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Dmitrij [34]
2 years ago
9

If you collect 1.75 L of hydrogen gas during a lab experiment when the room temperature is 23oC and the barometric pressure is 1

05 kPa how many moles of hydrogen will you have
Chemistry
1 answer:
ziro4ka [17]2 years ago
8 0

Answer:

n=0.0747mol

Explanation:

Hello,

In this case, since we can consider hydrogen gas as an ideal gas, we check the volume-pressure-temperature-mole relationship by using the ideal gas equation:

PV=nRT

Whereas we are asked to compute the moles given the temperature in Kelvins, thr pressure in atm and volume in L as shown below:

n=\frac{105kPa*\frac{0.009869atm}{1kPa}*1.75L}{0.082\frac{atm*L}{mol*K}*(23+273.15)K} \\\\n=0.0747mol

Best regards.

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2 years ago
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

7 0
3 years ago
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Product of mixing acids and bases describes salt is a physical property.

Product of mixing acids and bases

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When an acid and a base are put together, they respond to kill the corrosive and base properties, creating a salt which portrays the physical property. The physical properties of table salt will be: Salt is a white cubic gem. At the point when the salt is unadulterated it clear.

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