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mojhsa [17]
3 years ago
9

Use the solubility graph to answer the following questions.

Chemistry
1 answer:
Akimi4 [234]3 years ago
4 0

The required amount of KClO₃ to prepare a saturated solution is 48 grams.

<h3>What is solubility?</h3>

The ability of a material, the solute, to create a solution with another substance, the solvent, is known as solubility.

Given graph is plotted between the temperature and solubility of the substance at 100g of water, so for the given points calculation will be:

  • According to the graph for the KClO₃ we require almost 48g of KClO₃ to prepare a saturated solution.
  • From the graph it is clear that at 40°C 40g of KCl will dissolve in 100g of water then amount of KCl which will dissolve in 68g of water will be calculated as:

x = (40)(68) / (100) = 27.2g

  • On the basis of the graph, if 75 grams of calcium chloride is added to 100 g of water at 25°C then the obtained solution is the unsaturated as they have les solute then the limit.
  • Based on the graph cerium sulfate behaves like a precipitate which will not dissolve at any temperature and present in the physical state.
  • We can add more 30g to make the KCl solution saturated.

Hence, we require almost 48g of KClO₃ to prepare a saturated solution of KClO₃.

To know more about solubility, visit the below link:

brainly.com/question/16903071

#SPJ1

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Answer : The enthalpy change for the reaction is, 97.7 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main chemical reaction is,

2C(s)+3H_2O(g)\rightarrow CH_4(g)+CO_2(g)    \Delta H=?

The intermediate balanced chemical reaction will be,

(1) C(s)+H_2O(g)\rightarrow CO(g)+H_2(g)     \Delta H_1=131.3kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=41.2kJ

(3) CH_4(g)+H_2O(g)\rightarrow 2H_2(g)+CO(g)    \Delta H_3=206.1kJ

Now we are multiplying reaction 1 by 2 and reversing reaction 3 and then adding all the equations, we get :

(1) 2C(s)+2H_2O(g)\rightarrow 2CO(g)+2H_2(g)     \Delta H_1=2\times 131.3kJ=262.6kJ

(2) CO(g)+H_2O(g)\rightarrow CO_2(g)+H_2(g)    \Delta H_2=41.2kJ

(3) 2H_2(g)+CO(g)\rightarrow CH_4(g)+H_2O(g)    \Delta H_3=-206.1kJ

The expression for enthalpy of main reaction will be,

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(262.6)+(41.2)+(-206.1)

\Delta H=97.7kJ

Therefore, the enthalpy change for the reaction is, 97.7 kJ

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