Which of the following techniques is most appropriate for the recovery of solid KNO3 from an aqueous solution of KNO3?

Nitrogen has a normal boiling point of 77.4 K and a melting point (at 1 atm) of 63.2 K. Its critical temperature is 126.2 K, and its critical pressure is 2.55 * 104 torr. It has a triple point at 63.1 K and 94.0 torr.

<h3>What is the triple point?</h3>

The temperature and pressure at which the solid, liquid, and vapour phases of a pure substance can coexist in equilibrium.

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).

- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.

- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.

- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.

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Which one of the following statements BEST explains what is meant by the “dual nature” of the electron? A : An electron may be e

erica [24]

D. An electron may acr with either particle like or wave like

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3 years ago

Read 2 more answers

Consider the following reaction:

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Answer:

Question 1: Both the rate of reaction of H₂O₂ and the rate of formation of H₂O are:

$6.6\cdot 10^{-3}mol/(liter.s)$

Question 2: 0.60 moles of O₂ are formed in the first 50 s of reaction.

Explanation:

The coefficients of the balanced chemical equation tells the relation of formation of the reactants into the products, which is also the relation between their rates of reaction.

The balanced chemical equation (given) is:

$2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)$

Then, in the same time that 2 moles of H₂O₂ react, 2 moles of H₂O and 1 mol of O₂ are formed.

1. Question 1.

That means that the rate of formation of O₂ is half the rate of reaction of H₂O₂ and half the rate of formation of H₂O.

Hence, if the instantaneous rate of formation of O₂ is 3.3×10⁻³moles/(liters×seconds), then the rate of reaction of H₂O₂ and the rate of formation of H₂O is twice:

$2\times 3.3\cdot 10^{-3}mol/(liter.s)=6.6\cdot 10^{-3}mol/(liter.s)$

2. Question 2.

First you must find how much the concentration has changed.

You can read the initial concentration of H₂O₂ on the graph. It is the y-intercept. It is 1.0M.

You can read the approixmate concentration of H₂O₂ at the time 50 s. It is about 0.2M.

Then the change in concentration in the first 50 s of reaction is about 1.0M - 0.2M = 0.8M.

Now you can convert the concentration 0.8M into number of moles, using the volume (1.5 liters), assuming the liquid volume does not change, and the molarity definition (molarity is the number of moles of solute in one liter of solution)

$\Delta [H_2O_2]=\Delta moles/Volume(liters)$

$0.8M=\Delta moles/1.5liter\\\\\Delta moles=0.8M\times 1.5liter=1.2mol$

Hence, 1.2 mol of H₂O₂ have reacted in the first 50s of reaction.

From the coefficients of the chemical equation, the number of moles of O₂ formed is half the number of moles of H₂O₂ that react.

Thus, 1.2mol/2 = 0.60 moles of O₂ are formed in the first 50 s of reaction.

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3 years ago