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Arada [10]
3 years ago
15

A 10.0 L balloon contains helium gas at a pressure of 660 mmHg . What is the final pressure, in millimeters of mercury, of the h

elium gas at each of the following volumes, if there is no change in temperature and amount of gas?
Physics
1 answer:
Ksivusya [100]3 years ago
6 0

Answer:

(a)= 264mmHg

(b)= 2000mmHg

(c)474.82mmHg

(d)= 511.63mmHg

Explanation:

the question deals with boyles law, which states that the volume of a given mass of gas at constant temperature is inversely proportional to its pressure

V ∝ 1/P

P₁V₁ = P₂V₂

making V₂ as the subject of formular

P₂ = P₁V₁/ V₂

with a volume of  25.0L

P₂ = 660×10 / 25

= 264mmHg

with a volume of 3.30 L

P₂  = 660 × 10 / 3.30

= 2000mmHg

with a volume of 13900 mL

= 13.9L

P₂  =660× 10 / 13.9

474.82mmHg

with a volume of 12900 mL

P₂ =660×10 / 12.9

= 511.63mmHg

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Two liquids, A and B, have equal masses and equal initial temperatures. Each is heated for the same length of time over identica
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Answer:

So the specific heat of the liquid B is greater than that of A.

Explanation:

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Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?
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Answer:

x-component=-9.3 m

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