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Natali5045456 [20]
2 years ago
14

Chlorine gas reacts with solid phosphorus to form phosphorus trichloride gas. Write a balanced chemical equation for this reacti

on.​
Chemistry
1 answer:
Oduvanchick [21]2 years ago
7 0

Explanation:

First write down the basic chemical equation with the reactants and products,

\text{P}_4+\text{ Cl}_2\to\text{ PCl}_3

Balancing the chemical equation*,

\text{P}_4 (s) + 6 \text{Cl}_2 (g) \to 4 \text{PCl}_3 (l)

\color{blue}\text{Note: To balance any chemical equations, we should ensure the number of atoms of each elements are the same on both the product and reactant side.}

\color{blue}\text{The phosphorous molecule has 4 P, but the }\text{PCl}_3\text{ molecule on the product side only has one P atom. Therefore, we need 4 }\text{PCl}_3\text{ molecules to balance.}

\color{blue}\text{But that will mean we have }4\cdot 3=12\text{Cl atoms on the product side. So we need 6 }\text{Cl}_2 \text{ molecules on the reactant side.}

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Answer:

<em>A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being</em> <u>supersaturaded</u>

Explanation:

This question is about solubility.

Regarding solubility, the solutions may be classified as:

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  • Saturated: the concentration is the maximum permitted at the given temperature, under normal conditions.

  • Supersaturated: the concentration has overcome the maximum permitted at the given temperature. This is possible only under special conditions and is a very unstable state.

Each substance has its own, unique solubility properties. So, in order to tell the state of the solution you need to compare with either solubility tables, or solubility curves; or run you own experiments.

  • In internet you can find the solubility curve of NaNO₃ showing the solubility for a wide range of temperatures.

  • In such curve the solubility of NaNO₃ at 50°C is about 115 g of NaNO₃ per 100 g  of water.

  • Hence, do the proportion to determine the amount of solute that can be dissolved in 50 grams of water at 50°CÑ

       115 g NaNO₃ / 100 g H₂O = x / 50 g H₂O  ⇒ x =  57.5 g NaNO₃

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Answer:

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