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taurus [48]
1 year ago
11

What pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c?

Physics
1 answer:
Ronch [10]1 year ago
5 0

The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

Therefore, option A is correct option.

Given,

Mass m = 14g

Volume= 3.5L

Temperature T= 75+273 = 348 K

Molar mass of CO = 28g/mol

Universal gas constant R= 0.082057L

Number of moles in 14 g of CO is

n= mass/ molar mass

= 14/28

= 0.5 mol

As we know that

PV= nRT

P × 3.5 = 0.5 × 0.082057 × 348

P × 3.5 = 14.277

P = 14.277/3.5

P = 4.0794 atm

P = 4.1 atm.

Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

learn more about pressure:

brainly.com/question/22613963

#SPJ4

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Explanation:

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Explanation:

<u>Question 6:</u>

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

  • The density of the substance in Crown A;

Density = mass ÷ volume = \frac{1930}{100} = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

  • The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043  g/cm³ ≈ 10.5 g/cm³  (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

  • The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

<h3><u>The density of the mixture:</u></h3><h3 />

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to \frac{29.8}{2} = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

<u>Question 7:</u>

<u />

  • An empty masuring cylinder has a mass of 500 g.
  • Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.
  • The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷  volume = 350 g ÷ 100 cm³ = 3.5g/cm³

<u>Question 8:</u>

<u />

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

\frac{0.02 * 1}{0.001} =  20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

\frac{0.02 * 1,000,000}{1} = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

<u>Question 9:</u>

<u />

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

<u>Question 10:</u>

<u />

  • A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
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The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy; \frac{20 * 1}{1} = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density ×  volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g

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