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Digiron [165]
3 years ago
9

Question 4 A mixture of xenon and oxygen gas is compressed from a volume of to a volume of , while the pressure is held constant

at . Calculate the work done on the gas mixture. Round your answer to significant digits, and be sure it has the correct sign (positive or negative).
Chemistry
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer:

W = - 1218 atm L

Explanation:

Complete question

A mixture of krypton and oxygen gas is expanded from a volume of 23.0L to a volume of 81.0L, while the pressure is held constant at 21.0atm. Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.

Solution

As we know that -

Work done is equal to the product of pressure applied and change in the difference of volume of two liquids.

This can be represented mathematically as follows -

W = -P*(V_2-V_1)\\

Where W represents the work done

P represents the change in pressure

V_2 is the final volume  and

V_1 is the initial volume

Substituting the given values in above equation, we get -

W = - 21 * (81 -23)\\

W = - 1218 atm L

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Rasek [7]

Number of moles = 8.3 x 10⁻¹³

<h3>Further explanation </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³  particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's =  6.02.10²³

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\tt mol=\dfrac{5\times 10^{11}}{6.02\times 10^{23}}=8.3\times 10^{-13}

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julia-pushkina [17]

Answer:

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Explanation:

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When the following equation is balanced with the lowest whole number coefficients possible, what is the coefficient in front of
Anastaziya [24]
Given: C3H8(g) + O2(g) ----> CO2 (g) + H2O (g)

Step : Put a 3 in front of CO2 (g) to balance C

=> C3H8(g) + O2(g) ----> 3CO2 + H2O to balance H

Step 2: Put a 4 in front of H2O

=> C3H8 (g) + O2(g) -----> 3CO2 (g) + 4H2O (g)

Step 3: Given that there are 3*2 + 4 = 10 O to the right side, put a 5 in front of O2 to balance O:

=> C3H8(g) + 5O2(g) -----> 3CO2(g) + 4H2O(g)

You can verify that the equation is balanced.

So, the answer is that the coefficient in front of O2 is 5.
7 0
3 years ago
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3 years ago
Given 700 ml of oxygen at 7 ºC and 106.6 kPa pressure, what volume does it take at 27ºC and 66.6 kPa pressure
o-na [289]

Answer:

The volume is 1.2L

Explanation:

Initial volume (V1) = 700mL = 0.7L

Initial temperature (T1) = 7°C = (7 + 273.15)K = 280.15K

Initial pressure = 106.6kPa = 106600Pa

Final temperature (T2) = 27°C = (27 + 273.15)K = 300.15K

Final pressure (P2) = 66.6kPa = 66600Pa

Final volume (V2) = ?

To solve this question, we need to use combined gas equation which is a combination of Boyle's law, Charles Law and pressure law.

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solve for V2 by making it the subject of formula,

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V2 = (P1 × V1 × T2) / (P2 × T1)

V2 = (106600 × 0.7 × 300.15) / (66600 × 280.15)

V2 = 22397193 / 18657990

V2 = 1.2L

The final volume of the gas is 1.2L

6 0
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