Chemistry, 25.05.2021 06:20 lpslover26237

You will need to copy out the equation TWICE. MAP the first copy of the equation against titanium (Ti) and the second copy against oxygen gas (O2). Then create a PAIR of stoichiometric setups which predict the THEORETICAL YIELD of titanium oxide (Ti2O3) from each reactant. 7. Based on your answer to question 6, is titanium (Ti) or oxygen gas (O2)
LIMITING? How many grams of the other reactant will remain in EXCESS?

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Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

Answer from: Quest

Bis the

Answer from: Quest

Standing in the sunlight and feeling the heat on your face (uv radiation is entering the skin, converting to infrared radiation, and is absorbed by your tissues)orstanding near a fire and feeling the heat on your skin (infrared radiation transfer)

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.

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You will need to copy out the equation TWICE. MAP the first copy of the equation against titanium (T...

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