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2 years ago

Which of these statements best describes the relationship between elements, compounds, and pure substances?

Chemistry

2 answers:

Tamiku [17]2 years ago

7 0

Answer:

Pure substances can be divided into two groups: elements and compounds.

Explanation:

Pure substances can be recognized because they have a definite constant composition, as can be seen in the formula, and distinct chemical properties.

For example, H₂O is a pure substance where the relation of 2 atoms of Hydrogen to 1 atom of Oxygen is always true. O₂ is also a pure substance because 1 molecule of O₂ always contains two atoms of Oxygen.

As we can see from above, pure substances can be formed by one or more types of atoms. Substances formed by 2 or more types of atoms are known as compounds (e.g. H₂O, CO₂). Substances formed by 1 type of atoms are known as elements (e.g. Na, O₂, O₃).

Compounds and elements are kinds of pure substances, so they inherit all properties from them (definite constant composition, distinct chemical properties).

olchik [2.2K]2 years ago

5 0

pure substances can be divided into two groups; elements and compounds

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3 years ago

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What mass, in grams, of CO2 and H2O is formed from 2.55 mol of propane?

oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

Being the molar mass of each compound:

C₃H₈: 44 g/mole
O₂: 16 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

C₃H₈: 1 mole* 44 g/mole= 44 grams
O₂: 5 moles* 16 g/mole= 80 grams
CO₂: 3 moles* 44 g/mole= 132 grams
H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

$mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}$

mass of CO₂= 336.6 grams

If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

$mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}$

mass of H₂O= 183.6 grams

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

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What volume in litres will 38gr of F2 occupy at 0.999 bar and 273 K

FrozenT [24]

Answer:

$V=22.68L$

Explanation:

Hello,

In this case, we use the ideal gas equation to compute the volume as shown below:

$PV=nRT\\\\V=\frac{nRT}{P}$

Nonetheless we are given mass, for that reason we must compute the moles of gaseous fluorine (molar mass: 38 g/mol) as shown below:

$n=38 g *\frac{x}{y} \frac{1mol}{38 g} =1mol$

Thus, we compute the volume with the proper ideal gas constant, R:

$V=\frac{1mol*0.083\frac{bar*L}{mol*K}*273K}{0.999bar} \\\\V=22.68L$

Best regards.

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