Answer:
An oxidising agent oxidises something else. Oxidation is loss of electrons (OIL RIG). That means that an oxidising agent takes electrons from that other substance.
Explanation:
When oxygen has an electronegativity of 3.5, and carbon has an electronegativity of 2.5, then the oxygen atom would have a slightly negative charge. The oxygen atom in the carbon monoxide molecule would pull more electrons to its side since it has higher electronegativity making it slightly negative and the carbon would have a slightly positive charge as it would contain less electrons. This results to the formation of a polar molecule. A polar molecule is made when the molecule contains a slightly positive end and a slightly negative end. It would have a net dipole which is a result of the partial opposing charges in the molecule.
Answer:
19.07 g mol^-1
Explanation:
The computation of the molecular mass of the unknown gas is shown below:
As we know that
where,
Diffusion rate of unknown gas = 155 mL/s
CO_2 diffusion rate = 102 mL/s
CO_2 molar mass = 44 g mol^-1
Unknown gas molercualr mass = M_unknown
Now placing these values to the above formula
After solving this, the molecular mass of the unknown gas is
= 19.07 g mol^-1
By crushing the salt, you are performing a physical change because you aren't altering the chemical makeup of the salt, just the physical form. Hope this helps! :)
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.
