Chemistry, 20.01.2020 23:31 larkinc2946

Identify all of the following that are chemical changes:
a. leaves turning color in the fall
b. dew forms on a blade of grass
c. ethanol evaporates
d. sugar dissolves in water
e. dry ice sublimes
f. charcoal burning

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Answer from: laskew37221

Answer : The correct options are, (a) and (f)

Explanation :

Chemical change : It is a change where changes occurs in the chemical composition and thus resulting in the formation of new compound or substance. It is an irreversible reaction.

The indication of chemical changes are :

Formation of gas bubbles.

Formation of precipitate.

Change in color.

Change in temperature.

Change in volume.

Change in smell or taste.

Evolution of heat, light and sound.

Physical change : It is a change in which shape and size of a substance changes. In this there is no change in the chemical composition of the substance and no new substance is forming. It is a reversible reaction.

The indication of chemical changes are :

Change in shape.

Change in form.

Change in size.

Change in phase.

(a) Leaves turning color in the fall : It is a chemical change.

(b) Dew forms on a blade of grass : It is a physical change because only changes takes place in the phase from vapor phase to liquid phase.

(c) Ethanol evaporates : It is a physical change because only changes takes place in the phase from liquid phase to vapor phase.

(d) Sugar dissolves in water : It is a physical change because only changes takes place in the phase from solid phase to liquid phase.

(e) Dry ice sublimes : It is a physical change because only changes takes place in the phase from solid phase to vapor phase.

(f) Charcoal burning : It is a chemical change. When charcoal burns, carbon reacts with oxygen to produce carbon dioxide and water as a product.

Hence, the correct options are, (a) and (f)

Answer from: Quest

particle p is an electron and particle q is a neutron

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.