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sweet [91]
3 years ago
8

A cyclopropane-oxygen mixture is used as an anesthetic. If the partial pressure of cyclopropane in the mixture is 334 mmHg and t

he partial pressure of the oxygen is 1.02 atm, what is the total pressure of the mixture in torr?
Chemistry
1 answer:
Svetach [21]3 years ago
8 0

Answer:

Total pressure = 1109.2 torr

Explanation:

Given data:

Partial pressure of cyclopropane = 334 mmHg

Partial pressure pressure of oxygen = 1.02 atm

Total pressure of mixture = ?

Solution:

Formula:

Total pressure = P of C₃H₆ + P of O₂

1 atm = 760 torr = 760 mmHg

1.02 atm × 760 torr / 1atm = 775.2 torr

Total pressure = P of C₃H₆ + P of O₂

Total pressure = 334 torr + 775.2 torr

Total pressure = 1109.2 torr

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Carbon tetrachloride, CCl4, once used as a cleaning fluid and as a fire extinguisher, is produced by heating methane and chlorin
Elena-2011 [213]

Answer:

1.882 g

Explanation:

Data Given

mass of Cl₂ = 33.4 g

mass of CH₄ = ?

Reaction Given:

                       CH₄+ 4Cl₂ --------→ CCl₄ + HCl

Solution:

First find the mass of CH₄ from the reaction that it combine with how many grams of Chlorin.

Look at the balanced reaction

                    CH₄     +     4Cl₂ --------→ CCl₄ + 4HCl

                    1 mol        4 mol

So 1 mole of CH₄ combine with 4 moles of Cl₂

Now

convert the moles into mass for which we have to know molar mass of CH₄ and Cl₂

Molar mass of Cl₂ = 2 (35.5)

Molar mass of  Cl₂  = 71 g/mol

mass of Cl₂

                mass in grams = no. of moles x molar mass

                mass of Cl₂ = 4 mol x 71 g/mol

                mass of Cl₂  = 284 g

Molar mass of CH₄= 12+ 4(1)

Molar mass of CH₄= 16 g/mol

mass of CH₄

                mass in grams = no. of moles x molar mass

                mass of CH₄= 1 mol x 16 g/mol

                mass of CH₄ = 16 g

So,

284 g of Cl₂  combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂  

Apply unity Formula

                           284 g of Cl₂  ≅ 16 g of methane ( CH₄ )

                           33.4 g of Cl₂  ≅ X g of methane ( CH₄ )

By cross multiplication

                          X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

                          X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

So

methane (CH₄) = 1.882 g

5 0
3 years ago
Why is it essential for a calorimetry to be an insulated (closed) system?
aksik [14]

Answer:

B.Thermal insulation minimizes energy loses to the atmosphere.

Explanation:

It is important because it helps to stop hit from transferring from the calorimeter to the environment. This would help to have an accurate measurement of the heat that was used in the chemical process. The greatest cause of error that happens in calorimetry is when heat is lost to the environment. To reduce this, you insulate the calorimeter and add a cover.

5 0
2 years ago
the most common source of copper (Cu) is the mineral chalcopyrite (CuFeS2). How many kilograms of chalcopyrite must be mined to
lara [203]
<span>0.650 kg... I think, 
But what are the answer choices?</span>
3 0
3 years ago
How many gallons of gasoline that is 5% ethanol must be added to 2,000 gallons of gasoline with no ethanol to get a mixture that
Amanda [17]

Answer:

V_1= 3000 gal

Explanation:

We have 3 solutions:

  • Solution 1 (with ethanol)
  • Solution 2 (no ethanol)
  • Final solution

V_f=V_1 + V_2

and for the ethanol:

V_f*0.03=V_1*0.05 + V_2*0

V_f=V_1 \frac{5}{3}

Combining:

V_1 \frac{5}{3}=V_1 + V_2

V_1 \frac{2}{3}= V_2

V_1= \frac{3}{2} V_2

If V2=2000 gal:

V_1= \frac{3}{2} 2000 gal

V_1= 3000 gal

8 0
3 years ago
How many molecules of carbon dioxide, CO2, are present in 388.1 grams?
swat32

Answer:

53.11× 10²³ molecules

Explanation:

Given data:

Number of molecules of CO₂ = ?

Mass of CO₂ = 388.1 g

Solution:

Formula:

Number of moles = mass/ molar mass

Molar mass of CO₂ = 12× 1 + 16×2

Molar mass of CO₂ = 44 g/mol

Now we will put the values in formula.

Number of moles = 388.1 g/ 44 g/mol

Number of moles = 8.82 moles

Now we will calculate the number of molecules 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.

1 mole = 6.022 × 10²³ molecules

8.82 mol × 6.022 × 10²³ molecules / 1 mol

53.11× 10²³ molecules

6 0
3 years ago
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