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3 years ago

A 1,70 atm, una muestra de gas ocupa 4,25 litros. Si la presión en el gas aumenta a 2.40 atm, ¿cuál será el nuevo volumen?

1 answer:

8 0

At 1.70 atm, a gas sample occupies 4.25 liters. If the pressure in the gas increases to 2.40 atm, what will the new volume be?

Answer:

3.01L

Explanation:

Given parameters:

Initial pressure, P1 = 1.7atm

Initial volume, V1 = 4.25L

Final pressure, P2 = 2.4atm

Unknown:

Final or new volume, V2 = ?

Solution:

To solve this problem, we use Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".

P1 V1 = P2 V2

P1 is the initial pressure

V1 is the initial volume

P2 final pressure

V2 final volume

1.7 x 4.25 = 2.4 x V2

V2 = 3.01L

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The diagram below shows the stars that are nearest to our solar system.

abruzzese [7]

Answer:

no u tried of the same dam

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Explanation:

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3 years ago

One way to measure g on another planet or moon by remote sensing is to measure how long it takes an object to fall a given dista

Jlenok [28]

Answer:

(a) 0.94 m/s²

(b) g (planet) = 0.096g

Explanation:

(a)

From Newton's equation of motion,

S = ut + 1/2gt²......................... equation 1

Making g the subject of equation 1

g =( S - ut)/t² ........................ equation 2

Where s = distance ( m), u = initial velocity (m/s), t = time (s), g = acceleration due to gravity (m/s²)

From the question, S = 12.02 m, t = 3.58 s, u= 0 ( at rest),

Substituting these values in equation 2

g = {12.02 -(0×3.58)}/3.58²

g = (12.02)/12.82

g = 0.94 m/s²

∴ The acceleration due to gravity on the planet = 0.94 m/s²

(b) g (planet)/g (earth) = 0.94/9.80

g (planet) = 0.096 g (earth).

The acceleration due to gravity of the planet in terms of the earth g is

g (planet) = 0.096g

5 0

3 years ago

The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur

Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

P₂ is the pressure in the hose

ρ is the density of the fluid

h₁ is the initial height

h₂ is the final height

v₁ is the initial velocity of the fluid

v₂ is the final velocity of the fluid and

g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

= (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

= (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

= (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

= 5.92 × 10⁵ Pa

= 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

#SPJ4

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2 years ago

A cube-shaped piece of copper has sides of 4cm each and it's density is

iris [78.8K]

Answer:

$64 cm^3$

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

The formula to calculate the density of a substance of mass (m) and volume (V) is:

$\displaystyle \rho=\frac{m}{V}$

We have a cube-shaped piece of copper of 4 cm of side length. The volume of the piece is:

$V=(4\ cm)^3=64\ cm^3$

Surprisingly, no other magnitude is required, thus the answer is:

$\mathbf{64 cm^3}$

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