The equation presented above is that of uranium reacting with fluorine forming uranium fluoride.
The chemical reaction can be balanced by carefully studying the equation and balancing the number of atoms of each of the element in both sides of the chemical reaction. That is,
<em> U(s) + 3F₂(g) --> UF₆(g)</em>
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:
T₂ = 84.375 K
Explanation:
Given data:
Initial volume = 3.3 L
Initial pressure = 2000 torr
Initial temperature = 225 K
Final temperature = ?
Final volume = 2.75 L
Final pressure = 900 torr
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
P₁V₁/T₁ = P₂V₂/T₂
T₂ = P₂V₂ T₁ /P₁V₁
T₂ = 900 torr× 2.75 L× 225 K / 2000 torr×3.3 L
T₂ = 556875 K/ 6600
T₂ = 84.375 K
Answer:
Explanation:
Hello there!
In this case, according to the described chemical reaction, it is possible to compute the theoretical mass of silver as mass via the 1:2 mole ratio of copper to silver and their atomic mass in the periodic table, in order to perform the following stoichiometric setup:
Next, given the actual yield of 120 g, we compute the percent yield via:
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