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Maharashtra SSC (Class 10) Unit 4A1: Types of Chemical Reactions — practice questions

Explore 13 free Unit 4A1: Types of Chemical Reactions questions in Maharashtra SSC (Class 10) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Unit 4A1: Types of Chemical Reactions questions and exercises with answers

Showing questions 1–13 of 13.

In the chemical equation CuSO$_4$(aq) + Zn(s) $\longrightarrow$ ZnSO$_4$(aq) + Cu(s), which substances are the reactants?Read the answer and worked solution →In a chemical equation, the symbol $\Delta$ written above the reaction arrow indicates that:Read the answer and worked solution →Balance the equation: NaOH + H$_2$SO$_4$ $\longrightarrow$ Na$_2$SO$_4$ + H$_2$O. The correct set of coefficients (NaOH, H$_2$SO$_4$, Na$_2$SO$_4$, H$_2$O) is:Read the answer and worked solution →The reaction 2Mg + O$_2$ $\longrightarrow$ 2MgO, in which magnesium burns in air to form a white powder, is an example of a:Read the answer and worked solution →Complete the displacement reaction: CuSO$_4$(aq) + Fe(s) $\longrightarrow$ ___ + ___Read the answer and worked solution →A process or reaction in which heat is absorbed from the surroundings is called:Read the answer and worked solution →When solid sodium hydroxide is dissolved in water the temperature of the solution rises. This dissolution is therefore a/an:Read the answer and worked solution →Consider the reaction CaCO$_3$(s) + heat $\longrightarrow$ CaO(s) + CO$_2$(g). Heat must be supplied for it to proceed. This reaction is:Read the answer and worked solution →Balance the decomposition of potassium chlorate: KClO$_3$ $\xrightarrow{\Delta}$ KCl + O$_2$. The correct coefficients (KClO$_3$, KCl, O$_2$) are:Read the answer and worked solution →According to the chapter, which definition of oxidation is correct?Read the answer and worked solution →In the reaction CuO + H$_2$ $\longrightarrow$ Cu + H$_2$O, which statement is correct?Read the answer and worked solution →The damage to metals caused by oxidation due to various atmospheric components is generally called:Read the answer and worked solution →Old or leftover cooking oil develops a foul odour and changed taste when left in air for a long time. This air-oxidation of oils and fats in food is called:Read the answer and worked solution →

Worked practice questions

In the chemical equation CuSO$_4$(aq) + Zn(s) $\longrightarrow$ ZnSO$_4$(aq) + Cu(s), which substances are the reactants?

  1. ZnSO$_4$ and Cu
  2. CuSO$_4$ and Cu
  3. ZnSO$_4$ and Zn
  4. CuSO$_4$ and Zn
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Correct answer: D. CuSO$_4$ and Zn

1. Reactants are the substances written on the left side of the arrow. 2. The arrow points towards the products, indicating the direction of the reaction. 3. Here the left side has CuSO$_4$(aq) and Zn(s), so these are the reactants. 4. ZnSO$_4$ and Cu, written on the right side, are the products. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 3 "Chemical Reactions and Equations", p.41_

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In a chemical equation, the symbol $\Delta$ written above the reaction arrow indicates that:

  1. the product is a gas
  2. the product is a precipitate
  3. heat is supplied from outside
  4. a catalyst is being used
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Correct answer: C. heat is supplied from outside

1. Certain reactions need heat to be supplied from outside to take place. 2. This is shown by the sign $\Delta$ written above the arrow that indicates the direction of the reaction. 3. A catalyst or temperature condition is written below or above the arrow as text, not as $\Delta$. 4. A gaseous product is shown by $\uparrow$ and a precipitate by $\downarrow$, so those options are wrong. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 3 "Chemical Reactions and Equations", p.41_

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Balance the equation: NaOH + H$_2$SO$_4$ $\longrightarrow$ Na$_2$SO$_4$ + H$_2$O. The correct set of coefficients (NaOH, H$_2$SO$_4$, Na$_2$SO$_4$, H$_2$O) is:

  1. 2, 2, 1, 2
  2. 1, 2, 1, 2
  3. 1, 1, 1, 1
  4. 2, 1, 1, 2
Show answer and solution

Correct answer: D. 2, 1, 1, 2

1. Start with Na: the product Na$_2$SO$_4$ has 2 Na, so place a factor 2 before NaOH. 2. Equation becomes 2NaOH + H$_2$SO$_4$ $\longrightarrow$ Na$_2$SO$_4$ + H$_2$O. 3. Now count H on the left: 2 (from 2NaOH) + 2 (from H$_2$SO$_4$) = 4; place factor 2 before H$_2$O to give 4 H. 4. Check O: left = 2 + 4 = 6, right = 4 (Na$_2$SO$_4$) + 2 (2H$_2$O) = 6. S = 1 each side. Balanced as 2, 1, 1, 2. 5. The set 1,1,1,1 leaves Na and H unequal; the sets beginning 2,2 and 1,2 over-count and unbalance the equation. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 3 "Chemical Reactions and Equations", p.44_

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