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3 years ago

If 3.6 g of aluminum completely reacts, how much Al2O3 (in grams) can be produced

Chemistry

1 answer:

marissa [1.9K]3 years ago

8 0

The amount in grams of Al₂O₃ produced is approximately 6.80 g.

Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:

AL + O₂ → Al₂O₃

Let's balance it

4AL + 3O₂ → 2Al₂O₃

4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,

Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,

Atomic mass of AL = 27 g

Molar mass of Al₂O₃ = 101.96 g/mol

4(27 g) of AL gives 2(101.96 g) of Al₂O₃

3.6 g of AL will give ?

cross multiply

mass of Al₂O₃ produced = 3.6 × 203.92 / 108 = 734.112 / 108 = 6.797

mass of Al₂O₃ produced = 6.80 g.

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It has been found by mass spectrometric analysis that in nature, the relative abundances of the various isotopic atoms of silico

pochemuha

Answer:

This means that the isotope of silicon with a mass number of 28 is by far the most common of these three isotopes.

Explanation:

The abundance of Si-28 is 92.23%. Si-29 is 4.68% and Si-30 is 3.09%.

Because most Si atoms have a mass of 28 amu, the average mass of all silicon atoms is very close to 28.

Here is a video which summarizes how to calculate average atomic mass from data about mass and relative abundance.

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3 years ago

What is the concept of galactic recycling?

frozen [14]

Galactic recycling is a natural phenomenon in which the stars expel some gas into the space so that it would mix with the interstellar medium. As a result, this would produce new younger stars. So, basically, the concept done here is the mass and energy conservation. In order to create new species, the energy must come from another source.

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3 years ago

A intramuscular medication is given at 5.00mg/kg of body weight. What is the dose in grams for a 180-lb patient?

Alex17521 [72]

0.408 gram for 180-lb patient

7 0

4 years ago

Read 2 more answers

A buffer contains 0.18 mol of propionic acid (C2H5COOH) and 0.26 mol of sodium propionate (C2H5COONa) in 1.20 L. What is the pH

Yuki888 [10]

Answer:

1) pH = 5.05

2) pH = 5.13

3) pH = 4.97

Explanation:

Step 1: Data given

Number of moles of propionic acid = 0.18 moles

Number of moles sodium propionate = 0.26 moles

Volume = 1.20 L

Ka = 1.3 * 10^-5 → pKa = 4.989

Step 2: Calculate concentrations

Concentration = moles / volume

[acid]= 0.18/ 1.2 =0.150 M

[salt]= 0.26/ 1.3 = 0.217 M

pH = 4.89 + log(0.217/0.150)=5.05

What is the pH of the buffer after the addition of 0.02 mol of NaOH?

moles acid = 0.18 - 0.02 = 0.16

[acid]= 0.16/ 1.2=0.133 M

moles salt = 0.26 + 0.02 = 0.28

[salt]= 0.28/ 12=0.233

pH = 4.89 + log 0.233/ 0.133 = 5.13

What is the pH of the buffer after the addition of 0.02 mol of HI?

moles acid = 0.18+ 0.02 = 0.20 moles

[acid]= 0.20/ 1.2 = 0.167 M

[salt]= 0.26 - 0.02= 0.24 moles

[salt]= 0.24/ 1.2 = 0.20 M

pH = 4.89 + log 0.20/ 0.167= 4.97

8 0

3 years ago

1 mole = 1000 millimoles (mmol); 1millimole = 1000 micromoles (µmol). if a solution contains 38231 µmol, what is that amount in

Mrrafil [7]

1 mmol --------------------- 1000 µmol
( mmol ) -------------------- 38231 µmol

mmol = 38231*1 / 1000

mmol = 38231/ 1000

=> 38.231 mmol

3 0

3 years ago