Assume that the test tube shown started out having 10.00 g of mercury(II) oxide. After heating the test tube briefly, you find 1
1 answer:
This problem is providing information about the initial mass of mercury (II) oxide (10.00 g) which is able to produce liquid mercury (8.00 g) and gaseous oxygen and asks for the resulting mass of the latter, which turns out to be 0.65 g after doing the corresponding calculations.
Initially, it is given a mass of 10.00 g of the oxide and 1.35 g are left which means that the following mass is consumed:
Now, since 8.00 grams of liquid mercury are collected, it is possible to calculate the grams of oxygen that were produced, by considering the law of conservation of mass, which states that the mass of the products equal that of the reactants as it is nor destroyed nor created. In such a way, the mass of oxygen turns out to be:
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Explanation:
The given data is as follows.
Vapour pressure of pure = 33.85 Torr
Temperature = 298 K
Mole fraction of = 122.36 torr
Therefore, calculate the vapor pressure of as follows.
Vapour pressure of = mole fraction of
x K of
= 0.050 x 122.36 Torr
= 6.118 Torr
So, vapor pressure of is 6.118 Torr
.
Now, calculate the vapor pressure of carbon tetrachloride as follows.
Vapour pressure of = mole fraction of
x Pressure of
= (1 - 0.050) × 33.85 Torr
= 32.1575 Torr
So, vapor pressure of is 32.1575 Torr .
Hence, the total pressure will be as follows.
= 6.118 Torr + 32.1575 Torr
= 38.2755 Torr
Therefore, composition of =
= 0.8405
Composition of is 0.8405
.
And, composition of =
= 0.1598
Composition of is 0.1598
.