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

7 0

3 years ago

S(s)+3F2(g)->SF6(g) how many mol of F2 are required to react completely with 2.30 mol of S?

Brilliant_brown [7]

Answer: There are 6.9 mol of $F_{2}$ are required to react completely with 2.30 mol of S.

Explanation:

The given reaction equation is as follows.

$S(s) + 3F_{2}(g) \rightarrow SF_{6}(g)$

Here, 1 mole of S is reaction with 3 moles of $F_{2}$ which means 1 mole of S requires 3 moles of $F_{2}$ .

Therefore, moles of $F_{2}$ required to react completely with 2.30 moles S are calculated as follows.

$1 mol S = 3 mol F_{2}\\2.30 mol S = 3 mol F_{2} \times 2.30 \\= 6.9 mol F_{2}$

Thus, we can conclude that there are 6.9 mol of $F_{2}$ are required to react completely with 2.30 mol of S.

3 0

3 years ago

How glass tube filed with acidified potassium dichromate detect drunken driver

hodyreva [135]

Answer:

See explanation

Explanation:

The reaction between alcohol and acidified potassium dichromate is a redox reaction. This reaction can be used to detect a drunken driver.

Alcohols can be oxidized to aldehydes, ketones and carboxylic acids depending on the structure of the alcohol. Primary alcohols yield adehydes and carboxylic acids while secondary alcohols are oxidized to ketones.

The colour of the acidified potassium dichromate turns from orange to green when exposed to alcohols from the breath of a drunken driver.

5 0

3 years ago