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lana66690 [7]
3 years ago
7

Using the balanced reaction below, drag and drop the terms into the correct location to solve the following problem:

Chemistry
1 answer:
k0ka [10]3 years ago
6 0

Answer:

4.5 moles of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

C2H5OH + 3O2 —> 2CO2 + 3H2O

From the balanced equation above,

1 moles of C2H5OH reacted to produce 3 moles of H2O.

Finally, we shall determine the number of mole of H2O produced by the reaction of 1.5 moles of C2H5OH. This can be obtained as follow:

From the balanced equation above,

1 moles of C2H5OH reacted to produce 3 moles of H2O.

Therefore, 1.5 moles of C2H5OH will react to produce = 1.5 × 3/ 1 = 4.5 moles of H2O.

Thus, 4.5 moles of H2O will be produced.

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Answer:

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Explanation:

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In Part III, the phenolphthalein indicator is used to monitor the equilibrium shifts of the ammonia/ammonium ion system. The phe
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The pink color in the solution fades. Some of the colored indicator ion converts to the colorless indicator molecule.

<h3>Explanation</h3>

What's the initial color of the solution?

\text{NH}_4\text{Cl} is a salt soluble in water. \text{NH}_4\text{Cl} dissociates into ions completely when dissolved.

\text{NH}_4\text{Cl} \; (aq)\to {\text{NH}_4}^{+} \; (aq) +{\text{Cl}}^{-} \; (aq).

The first test tube used to contain \text{NH}_4\text{OH}. \text{NH}_4\text{OH} is a weak base that dissociates partially in water.

\text{NH}_4\text{OH} \; (aq) \rightleftharpoons {\text{NH}_4}^{+}  \;(aq)+ {\text{OH}}^{-} \; (aq).

There's also an equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l).

\text{OH}^{-} ions from \text{NH}_4\text{OH} will shift the equilibrium between \text{OH}^{-} and {\text{H}_3\text{O}}^{+} to the right and reduce the amount of {\text{H}_3\text{O}}^{+} in the solution.

The indicator equilibrium will shift to the right to produce more {\text{H}_3\text{O}}^{+} ions along with the colored indicator ions. The solution will show a pink color.

What's the color of the solution after adding NH₄Cl?

Adding \text{NH}_4\text{Cl} will add to the concentration of {\text{NH}_4}^{+} ions in the solution. Some of the {\text{NH}_4}^{+} ions will combine with \text{OH}^{-} ions to produce \text{NH}_4\text{OH}.

The equilibrium between  \text{OH}^{-} and {\text{H}_3\text{O}}^{+} ions will shift to the left to produce more of both ions.

{\text{OH}}^{-}\;(aq) + {\text{H}_3\text{O}}^{+} \;(aq) \to 2\; \text{H}_2\text{O} \;(l)

The indicator equilibrium will shift to the left as the concentration of {\text{H}_3\text{O}}^{+} increases. There will be less colored ions and more colorless molecules in the test tube. The pink color will fade.

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What is the osmotic pressure of a solution made from 22.3 g of methanol (MM = 32.04 g/mol) that was added to water to make 321 m
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Answer: The osmotic pressure of a solution is 53.05 atm

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT

where,

\pi = osmotic pressure of the solution = ?

i = Van't hoff factor = 1 (for non-electrolytes)

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Volume of solution = 321 mL

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

T = temperature of the solution = 25^oC=[273+25]=298K

Putting values in above equation, we get:

\pi=1\times \frac{22.3\times 1000}{32.04\times 321}\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K

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