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3 years ago

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Two gas jars are connected to each other, but they are separated by a closed valve. One gas jar contains oxygen, and the other i

s empty. What will happen when the valve is opened?

1 answer:

Yanka [14]3 years ago

3 0

<h2>Answer:</h2>

When the valve is opened the gas from the filled jar will moved through the valve toward the empty jar. After some time the empty jar will filled with oxygen also. This process is called Diffusion.

Diffusion is the net movement of molecules or atoms from a region of high concentration to a region of low concentration as a result of random motion of the molecules or atoms.

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Given the chemical formulas Na2O, MgO, and Al2O3, predict the formula for radioactive francium oxide, Fr?O?.

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Francium oxide Fr₂O

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Explanation:

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An experiment was conducted to see which metal pot would be better for cooking food quickly. The table below shows how quickly s

KengaRu [80]

Answer : It is a conclusion.

Explanation:

Conclusion: It is a judgement or decision which can be attained by reasoning.

From the observation table given we can conclude that :

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After analyzing the observations and results in an experiment conclusion was made that copper pots are best for cooking foods.

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2 years ago

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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

= $\frac{2.74}{6000}*100$ %

= 0.0456667 %

= 0.046% to three decimal places

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