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nadezda [96]
4 years ago
13

When 2,4-hexadiene is protonated, a resonance-stabilized secondary allylic carbocation is formed. Draw the curved arrow notation

below that shows the movement of electrons between the two major resonance structures.

Chemistry
1 answer:
NemiM [27]4 years ago
4 0

Answer:see the image for the structure of the allylic carbocation

Explanation:

When 2,4-hexanediene is protonated at the 5-position as shown in the image, a carbonation is formed which can be stabilized by resonance. The positive charge can be found on carbon 4 or carbon 2 due to resonance of the allylic carbonation. The both structures are shown and electron movements were depicted using curved arrow notation in the structures shown in the image attached.

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A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
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Approximately 75%.

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There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

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Calculate the mass ratio of Ca in a pure sample of CaCO₃:

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Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

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