Chemistry, 30.08.2019 01:30 janeou17xn

Plastic, rocks, and steel are all matter. they are made of tiny particles that can’t be broken down into smaller pieces. these particles are called:

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Answer from: Aerial14

The tiny unbreakable particles which are the building blocks of plastic, rocks and steel are termed as atom.

Explanation:

It is known that every matter is built by atoms bonded together in different structure. The atoms are present in solids, liquids and gases, the only difference is their binding strength and bond length.

The solids experience the highest binding strength while gases experience the lowest binding strength. The solids are also classified in different categories based on the arrangement of atoms to build the specified solid.

In this case, plastics are made of atoms arranged in polymer chain or network like structure, stone has different bond type between the atoms used for building of it, the same case goes for steel, the bond type.

And the arrangement of atoms are having different crystal structure. But the main building blocks or the tiny particles which cannot be broken down into smaller pieces are termed as atoms.

Answer from: Quest

Answerit curves up and levels out.explanationwe assume that the container contains absolutely no ammonia to begin with as it states that "nitrogen gas and hydrogen gas" react to "produce ammonia."at dynamic equilibrium, the product and reactant concentrations are equal.this means that, starting from a concentration of zero, [nh₃] must curve "up" from zero and then level out to a constant concentration.

Answer from: Quest

Answerthe half-reaction having the greatest tendency to occur at the anode: i₂(aq) + 2e⁻ ⇄ 2i⁻ ; e = +0.53 v====== $\begin{aligned} \text{i}_2\text{\footnotesize (aq)}+2\,\text{e}^{-} & \rightleftharpoons 2\,\text{i}^{-}\text{\footnotesize (aq)}& \text{e} & = +0.53\text{ v} \\ \text{co}^{3+}\text{\footnotesize (aq)} + \text{e}^{-} & \rightleftharpoons \text{co}^{2+}\text{\footnotesize (aq)} & \text{e} & = +1.80\text{ v} \\ \text{o}_2\text{\footnotesize (g)}+4\,\text{h}^{+}\text{\footnotesize (aq)} + 4\,\text{e}^{-} & \rightleftharpoons 2\,\text{h}_2\text{o}\text{\footnotesize(l)}& \text{e} & = +1.23\text{ v} \\ \end{aligned}$ oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be i₂(aq) + 2e⁻ ⇄ 2i⁻ e = +0.53 vbecause its reduction potential is the lowest out of the three.