Chemistry, 25.09.2019 06:30 ohshushan2

In a balanced chemical reaction, the total mass of the products always equals the

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Answer from: kingkuka1700

mass of reactants

Explanation: According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

Reactants of a chemical reaction is defined as the substances which take part in a chemical reaction and undergoes a chemical change. Reactants are written on left side of the arrow.Products are defined as the substance which are formed at the end of the chemical reaction.Products are written on the right side of the arrow.

For Example: $C+O_2\rightarrow CO_2$

Mass of Carbon = 12 g

Mas of Oxygen = 32 g

Mass of Carbon dioxide = 12+2(16)= 44 g

Thus mass of products is equal to mass of reactants in a balanced chemical equation.

Answer from: Quest

< > according to newton's third law of motion, forces always act in equal but opposite pairs. another way of saying this is for every action, there is an equal but opposite reaction.< >

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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