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3 years ago

What is the number of carbon atoms in the ring portion of the haworth structure of glucose?

Chemistry

1 answer:

Anna35 [415]3 years ago

8 0

I have attached the picture ot the H<span>aworth structure of glucose.

You can count in total 6 carbon atoms but only 5 are in the ring portion of the structure.

Therefore, the answer is 5.
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3 years ago

Read 2 more answers

In the equation below

yulyashka [42]

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

$\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)} + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}$

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)

The ratio between NH₃ and Cu is 2:3.
The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

The amount of N₂ produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}$

The amount of Cu produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}$

And the amount of H₂O produced:

$\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}$

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

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