<h3>
Answer:</h3>
0.144 moles
<h3>
Explanation:</h3>
- The relationship between mass of a compound, number of moles and molar mass of the compound is given by;
- Number of moles = Mass ÷ Molar mass
- Molar mass is equivalent to the relative formula mass of the compound that is calculated the atomic masses of the elements making the compound.
In this case;
Our compound, KClO3 will have a molar mass of;
= 39 + 35.5 + 4(16)
= 138.5 g/mol
Mass of KClO3 is 20 g
Therefore;
Number of moles = 20 g ÷ 138.5 g/mol
= 0.144 moles
Thus, the number of moles in 20 g of KClO3 is 0.144 moles
1. its temperature will rise continuously until it melts
I don't believe that any of the other answers are correct because it can not stay at a certain temperature if it is melting
Answer:
b) The dehydrated sample absorbed moisture after heating
Explanation:
a) Strong initial heating caused some of the hydrate sample to splatter out.
This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).
b) The dehydrated sample absorbed moisture after heating.
Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.
c) The amount of the hydrate sample used was too small.
It will create some errors but they do not create a difference of 13% difference as stated in the problem.
d) The crucible was not heated to constant mass before use.
Here the error is small.
e) Excess heating caused the dehydrated sample to decompose.
Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.
Answer:

Explanation:
Hello,
In this case, since the 3% by volume is represented as:

By using the ideal gas equation we compute the density of CO:

Then we apply the conversion factors as follows:

Regards.