
How many cubic centimeters would you need to have 1 mole of aluminum?
Answer
544.8k+ views
Hint: The mole is a worth that can be contrasted with particles (atoms and molecules), volume and mass. The mole is equivalent to \[6.02 \times {10^{23}}\] atoms, which is known as Avogadro's number. The mole is equivalent to \[22.4\] Liters of gas at standard temperature and pressing factor (STP). Furthermore, the mole is equivalent to the molar mass (g) of the component or particle.
Complete step by step answer:
The component aluminum has a molar mass of \[26.99amu\] . In this way, a 1 mole test of aluminum would have a mass of \[26.99\] grams.
When we know the mass of the mole of aluminum we can utilize the thickness to figure the volume. The density condition is
\[Density = \dfrac{{Mass}}{{Volume}}\]
Where D = thickness, m = mass and v = volume.
The standard thickness of aluminum is \[2.70{\text{ }}gm/c{m^3}\].
The volume can be calculated as follows,
$V = \dfrac{{26.99g}}{{2.70g/c{m^3}}} = 9.99c{m^3}$
Note:
Now we can discuss the uses of the Mole:
We must need to remember that the mass of a mole of substance is known as the molar mass of that substance. The molar mass is utilized to change grams of a substance over to moles and is utilized regularly in science. The molar mass of a component is found on the intermittent table, and it is the component's nuclear load in grams/mole (g/mol). In the event that the mass of a substance is known, the quantity of moles in the substance can be determined. Changing over the mass, in grams, of a substance to moles requires a transformation factor of (one mole of substance/molar mass of substance).
We need to know that the mole idea is additionally relevant to the creation of synthetic mixes. For example, think about methane. This particle and its atomic recipe show that per mole of methane there is one mole of carbon and four moles of hydrogen.
Complete step by step answer:
The component aluminum has a molar mass of \[26.99amu\] . In this way, a 1 mole test of aluminum would have a mass of \[26.99\] grams.
When we know the mass of the mole of aluminum we can utilize the thickness to figure the volume. The density condition is
\[Density = \dfrac{{Mass}}{{Volume}}\]
Where D = thickness, m = mass and v = volume.
The standard thickness of aluminum is \[2.70{\text{ }}gm/c{m^3}\].
The volume can be calculated as follows,
$V = \dfrac{{26.99g}}{{2.70g/c{m^3}}} = 9.99c{m^3}$
Note:
Now we can discuss the uses of the Mole:
We must need to remember that the mass of a mole of substance is known as the molar mass of that substance. The molar mass is utilized to change grams of a substance over to moles and is utilized regularly in science. The molar mass of a component is found on the intermittent table, and it is the component's nuclear load in grams/mole (g/mol). In the event that the mass of a substance is known, the quantity of moles in the substance can be determined. Changing over the mass, in grams, of a substance to moles requires a transformation factor of (one mole of substance/molar mass of substance).
We need to know that the mole idea is additionally relevant to the creation of synthetic mixes. For example, think about methane. This particle and its atomic recipe show that per mole of methane there is one mole of carbon and four moles of hydrogen.
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