Chemistry, 13.02.2022 04:40, dayanaraa61

How many moles are in 100.0g of each of the following compounds? a. dinitrogen oxide (N2O)
b. methanol (CH3OH)

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answered: potternatalie90

a. dinitrogen oxide (N2O)

O=16.00g

N=14.00g * 2 = 28.00g

So NO2 = 44.00g

You have 100g of NO2, you need moles.

So:

100g of NO2 x 1 mol NO2/44g of NO2 = your answer

b. All right. So let's go ahead and convert our 100 grams of stuff. Two moles. So starting with, we have 100 grams, uh, and 20 times we know we're gonna have one more on the top. We need the molar mass. So if we look at our periodic table, nitrogen has a molar mass of 14 point. Oh, sevens will multiply that by two. Since we have two nitrogen in our compound plus 15.999 your period table might be slightly different. That's okay. Just stay consistent with your numbers. If we add those together, we should get 44 point no. 13 grams for everyone. Mole. That's why it's ah conversion factor of the inner calculator and get 2.27 moles into right. And then, um, we have 100 grams of other substance teach your each times we know our one mole from the conversion factor will be on the tops. Wouldn't do that now and then. Carbon 12.1 Um, options. 15.99 and then hydrogen is one point 08 So I had all those together multiply by our sub scripts and get 32 point go for two calculate. That would be 3.12 moles of C H 30 H for the same mass. What this means is that CH. 30 H husband It contains more moles than in 20

answered: Guest

yes the answer is total mass in unchanged : )

answered: Guest

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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How many moles are in 100.0g of each of the following compounds? a. dinitrogen oxide (N2O)
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