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MrRa [10]
2 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]2 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]2 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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If a pharmacist added 12 g of azelaic acid diluent should be used to prepare 8 fluidto 50 g of an ointment containing 15% ounces
natta225 [31]
Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-<span>azelaic acid) = 12g.
mr</span>₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
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7 0
2 years ago
A pan containing 20.0 grams of water was allowed to cool from a temperature of 95.0 °C. If the amount of heat released is 1,200
Sedbober [7]

Answer:

81°C.

Explanation:

To solve this problem, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released from water (Q = - 1200 J).

m is the mass of the water (m = 20.0 g).

c is the specific heat capacity of water (c of water = 4.186 J/g.°C).

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).

∵ Q = m.c.ΔT

∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).

(- 1200 J) = 83.72 final T - 7953.

∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.

<em>So, the right choice is: 81°C.</em>

7 0
2 years ago
A 32.5 g piece of aluminum (which has a specific heat capacity of 0.921 J/g°C) is heated to 82.4°C and dropped into a calorimete
N76 [4]

Answer:

The mass of water = 219.1 grams

Explanation:

Step 1: Data given

Mass of aluminium = 32.5 grams

specific heat capacity aluminium = 0.921 J/g°C

Temperature = 82.4 °C

Temperature of water = 22.3 °C

The final temperature = 24.2 °C

Step 2: Calculate the mass of water

Heat lost = heat gained

Qlost = -Qgained

Qaluminium = -Qwater

Q = m*c*ΔT

m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)

⇒with m(aluminium) = the mass of aluminium = 32.5 grams

⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C

⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C =  -58.2 °C

⇒with m(water) = the mass of water = TO BE DETERMINED

⇒with c(water) = 4.184 J/g°C

⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C

32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9

-1742.1 = -7.95m

m = 219.1 grams

The mass of water = 219.1 grams

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