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horsena [70]
2 years ago
7

A sample of oxygen occupies 20.1 liters under a pressure of 1520 torr at 25.0o What volume would it occupy at 25.0oC if the pres

sure were decreased to 760.0 torr?
Chemistry
1 answer:
Zolol [24]2 years ago
4 0

Answer:

The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L

Explanation:

Boyle's law establishes the relationship between the pressure and the volume of a gas when the temperature is constant, so that the pressure of a gas in a closed container is inversely proportional to the volume of the container. That is, if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Considering an initial state 1 and a final state 2, it is true:

P1* V1= P2*V2

In this case:

  • P1= 20.1 L
  • V1= 1520 torr
  • P2= 760 torr
  • V2= ?

Replacing:

20.1 L* 1520 torr= 760 torr* V2

Solving:

V2=\frac{20.1 L* 1520 torr}{760 torr}

V2= 40.2 L

<em><u>The volume that the sample of oxygen would occupy at 25 ° C if the pressure were reduced to 760.0 torr is 40.2 L</u></em>

<em><u></u></em>

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What volume, in mL, of carbon dioxide gas is produced at STP by the decomposition of 0.242 g calcium carbonate (the products are
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Answer:

54.21 mL.

Explanation:

We'll begin by calculating the number of mole in 0.242 g calcium carbonate, CaCO3.

This is illustrated below:

Mass of CaCO3 = 0.242 g

Molar mass of CaCO3 = 40 + 12 +(16x3) = 40+ 12 + 48 = 100 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 0.242/100

Mole of CaCO3 = 2.42×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

CaCO3 —> CaO + CO2

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole CaO and 1 mole of CO2.

Next, we shall determine the number of mole of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole of CO2.

Therefore,

2.42×10¯³ mole of CaCO3 will also decompose to produce 2.42×10¯³ mole of CO2.

Therefore, 2.42×10¯³ mole of CO2 were obtained from the reaction.

Finally, we shall determine volume occupied by 2.42×10¯³ mole of CO2.

This can be obtained as follow:

1 mole of CO2 occupies 22400 mL at STP.

Therefore, 2.42×10¯³ mole of CO2 will occupy = 2.42×10¯³ x 22400 = 54.21 mL

Therefore, 54.21 mL of CO2 were obtained from the reaction.

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What is the VSEPR model of CB4I?
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Answer:

Molecular geometry Vsepr

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Explanation:

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What is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa
Nezavi [6.7K]

60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.

Explanation:

Data given:

V1 = 75 ml

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P1 = 91 KPa

V2  =?

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

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putting the values in the equation of Gas Law:

V2 = \frac{P1V1T2}{P2T1}

V2 = \frac{91 X 75 X 273.15}{303.15 X 101.3}

V2 = 60.7 ml

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