68 g of KClO3 must be heated to produce as much O2 as would be obtained from 50 g of HgO.
Crystals of sugar are obtained.
The reaction between H2SO3 and KClO3 does not occur. H2SO3 is a weak acid, and KClO3 is a salt. There is no precipitation, gas evolution, or significant heat release that would indicate a chemical reaction taking place between these two substances.
6g
It would evaporate
I dont
2 grams of Oxygen can be obtained from 5 grams of KClO3 (only if the "CL" means "Cl", which is Chlorine! Remember that only the first letter of the atomic symbol is capitalized.)
2KClO3 --> 2KCl + 3O2For every 3 moles of oxygen gas produced, 2 moles of potassium chlorate are used.6 moles O2 * (2 moles KClO3 reacted / 3 moles O2 produced) = 4 moles KClO3
Crystals of sugar are obtained.
The reaction between H2SO3 and KClO3 does not occur. H2SO3 is a weak acid, and KClO3 is a salt. There is no precipitation, gas evolution, or significant heat release that would indicate a chemical reaction taking place between these two substances.
no. what would give you that idea?
A 1.80-gram mixture of potassium chlorate, kclo3, and potassium chloride, kcl, was heated until all of the kclo3 had decomposed the liberated oxygen, after drying, occupied 405 ml at 25C when the barometric pressure was 745 torr. This is the problem and the questions were... a. How many moles of O2 were produced? b. What percent of the mixture was KClO3? KCl? Please help!!
2 to 3, because of the balanced equation:2 KClO3 --> 2 KCl + 3 O2
It would produce a purple colour.
Francium is a highly reactive alkali metal that would produce a bluish glow when heated. However, due to its extreme rarity and radioactivity, it has not been studied extensively enough to confirm the exact color of flame it would produce.
Alls you do to find a molar mass is add up all of the atomic masses. Potassium=39.09 Chlorine= 35.453 Oxygen(3)=15.999. So KClO3 would equal 271.65g to a mol? Then I think to find the number of atoms of each you would take the atomic mass * avacodo's number so it would be for example chlorine. 1gCl(35.453/1)(6.022*10^23/1) However I might be wrong on that part.
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2.52 L