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Paha777 [63]
3 years ago
14

How many atoms are in 8.28 moles of aluminum?

Chemistry
1 answer:
Lilit [14]3 years ago
4 0

Answer:

49.86 × 10²³  atoms of Al

Explanation:

Given data:

Number of moles of Al = 8.28 mol

Number of atoms = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

For 8.28 moles of Al:

1 mole = 6.022 × 10²³  atoms of Al

8.28 mol×6.022 × 10²³  atoms / 1mol

49.86 × 10²³  atoms of Al

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The limiting reactant, O2, can form up to 2.7 mol Al2O3. What mass of Al2O3 forms?
Margaret [11]

Answer:

280 g Al₂O₃

Explanation:

To find the mass, you need to multiply the given value by the molar mass. This will cause the conversion because the molar mass exists as a ratio; technically, the ratio states that there are 101.96 grams per every 1 mole Al₂O₃. It is important to arrange the ratio in a way that allows for the cancellation of units. In this case, the desired unit (grams) should be in the numerator. The final answer should have 2 sig figs to reflect the given value (2.7 mol).

Molar Mass (Al₂O₃): 101.96 g/mol

2.7 moles Al₂O₃          101.96 g
------------------------  x  -------------------  = 275 g Al₂O₃  = 280 g Al₂O₃
                                     1 mole

5 0
2 years ago
An internal combustion engine relies primarily on the efficient production of energy by the combination of oxygen (02) and gasol
jeka57 [31]
Answer is: <span>the exact ratio of oxygen to octane for is 12.5 : 1.
</span>Balanced chemical reaction: C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O or multiply by 2:
2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O.
There same number of atoms on both side of balanced chemical reaction: eight carbon atoms, eighteen hydrogen atoms and twenty five oxygen atoms.

7 0
3 years ago
Which is a problem associated with the use of trees for biomass? I. Potential deforestation II. Burning fossil carbon III. Poten
BigorU [14]

Answer:

Both Option (I) and Option (III)

Explanation:

Biomass is usually defined as those organic molecules that are obtained from plants and animals, in order to produce energy. It is a renewable resource. Some of the examples of biomass include wood, crop, animal fossil fuel.

A large number of trees are being cut down to construct these biomass energies such as woods. This results in the deforestation process where the forested areas are converted into empty dry soils. This results in the increasing rate of erosion because the prevailing wind and flowing water can easily carry away the topsoil and loses the fertility of the soil.

Thus, the correct answers are both options (I) and (III).

7 0
4 years ago
What is the mass of sodium (Na) in 50 grams of table salt (NaCl)? Show your work.
Margaret [11]

Answer:

19 g

Explanation:

Data Given:

Sodium Chloride (table salt) = 50 g

Amount of sodium (Na) = ?

Solution:

Molecular weight calculation:

NaCl = 23 + 35.5

NaCl = 58.5 g/mol

Mass contributed by Sodium = 23 g

calculate the mole percent composition of sodium (Na) in sodium Chloride.

Since the percentage of compound is 100

So,

Percent of sodium (Na) = 23 / 58.5 x 100

Percent of sodium (Na) = 39.3 %

It means that for ever gram of sodium chloride there is 0.393 g of Na is present.

So,

for the 50 grams of table salt (NaCl) the mass of Na will be

mass of sodium (Na) = 0.393 x 50 g

mass of sodium (Na) = 19 g

8 0
4 years ago
Read 2 more answers
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
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