Which observation illustrates the law of conservation of mass?
2 answers:
Answer: When 3 g of carbon reacts with 8 g of oxygen, 11 g of carbon dioxide is produced.
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Given: mass of carbon = 3 g
mass of oxygen = 8 g
Mass of products = Mass of carbon dioxide = 11 g
Mass or reactant = mass of carbon + mass of oxygen = 3 g + 8 g = 11g
Thus the total mass of reactants is equal to the total mass of products.
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Answer : The rate of effusion of sulfur dioxide gas is 52 mL/s.
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
..........(1)
where,
= rate of effusion of nitrogen gas =
= rate of effusion of sulfur dioxide gas = ?
= molar mass of nitrogen gas = 28 g/mole
= molar mass of sulfur dioxide gas = 64 g/mole
Now put all the given values in the above formula 1, we get:
Therefore, the rate of effusion of sulfur dioxide gas is 52 mL/s.