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Alja [10]
2 years ago
6

Why are metal containers not used for storing acids​

Chemistry
1 answer:
kaheart [24]2 years ago
3 0

<em><u>Answer </u></em><em><u>:</u></em><em><u> </u></em><em><u>M</u></em><em><u>etal containers</u></em><em><u> </u></em><em><u>are</u></em><em><u> not</u></em><em><u> </u></em><em><u> </u></em><em><u>used for storing acid</u></em><em><u> because most of the time acid reacts with almost every metal and produces </u></em><em><u>salts</u></em><em><u> or oxid</u></em><em><u>e</u></em><em><u>s</u></em><em><u> </u></em><em><u>which alters the acid characteristics making it useless</u></em><em><u> </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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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.

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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

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

C= 12-2.4

C =9.6

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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 %

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