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marusya05 [52]
2 years ago
9

Calculate the amount of hcn that gives the lethal dose in a small laboratory room measuring 12.0 ft×15.0 ft×8.50ft.

Mathematics
1 answer:
Anarel [89]2 years ago
7 0
Dimensions of the room in cm = 2.54 x 12 by 15 x 2.54 by 2.54 x 8.5 = 30.48 by 38.1 by 21.59

Volume of the room in cubic cm = 30.48 x 38.1 x 21.59 cubic cm = 25,072.21 cubic cm

Given that the density of air at room temperature is 0.00118g/cm^3, thus the mass of air in the room = 25,072.21 x 0.00118 = 29.59 g = 0.0296 kg

Given that the lethal dose of HCN is approximately 300 mg HCN per kilogram of air when inhaled, thus the <span>amount of HCN that gives the lethal dose in the small laboratory room is given by 300 x 0.0296 = 8.88 mg.</span>
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Answer:

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3 0
2 years ago
4t+3c=$81.00<br> 10t+3c=$135.00
muminat
\left \{ {{4t+3c=81.00\:(I)} \atop {10t+3c=135.00\:(II)}} \right.
<span>Simplify the first equation by (-1)
</span>\left \{ {{4t+3c=81.00\:*(-1)} \atop {10t+3c=135.00\:\:\:\:\:}} \right.
<span>Once simplified, cancel the opposing terms.
</span>\left \{ {{-4t-\diagup\!\!\!\!3c=-81.00} \atop {10t+\diagup\!\!\!\!3c=135.00}} \right.
Now find the value of "t".
\left \{ {{-4t=-81.00} \atop {10t=135.00}} \right.
-------------------------------
6t = 54.00
t =  \frac{54.00}{6}
\boxed{\boxed{t = 9.00}}\end{array}}\qquad\quad\checkmark

Now find the value of "c", replace the found value of "t" in the first equation:
4t+3c=81.00\:(I)
4*9+3c=81.00
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c =  \frac{45.00}{3}
\boxed{\boxed{c = 15.00}}\end{array}}\qquad\quad\checkmark

Answer:
<span>The values ​​of "t" and "c" satisfying the linear system are successively: $ 9.00 and $ 15.00</span>
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Answer:

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Step-by-step explanation:

To determine the interest that Preston McCord could earn by investing $ 1,400 in an account that pays 3.2% annual interest compounded monthly, leaving said money invested for a period of 6 months, it is necessary to perform the following calculation:

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