Answer:
ΔH rx = -43.5 kJ / mol
Explanation:
In water, Xdissolves thus:
X(s) + H₂O(l) → X(aq) + H₂O(aq)
It is possible to find the heat in dissolution process using coffee cup calorimeter equation:
Q = -m×C×ΔT
<em>Where Q is heat, m is mass of solution (35.0g -density 1g/mL- + 2.20g = 37.2g), C is specific heat of solution (4.18J/g°C), and ΔT is change in temperature (26.0°C-15.0°C = 11.0°C)</em>
Replacing:
Q = -37.2g×4.18J/g°C×11.0°C
Q = -1710J = -<em>1.71kJ</em>
As enthalpy is the change in heat per mole of reaction, moles of X that reacted were:
2.20g X × (1mol / 56.0g) = <em>0.0393 moles</em>
As heat produced per 0.0393moles was -1.71kJ, heat per mole of X is:
-1.71kJ / 0.0393mol = -<em>43.5 kJ / mol = ΔH rx</em>
Explanation:
protons.
An acid can also be thought of as a chemical that can neutralize a base. Similarly, a base can neutralize an acid.
Acids turn litmus paper red, while bases make litmus paper turn blue.
Some examples of acids are sulfuric acid, hydrochloric acid, nitric acid, and so on. Some examples of bases are sodium hydroxide, potassium hydroxide, and so on.
Acids generally taste sour, while bases have a bitter taste.
Alkalis are the bases that are water-soluble, which means that they dissolve in water. In other words, not all bases are water-soluble, and only the water-soluble bases are known as alkalis. An example of an alkali is sodium hydroxide. It is a base because it can neutralize an acid, and because it is water-soluble, it is an alkali. An example of a base that is not alkali is copper oxide. This chemical can neutralize an acid, but it is insoluble in water.
In other words, all alkali are bases but not all bases are alkalis.
Also, an alkali has a hydroxide group, while a base has an oxide group in it.
The solution to your problem is as follows:
<span>2 KClO3 → 2 KCl + 3 O2
</span><span>MW of KCL = 39.1 + 35.35 = 74.6 g/mole
62.6/74.6 = 0.839 moles KCl produced
0.839 moles KCl x 3O2/2KCl x 32g/mole O2 = 40.27g of O2 was produced
</span>
Therefore, there are <span>40.27g of O2 produced during the reaction.
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