multiple choice chemistry the s-block elements (alkali and alkaline earth metals) some important compounds of calcium some important compounds of magnesium and calcium compounds of s block elements

In the hardening stage of plaster of paris, the compound formed is :

  1. $CaSO _{4}$
  2. orthorhombic $CaSO _{4} . 2H _{2}O$
  3. $CaSO _{4} . H _{2}O$
  4. monoclinic $CaSO _{4} . 2H _{2}O$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\underset{\text{plaster of paris}}{CaSO _4.1/2H _2O} \xrightarrow[\text{setting}]{H _2O} \underset{\begin{matrix}\text{gypsum}\\text{orthorhombic}\\text{dihydrate}\end{matrix}}{CaSO _4.2H _2O} \xleftarrow[\text{hardening}]{} \underset{\begin{matrix} \text{gypsum} \\text{monoclinic dihydrate}\end{matrix}}{CaSO _4.2H _2O}$
The plaster of paris absorbs water to form orthorhombic calcium sulfate dihydrate which sets to form a hard mass containing monoclinic $CaSO _4.2H _2O$.

Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

Which of the following gives blue flame when burn and is not very soluble in water?

  1. $\displaystyle { O } _{ 2 }$
  2. $\displaystyle { CO } _{ 2 }$
  3. $\displaystyle { N } _{ 2 }$
  4. $\displaystyle { H } _{ 2 }{ SO } _{ 4 }$
  5. $\displaystyle He$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Nitrogen burns with a blue flame and is not very soluble in water.

Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

Which of the following forms a white precipitate when added to a solution of $NaCl$?

  1. $N _2$
  2. $KI$
  3. $CCl _4$
  4. $AgNO _3$
  5. $CaCO _3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Silver nitrate $\displaystyle AgNO _3$ forms a white precipitate of $AgCl$ when added to a solution of $NaCl$. 

$\displaystyle AgNO _3 + NaCl \rightarrow AgCl \uparrow + NaNO _3$

Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

An orange solid (X) on heating, gives a colourless gas (Y) and a only green residue (Z). Gas (Y) on treatment with Mg, produceds a white solid substance................

  1. $Mg _{3}N _{2}$
  2. $MgO$
  3. $Mg _{2}O _{3}$
  4. $MgCl _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$(NH _4) _2Cr _2O _7  \rightarrow N _2 + Cr _2O _3 + 4H _2O$
$3Mg + N _2 \rightarrow Mg _3N _2$
Orange solid is $(NH _4) _2Cr _2O _7$
Colourless gas is $N _2$
Green residue is $Cr _2O _3$

Multiple choice chemistry separation of matter sedimentation, decantation and filtration other methods of separation separation methods

Evaporation of solution of $CuSO _4$ helps in:

  1. making it more concentrated
  2. crystallization of $CuSO _4$
  3. evaporation of salt $CuSO _4$
  4. both $A$ and $B$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

On evaporation of $CuSO _4$ solution, the amount of solvent will decrease and its concentration increases and it will leads to the crystallization of $CuSO _4$.