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Chemical Reactions and Equations Test - 22

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Chemical Reactions and Equations Test - 22
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  • Question 1
    1 / -0
    Which of the following is not an example of single displacement reaction?
    Solution
    $$\displaystyle AgNO_3+NaCl\rightarrow AgCl+NaNO_3$$  is not an example of single displacement reaction. 

    It is an example of a double displacement reaction. It is a chemical process involving the exchange of bonds between two non-reacting chemical species.
    Hence, option $$\text{C}$$ is correct.
  • Question 2
    1 / -0
    A Science teacher wrote $$3$$ statements about rancidity:
    (i) When fats and oils are reduced, they become rancid.
    (ii) In chips packet, rancidity is prevented by oxygen.
    (iii) Rancidity is prevented by adding antioxidants.

    Select the correct option.
    Solution
    The oils and fats are slowly oxidised to certain bad smelling compounds which release a foul smell. This is known as Rancidity.

    To prevention of rancidity:
    (1) By adding anti-oxidants.
    (2) In the chips packet, it is prevented by Nitrogen gas.
  • Question 3
    1 / -0
    $$BaCl_{2}(aq)+Na_2SO_4(aq)\rightarrow BaSO_{4}(s)+2NaCl (aq)$$
    Above reaction involves which type of reaction :
    (a) Displacement
    (b) Precipitation
    (c) Combination
    (d) Double displacement
    Solution
    $$\displaystyle BaCl_{2}(aq)+Na_2SO_4(aq)\rightarrow BaSO_{4}(s)+2NaCl (aq)$$ 

    The types of reactions are (b) Precipitation and (d) Double displacement. 

    A precipitate of $$\displaystyle Barium \: sulphate $$ is obtained. 

    In double displacement reaction, two compounds react, and the positive ions (cation) and the negative ions (anion) of the two reactants switch places, forming two new compounds or products.

    So, the correct option is $$D$$.
  • Question 4
    1 / -0

    Directions For Questions

    Fill in the blanks with the given word box.
    bubbles, reaction, gases, condensation, colour, reactants, reversed, word, element, compound, products, boiling.

    ...view full instructions

    The substances you start with are called ______ and after the chemical change, what is formed is called the _______.
    Solution
    A reactant is a substance that is present at the start of the chemical reaction or substances present to the left of the arrow in a chemical equation. A product is a substance that is present at the end of a chemical reaction or substance to the right of the arrow is called products.

    For example,
    $$Zn + S \rightarrow ZnS$$

    In the above equation, zinc and sulfur are reactants that combine chemically to form the product zinc sulfide.
  • Question 5
    1 / -0
    Study the given figure carefully.

    Which of the following reactions explains the above change most appropriately?

    Solution
    The blue coloured solution is $$CuSO_4$$ solution. Iron $$(Fe)$$ is more reactive than copper $$(Cu)$$. So, $$Fe$$ displaces $$Cu$$ from the $$CuSO_4$$ solution and forms $$FeSO_4$$ solution which is green in color. Brown-red deposition in a test tube is of copper. Such types of reactions are called displacement reactions.
  • Question 6
    1 / -0
    Name the type of following chemical reaction.
    $$2H_2O_2 \rightarrow 2H_2O + O_2$$
    Solution
    A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more elements or new compounds. These reactions often involve an energy source such as heat, light, or electricity that breaks apart the bonds of compounds.
    $$2H_2O_2 \rightarrow 2H_2O + O_2$$
  • Question 7
    1 / -0
    What happens when a piece of zinc metal is added to copper sulphate solution?
    Solution
    When zinc metal is added to a solution of copper sulphate ($$CuSO_{4}$$), the blue colour of the latter changes to colourless as zinc sulphate ($$ZnSO_{4}$$) is formed. 

    In this displacement reaction, zinc being more active metal than copper displaces it to form zinc sulphate, as shown below:

    $$ Zn(s)  +  CuSO_{4}(aq)  \rightarrow   ZnSO_{4}(aq)  +  Cu(s)$$
                     (blue)               (colourless)
  • Question 8
    1 / -0
    In which of the following chemical equations, the abbreviations represent the correct states of the reactants and products involved at reaction temperature?
    Solution
    Hydrogen and oxygen are correctly mentioned in the gaseous state and the product water in a liquid state.

    2$$H_{2}(g)    +    O_{2}(g)     \rightarrow     2H_{2}O(l)$$

    Hence, option C is correct.
  • Question 9
    1 / -0
    Which among the following is(are) double displacement reaction(s)?
    (i) $$Pb+CuCl_{2} \rightarrow PbCl_{2}+ Cu$$
    (ii) $$Na_{2}SO_{4} + BaCl_{2} \rightarrow  BaSO_{4} + 2NaCl$$
    (iii) $$C + O_{2} \rightarrow CO_{2}$$
    (iv) $$CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$$
    Solution
    Double displacement reactions are those in which ions of the reactants are exchanged to form new compounds as products. 
    In option $$(ii)$$, 
    $$Na^{+}$$ion combines with $$Cl^{-}$$ ion and $$SO_{4}^{2-}$$ ion combines with $$Ba^{2+}$$ ion to form two new compounds as sodium chloride ($$NaCl$$) and barium sulphate ($$BaSO_{4}$$) respectively.
  • Question 10
    1 / -0
    In the double displacement reaction between aqueous potassium iodide and aqueous lead nitrate, a yellow precipitate of lead iodide is formed. While performing the activity if lead nitrate is not available, which of the following can be used in place of lead nitrate?
    Solution
    $$2KI(aq)  +  Pb(NO_{3})_{2}(aq)  \rightarrow   PbI_{2}(s)  +  2KNO_{3}$$
    In the above double displacement reaction, potassium Iodide(KI) and lead nitrate ($$Pb(NO_{3})_{2}$$) dissociate in their aqueous states to form ions. The lead ($$Pb^{2+}$$) ions combine with the iodide ($$I^{-}$$) ions to form precipitates of lead iodide ($$PbI_{2}$$). If Lead sulphate is used in place of lead nitrate, no precipitates of lead iodide will be formed, because lead sulphate being insoluble in water does not give $$Pb^{2+}$$ ions which can combine with $$I^{-}$$ to form $$ PbI_{2}$$. On the other hand, lead acetate dissociates in aqueous state to give $$Pb^{2+}$$ ions and $$CH_{3}COO^{-}$$ ions . Therefore potassium iodide combines with lead acetate to form precipitates of lead iodide and potassium acetate.
    $$2KI(aq)  +  Pb(CH_{3}COO)_{2}(aq)  \rightarrow   PbI_{2}(s)  +  2CH_{3}COOK$$
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