<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>
The balanced equation for the above reaction is as follows;
3NO₂ + H₂O --> 2HNO₃ + NO
stoichiometry of NO₂ to NO is 3:1
molar volume is where 1 mol of any gas occupies a volume of 22.4 L
volume of gas is directly proportional to number of moles of gas.
therefore stoichiometry can be applied for volume as well.
volume ratio of NO₂ to NO is 3:1
volume of NO₂ reacted - 854 L
therefore volume of NO formed - 854 L /3 = 285 L
volume of NO formed - 285 L
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask.
1mmol = 10^-3 mol
Therefore 4.10*10^-5mmol = 4.10*10^-8mol
molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol
You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below)
But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g
Mass is = 9.75*10^-7 grams
1µg = 10^-6 g
You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4
(*see below) at this point you could have said:
1µg = 10^-6 g therefore you have a solution of 6.29µg per litre,
155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
<span>Many scientific investigations have provided evidence to support this as the best explanation of the data</span>
