Here we have to calculate the heat required to raise the temperature of water from 85.0 ⁰F to 50.4 ⁰F.
10.857 kJ heat will be needed to raise the temperature from 50.4 ⁰F to 85.0 ⁰F
The amount of heat required to raise the temperature can be obtained from the equation H = m×s×(t₂-t₁).
Where H = Heat, s =specific gravity = 4.184 J/g.⁰C, m = mass = 135.0 g, t₁ (initial temperature) = 50.4 ⁰F or 10.222 ⁰C and t₂ (final temperature) = 85.0⁰F or 29.444 ⁰C.
On plugging the values we get:
H = 135.0 g × 4.184 J/g.⁰C×(29.444 - 10.222) ⁰C
Or, H = 10857.354 J or 10.857 kJ.
Thus 10857.354 J or 10.857 kJ heat will be needed to raise the temperature.
The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1.
<span>Answer: option (1) solubility of the solution increases.
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<span>Justification:
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<span>The solubility of substances in a given solvent is temperature dependent.
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<span>The most common behavior of the solubility of salts in water is that the solubiilty increases as the temperature increase.
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<span>To predict with certainty the solubility at different temperatures you need the product solubility constants (Kps), which is a constant of equlibrium of the dissolution of a ionic compound slightly soluble in water, or a chart (usually experimental chart) showing the solubilities at different temperatures.
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<span>KClO₃ is a highly soluble in water, so you do not work with Kps.
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<span>You need the solubility chart or just assume that it has the normal behavior of the most common salts. You might know from ordinary experience that you can dissolve more sodium chloride (table salt) in water when the water is hot. That is the same with KClO₃.
</span><span>The solubility chart of KlO₃ is almost a straight line (slightly curved upward), with positive slope (ascending from left to right) meaning that the higher the temperature the more the amount of salt that can be dissolved.</span>
<span>136.14 g/mol </span><span><span>Calcium sulfate, Molar mass</span></span>
