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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Answer:
The Ideal gas law
Explanation:
From the given question, we have:
V
where each variable has its usual meaning.
Thus,
V =
where R is the ideal gas constant
cross multiply to have;
PV = nRT
This implies that the volume of the gas is directly proportional to the number of moles of the gas.
Therefore, the law can be used to determine the relationship between the volume and number of moles is the ideal gas law.