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rusak2 [61]
2 years ago
11

Phosphoric acid is neutralized by potassium hydroxide according to the following reaction:

Chemistry
1 answer:
rusak2 [61]2 years ago
5 0
Balance the reaction first:

3KOH + H3PO4 —> K3PO4 + 3H2O

So for every mol of H3PO4, you need 3 mol of OH- to fully neutralize the acid, since H3PO4 is polyprotic.

0.0200 L KOH • (2.000 mol KOH / L KOH) • (1 mol H3PO4 / 3 mol KOH) = 0.0133 mol H3PO4

Divide this by the volume of H3PO4 to get the concentration.

0.0133 mol H3PO4 / 0.0250 L = 0.532 M H3PO4
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The individual particles in an ionic solid are held together as a result of:
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electrostatic force

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Ionic bonds happened between?
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\large \boxed{\sf metal \ and \ non \ metal}

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2 years ago
A solution containing 60 grams of nano3 completely dissolved in 50. Grams of water at 50°c is classified as being
Igoryamba

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

  • <u>Conclusion</u>: 50 grams of water can contain 57.5 g of NaNO₃ dissolved; so, <em>a solution containing 60 g of NaNO₃ completely dissolved in 50 grams of water is supersaturated.</em>

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3 years ago
A b c d e f g h i j k l m n o p q r s t u v w x y z
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It’s B. Substitution hope this helps
6 0
3 years ago
Read 2 more answers
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