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3 years ago

Activated charcoal is used in gas masks. What physical property is related to this use?

Chemistry

1 answer:

valina [46]3 years ago

8 0

Answer:

physical property is related to the use of Activated charcoal in gas masks is Physical adsorption

Explanation:

Respirators in the gas masks has a 2-stage inhalation protection process . In the first step, particle filter is used that behaves like a physical barrier that removes bacteria from air which is inhaled. When the bacteria is inhaled, it gets entangled in the filter fibres preventing it from being inhaled into the lungs.

In the next stage , activated charcoal is used .This layer with highly porous charcoal attracts and binds to toxins. Through adsorption process, a solid or a liquid can trap particles on its surface. Toxins that pass through the charcoal become bonded to its surface, preventing inhalation of the toxins.

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What is the volume of 40g of sugar given it’s listed density

lora16 [44]

Answer:

25.157 cm³

Explanation:

Data Given:

Mass of Sugar (m) = 40g

Density of sugar given in literature = 1.59 g/cm³

Volume of Sugar = ?

The formula will be used is

d = m/v ........................................... (1)

where

D is density

m is the mass

v is the volume

Rearrange the Equation (1)

d x v = m

v = m/ d ................................................ (2)

put the given values in Equation (2)

v = 40g / 1.59 g/cm³

v = 25.157 cm³

volume of 40 g of sugar = 25.157 cm³

8 0

3 years ago

Anh measured the temperature of a pond near his house. Before he left for school, the water in the pond was 18 degrees celsius.

Hitman42 [59]

As the temperature increases, B) the molecules started moving faster.

Explanation:

The temperature of a substance is a measure of the average kinetic energy of the particles in a substance. In particular, it can be found that the temperature is directly proportional to the average kinetic energy of the particles:

$T\propto KE$

The kinetic energy of a particle is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its speed

This means that the higher the temperature of a substance, the greater the speed of the particles in the substance.

Therefore, if we apply this concept to this problem, we infer that as the temperature of the water in the pond gets higher, the speed of the molecules inside the water increases, which means that the molecules are moving faster.

Therefore, the correct answer is

B) the molecules started moving faster.

Learn more about temperature:

brainly.com/question/1603430

brainly.com/question/4370740

#LearnwithBrainly

3 0

4 years ago

The orbital radii of four planets in our solar system is shown in the following table. Orbital Radii Planet Orbital Radii (AU) W

nalin [4]

Answer:

w and x

Explanation:

6 0

3 years ago

A student jumps of a sled toward the west after it stops at the bottom of an icy hill. Based on the law of action-reaction, in w

Delicious77 [7]

Answer:

East

Explanation:

Given Newton's third law of motion; "Action and reaction are equal and opposite", when a student jumps off a sled toward the west after it stops at the bottom of an icy hill, the sled will move in the East direction.

This is because, the force exerted on the sled is a reaction force and is opposite in direction to the force that thrusts the boy westward though equal in magnitude with the former.

3 0

3 years ago

Naphthalene, C10H8, melts at 80.2Â°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8Â°C and 5.3 kPa at 119.3Â°C, use th

sweet-ann [11.9K]

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

P₁ = 1.3 kPa

P₂ = 5.3 kPa

T₁ = 85.8°C = 358.96 K

T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)

ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)

ΔHv = 49111.12 J/molK

(b) Normal boiling point means that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, solving for T₂:

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)

1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]

1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]

1/T₂ = 2.049 * 10⁻³ K⁻¹

T₂ = 488.1 K = 214.94 °C

(c) The enthalpy of vaporization was calculated in part (a), and it does not vary depending on temperature, meaning that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK.

3 0

3 years ago