An intensive property does not change when you take away
some of the sample. The procedures that a student could use to examine the
intensive property of a rectangular block of wood are the hardness, color,
density and molecular weight.
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Understanding how much of a product is produced in a reaction is referred to as Stoichiometrically understanding the reaction.
Stoichiometry is the calculation of the quantitative relationships between reactants and products in a chemical reaction. The first to talk about stoichiometry was Jeremias Benjamin Ritcher, who said that "Stoichiometry is the science that measures the quantitative proportions or mass ratios of chemical elements that are involved in a chemical reaction".
To calculate how much of a product is produced in a reaction, Stoichiometry is used, applying the law of conservation of mass. That means that the amount of product can be calculated from the amounts of reactants if they are known.
Answer:
Compounds
Explanation:
Substances in which atoms of two or more elements are chemically combined are called compounds.
A compound is a substance formed when two or more elements are chemically joined.
Considering the ideal gas law, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P× V = n× R× T
In this case, you know:
- P= 2 atm
- V= ?
- n=
being 2g/mole the molar mass of H2, that is, the amount of mass that a substance contains in one mole. - R= 0.082

- T= 353 K
Replacing:
2 atm× V = 4.745 moles× 0.082
× 353 K
Solving:
V = (4.745 moles× 0.082
× 353 K)÷ 2 atm
<u><em>V= 68.67 L</em></u>
Finally, a sample weighing 9.49 g occupies 68.67 L at 353 K and 2.00 atm.
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