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MrRa [10]
3 years ago
13

In the manufacture of steel, pure oxygen is blown through molten iron to remove some of the carbon impurity. if the combustion o

f carbon is efficient, carbon dioxide (density = 1.80 g/l) is produced. incomplete combustion produces the poisonous gas carbon monoxide (density - 1.15 g/l) and should be avoided. if you measure a gas density of 1.77 g/l, what can you conclude?
Chemistry
2 answers:
Schach [20]3 years ago
8 0

In the extraction process of steel, one of the step is purification of the iron used to make the steel. In which pure oxygen is blown on the steel at high temperature so that the carbon percentage present in the steel can be thrown out in the form of gas. The process occurs at high temperature which is called combustion process. The reaction occurs can be shown as- C(s)+O_{2}→CO_{2} (g) + CO (g). In presence of excess oxygen, the produced carbon mono oxide (CO) converts to carbon di-oxide. The reaction is CO(g) + O_{2}(g) → CO_{2} (g). From the density of the evolved gas one could identify the gas. If the gas density is 1.77g/L which is very close to the standard density of CO_{2} i.e. 1.80g/L, the gas is carbon dioxide only.    

Oksi-84 [34.3K]3 years ago
4 0

Answer:

We can conclude that the gas evolved was carbon dioxide.

Explanation:

On an efficient combustion that is on complete combustion of carbon, carbon dioxide is produced.

C+O_2\rightarrow CO_2 (Complete combustion)

But during incomplete combustion of carbon results in formation of carbon monoxide (poisons gas).

2C+O_2\rightarrow 2CO (Incomplete combustion)

Theoretical value density of carbon dioxide = 1.80 g/L

Theoretical value density of carbon monoxide = 1.15 g/L

Experimental measured density of the gas = 1.77 g/L

1.77 g/L ≈ 1.80 g/L

Since, the experimental measured density of the gas is more closer to theoretical value of density of carbon dioxide from which we can conclude that  'the gas evolved after the combustion of carbon impurity was carbon dioxide'.

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2 years ago
For each of the following units and concentration values, mention if they are parts per million (ppm), parts per billion (ppb) o
arsen [322]

Answer:

a) ppm

b) ppm

c) ppb

d) ppt

e) ppb

Explanation:

a) You know that 1000 g are 1 kg, and 1000 kg are 1 ton, so (1000)*(1000) g are 1 ton, so 1,000,000 grams are one ton.

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c) You know that 1000 ug are 1 mg, so with the b), we just need to multiply the answer by 1000, so 1,000,000,000 ug are 1 liter.

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7 0
3 years ago
The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f
frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 4.60 atm = 7.59 \times 10^{-3} M

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

\frac{7.59 \times 10^{-3} mol}{L} \times  1.1 L \times \frac{44.01 g}{mol}  = 0.367 g

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 1.28 atm = 2.11 \times 10^{-3} M

\frac{2.11 \times 10^{-3} mol}{L} \times  1.1 L \times \frac{44.01 g}{mol}  = 0.102 g

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

m = 0.367 g - 0.102 g = 0.265 g

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

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2 years ago
How many atoms in 5.01 mole of lead?
leva [86]

Answer:

30.17 × 10²³ atoms

Explanation:

Given data:

Number of moles of lead = 5.01 mol

Number of atoms = ?

Solution:

Avogadro number:

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

In given question:

1 mole = 6.022 × 10²³ atoms

5.01 mol × 6.022 × 10²³ atoms / 1 mol

30.17 × 10²³ atoms

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