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Chapter 6: Problem 9

In a chemical reaction, the total number of atoms present after the reaction is complete is (larger than/smaller than/the same as) the total number of atoms present before the reaction began.

Short Answer

Expert verified
The total number of atoms present after the reaction is complete is \(the\ same\ as\) the total number of atoms present before the reaction began.

Step by step solution

01

Understand the Law of Conservation of Mass

The Law of Conservation of Mass states that in a chemical reaction, the total mass of the reactants is always equal to the total mass of the products. This means that the atoms cannot be created or destroyed during a chemical reaction 鈥� they are merely rearranged to form new substances.
02

Apply the Law to our Exercise

Since no atoms are created or destroyed in a chemical reaction, the total number of atoms present after the reaction must be the same as the total number of atoms before the reaction began. This principle originates from the Law of Conservation of Mass, which ensures that the mass and the number of atoms remain constant in a chemical reaction. So the correct answer is: The total number of atoms present after the reaction is complete is \(the\ same\ as\) the total number of atoms present before the reaction began.

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Most popular questions from this chapter

Balance each of the following chemical equations. a. \(\mathrm{KO}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{KOH}(a q)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) b. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) c. \(\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{PCl}_{5}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(a q)+\mathrm{HCl}(g)\) e. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) f. \(\operatorname{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}_{2}(g)\) g. \(\operatorname{MoS}_{2}(s)+\mathrm{O}_{2}(g) \rightarrow \operatorname{MoO}_{3}(s)+\mathrm{SO}_{2}(g)\)h. \(\operatorname{Fe} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightarrow \mathrm{Fe}\left(\mathrm{HCO}_{3}\right)_{2}(a q)\)

Ozone gas is a form of elemental oxygen containing molecules with three oxygen atoms, \(\mathrm{O}_{3}\). Ozone is produced from atmospheric oxygen gas, \(\mathrm{O}_{2},\) by the highenergy outbursts found in lightning storms. Write the unbalanced equation for the formation of ozone gas from oxygen gas.

Crude gun powders often contain a mixture of potassium nitrate and charcoal (carbon). When such a mixture is heated until reaction occurs, a solid residue of potassium carbonate is produced. The explosive force of the gunpowder comes from the fact that two gases are also produced (carbon monoxide and nitrogen), which increase in volume with great force and speed. Write the unbalanced chemical equation for the process.

Balance each of the following chemical equations. a. \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) b. \(\mathrm{Ag}(s)+\mathrm{H}_{2} \mathrm{S}(g) \rightarrow \mathrm{Ag}_{2} \mathrm{S}(s)+\mathrm{H}_{2}(g)\) c. \(\operatorname{FeO}(s)+\mathrm{C}(s) \rightarrow \operatorname{Fe}(l)+\mathrm{CO}_{2}(g)\) d. \(\quad \mathrm{Cl}_{2}(g)+\mathrm{KI}(a q) \rightarrow \mathrm{KCl}(a q)+\mathrm{I}_{2}(s)\) e. \(\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow\) \(\mathrm{H}_{3} \mathrm{BO}_{3}(s)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) f. \(\mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)\) g. \(\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \rightarrow \mathrm{HCl}(g)+\mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) h. \(\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \rightarrow \mathrm{Si}(l)+\mathrm{CO}(g)\)

When hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{S},\) gas is bubbled through a solution of lead(II) nitrate, \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2},\) a black precipitate of lead(II) sulfide, PbS, forms, and nitric acid, \(\mathrm{HNO}_{3},\) is produced. Write the unbalanced chemical equation for this reaction.
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