Cooling curve is the plot of temperature versus time as the sample is allowed to cool. In a cooling curve, we start at a temperature greater than the boiling point. At this temperature, the sample is in gaseous state. At the boiling point, there is no change in temperature as the gaseous and liquid states are in equilibrium. As the temperature reduces further, the liquid starts to condense and at the melting point of the sample the liquid undergoes phase transition to solid state. At the melting temperature, a second plateau is observed as the temperature remains unchanged. At temperatures below the melting point, the sample exists as a solid.

So from the curve, the second plateau is observed at around -111 $^{0}C$ . This point represents the phase transition from liquid to solid state.

6 0

3 years ago

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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.

8 0

2 years ago

What is the maximum amount of ammonia that can be produced when 0.35 mol Nz reacts with 0.90 mol Hz?

professor190 [17]

Answer:

0.6 moles NH₃

Explanation:

The reaction that takes place is:

N₂ + 3H₂ → 2NH₃

First we determine the limiting reactant:

0.35 mol N₂ would react completely with (3*0.35) 1.05 moles of H₂. There are not as many H₂ moles, so H₂ is the limiting reactant.

Then we convert H₂ moles (the limiting reactant) to NH₃ moles, keeping in mind the stoichiometry of the reaction:

0.90 mol H₂ * $\frac{2molNH_3}{3molH_2}$ = 0.6 moles NH₃

5 0

3 years ago