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
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Therefore, there are <span>40.27g of O2 produced during the reaction.
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They speed up reactions by lowering activation energy. Many enzymes change shape when substrates bind.
Answer:
Bases (solutions with a high pH) feel slipper, have an -OH group, and are corrosive.
What do you want us to work out?
Answer:
86.3 g of N₂ are in the room
Explanation:
First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.
We apply the mole fraction:
Mole fraction N₂ = N₂ pressure / Total pressure
0.78 . 1 atm = 0.78 atm → N₂ pressure
Room temperature → 20°C → 20°C + 273 = 293K
Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K
(0.78 atm . 95L) /0.082 . 293K = n
3.08 moles = n
Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g
