Answer:
the boiling point will be 100° C for both.....as water boiling point is this only
Answer:
Decomposition of potassium chlorate yields potassium chloride and oxygen as:
2KClO
3
→2KCl+3O
2
Thus 2 moles of potassium chlorate yields 3 moles of oxygen gas.
2 moles of potassium chlorate =2×122.5=245g of potassium chlorate
At STP, the volume occupied by 1 mol of gas =22.4 dm
3
the volume occupied by three moles of a gas =3×22.4=67.2dm
3
Therefore, 245g of potassium chlorate yields 67.2dm
3
of oxygen gas
To liberate 6.72 dm
3
oxygen amount of potassium chlorate required is
=
67.2
245
×6.72=24.5g
Hence, 24.5 g of potassium chlorate required to liberate 6.72 dm
3
of oxygen at STP
<h2><u>Answer:</u></h2>
It wasn't an adjustment in the condition of issue on the grounds that the vitality in the can did not change. Additionally, since this was a physical change, the atoms in the can are as yet similar particles. No synthetic bonds were made or broken. You added enough vitality to make a stage change from strong to fluid.
The main changes recorded which don't include framing or breaking substance bonds would bubble and liquefying. Bubbling and liquefying are physical changes as opposed to synthetic changes, so no new items are shaped.
