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Fynjy0 [20]
3 years ago
13

A driver blows 0.75 -L air bubble 10 m under water. As it rises to the surface, the pressure goes from 2.25 atm to 1.03 atm. Wha

t will be the volume of air in the bubble at the surface ?
Chemistry
1 answer:
kakasveta [241]3 years ago
3 0

Answer:

1.64 L

Explanation:

P₁V₁ = P₂V₂

P₁ = 2.25 atm

V₁ = 0.75 L

P₂ = 1.03 atm

V₂ = ?

(2.25 atm)(0.75 L) = (1.03 atm)V₂

[(2.25 atm)(0.75 L)]/(1.03 atm) = V₂

V₂ = 1.64 L

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Answer:

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8 0
2 years ago
A copper cylinder, 12.0 cm in radius, is 44.0 cm long. If the density of commoner is 8.90 g/cm3, calculate the mass in grams of
inna [77]

The mass of the copper cylinder is 177065.856g

Given:

Radius of the copper cylinder R=12cm

Height of the copper cylinder H=44cm

Density of the cylinder=8.90 \frac{g}{c m^{3}}

To find:

Mass of the copper cylinder

<u>Step by Step by explanation:</u>

Solution:

According to the formula, Mass can be calculated as

\rho=\frac{m}{v} and from this

m=\rho \times v

Where, m=mass of the cylinder

\rho =density of the cylinder

v=volume of the cylinder

And also cylinder is provided with radius and height value.

So volume of the cylinder is calculated as

v=\pi r^{2} h

Where \pi=3.14

r=radius of the cylinder=12cm

h=height of the cylinder=44cm

Thus, v=3.14 \times 12^{2} \times 44

v=3.14 \times 144 \times 44

v=19895.04 \mathrm{cm}^{3}

And we know that, m=\rho \times v

Substitute the known values in the above equation we get

m=8.90 \times 19895.04  

m=177065.856g or 177.065kg

Result:

Thus the mass of the copper cylinder is 177065.856g

4 0
3 years ago
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Jose times how long sugar takes to dissolve in warm water. he conducts four trials of his experiment. what should he conclude fr
mylen [45]

He can conclude that his experiment has very low precision.

<h3>What is Precision ?</h3>

Precision is defined as the degree of refinement with which an operation is performed or a measurement .

Precision is how close the exact answers are together.

As, the answers are increasing in time.

None of the answers are similar to one another.

Hence, He can conclude that his experiment has very low precision.

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