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Gnom [1K]
3 years ago
11

Nitric acid (specific gravity 1.42) costs $51.95 per liter. A student spills 5.112 kg of nitric acid in the laboratory, what is

the cost of the nitric acid wasted?
Mathematics
1 answer:
Leno4ka [110]3 years ago
5 0

Answer:

The cost of wasted nitric acid is $187.02

Step-by-step explanation:

Since it is given that specific gravity of Nitric Acid is 1.42 Thus the density of nitric acid is

\rho _{acid}=S.G\times \rho _{water}\\\\\\\rho_{acid}=1.42\times 1\frac{g}{cm^3}\\\\\rho_{acid}=1.42g/cm^3

Now from the basic relation of mass volume and density we have

\rho =\frac{Mass}{Volume}\\\\\therefore Volume=\frac{Mass}{\rho }

Since the mass of the nitric acid is given to be 5.112 kg = 5112 grams

Thus the volume of nitric acid is

Volume=\frac{5112}{1.42}=3600cm^3

Now we know that 1000 cubic centimeters euals 1 litrer

Thus the calculated volume in liters is

V_{liters}=\frac{3600}{1000}=3.6Liters

Thus the cost of acid spilled  equals

Cost=3.6\times 51.95=187.02

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The center of mass is mathematically given as

\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}

<h3>What is the center of mass.?</h3>

Determine the center of mass in one dimension:

Represent the masses at the respective distances.

\begin{|c|c|} Masses \ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ Located at \\\rho=x^{3}+x \cdot e^{-x} & \ \  \ \  x \in(0,1)$ \\\end

We calculate the total mass of the system.

\begin{aligned}m &=\int_{0}^{1} \rho \cdot d x \\& m =\int_{0}^{1}\left(x^{3}+x \cdot e^{-x}\right) \cdot d x \\&m =\left|\frac{x^{4}}{4}-(x+1) e^{-x}\right|_{0}^{1} \\&m =\left(\frac{5}{4}-\frac{2}{e}\right)\end{aligned}

Step 03: Calculate the moment of the system.

\begin{aligned}M &=\int_{0}^{1}(\rho \cdot x) \cdot d x \\& M=\int_{0}^{1}\left(x^{4}+x^{2} \cdot e^{-x}\right) \cdot d x \\&M =\left|\frac{x^{5}}{5}-\left(x^{2}-2 x+2\right) \cdot e^{-x}\right|_{0}^{1} \\&M=\left(\frac{11}{5}-\frac{5}{e}\right)\end{aligned}

we calculate the center of mass.

\begin{aligned}\bar{x} &=\left(\frac{M}{m}\right) \\& \bar{x}=\left\{\left(\frac{\left.11-\frac{5}{5}\right)}{\left(\frac{5}{4}-\frac{2}{e}\right)}\right\}\right.\\& \bar{x}=\left(\frac{11 e-25}{5 e}\right) \cdot\left(\frac{4 e}{5 e-8}\right) \\&\bar{x}=\left(\frac{44 e-100}{25 e-40}\right)\end{aligned}

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2 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
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Answer:

a) 0.283  or 28.3%

b) 0.130 or 13%

c) 0.4 or 40%

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Step-by-step explanation:

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z=\frac{X-M}{s} where

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Then

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