First convert grams to moles
using molar mass of butane that is 58.1 g
3.50g C4H10 x (1 mol
C4H10)/(58.1g C4H10) = 0.06024 mol C4H10 <span>
<span>Now convert moles to molecules by using Avogadro’s number
0.06024 mol C4H10 x (6.022x10^23 molecules C4H10)/(1 mol
C4H10) = 3.627x10^22 molecules C4H10
And there are 4 carbon atoms in 1 molecule of butane, so use
the following ratio:
3.627 x 10^22 molecules C4H10 x (4 atoms C)/(1 molecule
C4H10)
<span>= 1.45 x 10^23 atoms of carbon are present</span></span></span>
A) At 0 C water forms ice but as mentioned above F) water's greatest density occurs at 4 C and it decreases below 4 C so ice is lighter than 4C water, thus, at 0C ice comes to surface and acts as insulator thereby preventing lower water from freezing.
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:
No because there is not enough Potential Energy at Point 1 to make it all the way through point 5.
Explanation:
