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egoroff_w [7]
3 years ago
10

Natural gas is a mixture of many substances, primarily CH₄, C₂H₆, C3H8, and C4₄H₁₀. assuming that the total pressure of the gase

s is 1.48 atm and that their mole ratio is 04 : 4.0 : 1.5 : 0.50 calculate the partial pressure in atmospheres of each gas?
Chemistry
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

A. The partial pressure for CH4 = 0.0925atm

B. The partial pressure for C2H6 = 0.925atm

C. The partial pressure for C3H8 = 0.346atm

D. The partial pressure for C4H10 = 0.115atm

Explanation:

Total pressure = 1.48atm

Total mole = 0.4+4+1.5+0.5=6.4

A. Mole fraction of CH4 = 0.4/6.4 = 0.0625

The partial pressure for CH4 = 0.0625 x 1.48 = 0.0925atm

B. Mole fraction of C2H6 = 4/6.4 = 0.625

The partial pressure for C2H6 = 0.625 x 1.48 = 0.925atm

C. Mole fraction of C3H8 = 1.5/6.4 = 0.234

The partial pressure for C3H8 = 0.234 x 1.48 = 0.346atm

D. Mole fraction of C4H10 = 0.5/6.4 = 0.078

The partial pressure for C4H10 = 0.078 x 1.48 = 0.115atm

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Which of the following regions does NOT match its description?
Fantom [35]

The region that does not match its description is the fourth answer choice which is Abdominal region: spine

  • For the first answer choice - Cranial region: head

The cranial region encompasses the upper part of the head.

∴ The first answer choice matches its description.

  • For the second answer choice - Axillary region: armpits

The axillary region is an anatomical region under the shoulder joint where the arm connects to the shoulder. Therefore, it encompasses the armpits.

∴ The second answer choice matches its description.

  • For the third answer choice - Thoracic region: chest

The thoracic region runs from the base of the neck down to the abdomen. Therefore, it encompasses the chest

∴ The third answer choice matches its description

  • For the fourth answer choice - Abdominal region: spine

The abdominal region is divided into four quadrants which include are

1. Right upper quadrant fossa (RUQ)

2. Right lower quadrant fossa (RLQ)

3. Left lower quadrant fossa (LLQ)

4. Left upper quadrant fossa (LUQ)

It is also divided into nine (9) areas, all of which does not include the spine.

∴ The fourth answer choice does NOT match its description.

Hence, the region that does not match its description is the fourth answer choice which is Abdominal region: spine

Learn more here: brainly.com/question/11504303

8 0
1 year ago
500.0 mL of a gas is collected at 745.0 mm Hg. what what will the volume be at standard pressure
Irina-Kira [14]
This is an application of Boyle's law:
P₁V₁ = P₂V₂. we don't have to convert volume and pressure to standard forms. we can even use the pressure with mmHg
1 atm = 760 mmHg 
V₂ = P₁V₁ / P₂ = 745 x 500 / 760 = 490 ml
Note that here we assume constant temperature
6 0
3 years ago
Identify three reasons why
BARSIC [14]
The radioactive decay of unstable isotopes continually generates new energy within Earth's crust and mantle, providing the primary source of the heat that drives mantle convection. Plate tectonics can be viewed as the surface expression of mantle convection.
6 0
2 years ago
A eraser has a mass of 4g and a volume of 2cm3 what is it’s density
sladkih [1.3K]

Answer:

<h3>The answer is 2.0 g/cm³</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 4 g

volume = 2 cm³

We have

density =  \frac{4}{2}  \\

We have the final answer as

<h3>2.0 g/cm³</h3>

Hope this helps you

4 0
3 years ago
7. A 26.4-ml sample of ethylene gas, C2H4, has a pres-sure of 2.50 atm at 2.5°C. If the
never [62]

Answer: 1.87 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 2.50 atm

P_2 = final pressure of gas = ?

V_1 = initial volume of gas = 26.4 ml

V_2 = final volume of gas = 36.2 ml

T_1 = initial temperature of gas = 2.5^oC=273+2.5=275.5K

T_2 = final temperature of gas = 10^oC=273+10=283K

Now put all the given values in the above equation, we get:

\frac{2.50\times 26.4}{275.5}=\frac{P_2\times 36.2}{283}

P_2=1.87atm

The new pressure is 1.87 atm by using combined gas law.

6 0
2 years ago
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