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3 years ago

19. 4.0 L of a gas at 1.0 atm is compressed into a 0.85 L cylinder. What is the pressure of the compressed gas if

Chemistry

1 answer:

Neko [114]3 years ago

4 0

Answer:

The correct answer is d. 4.7 atm

Explanation:

We apply Boyle Mariotte's law, where for a certain mass of gas at constant temperature, the pressure and temperature vary inversely proportionally: P1xV1 = P2xV2

P2= (P1xV1)/V2= (1,0 atm x 4,0 L)/0,85L= <em>4, 705882353 atm</em>

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Answer:

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3 years ago

Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm

8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

$\frac{dV}{dt}=10\frac{cm^3}{s}$

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

$\frac{dh}{dt} = ?*\frac{dV}{dt}$

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We proceed to its differentiation:

$\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}$

Then, we compute $\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}$

Finally, at h=2:

$\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}$

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3 years ago

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Answer:

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Explanation:

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3 years ago

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Lelechka [254]

Answer:

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Explanation:

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