Question

matching 10 points
match the following synonyms.
atom
solution
molecule
molecular compound
homogeneous mixture
element
multiple choice 10 points
arresting cables on aircraft carriers are made of a high - grade steel with a density of 0.284 lb/in³. what volume, in l of steel would be required to make a 228 kg cable?
4.50 l
0.0116 l

Explanation:

Step1: Convert mass to pounds

We know that 1 kg = 2.20462 lb. So for a 228 - kg cable, the mass in pounds is $m = 228\times2.20462\ lb\approx502.653\ lb$.

Step2: Use the density - mass - volume formula

The density formula is $
ho=\frac{m}{V}$, where $
ho$ is density, $m$ is mass and $V$ is volume. We can re - arrange it to solve for volume $V=\frac{m}{
ho}$. Given $
ho = 0.284\ lb/in^{3}$, then $V=\frac{502.653\ lb}{0.284\ lb/in^{3}}\approx1770\ in^{3}$.

Step3: Convert cubic inches to liters

We know that 1 in³ = 0.0163871 L. So $V = 1770\ in^{3}\times0.0163871\ L/in^{3}\approx29.0\ L$. But if we assume there is a calculation error above and re - calculate:

1. First convert mass to pounds: $m = 228\ kg\times2.20462\ lb/kg=502.65336\ lb$.
2. Then find volume in cubic inches: $V=\frac{m}{

ho}=\frac{502.65336\ lb}{0.284\ lb/in^{3}}\approx1770\ in^{3}$.

1. Convert to liters: $V = 1770\ in^{3}\times\frac{0.0163871\ L}{1\ in^{3}}\approx29.0\ L$. If we made a wrong start and use the correct conversion steps more precisely:

• $m = 228\ kg$, $

ho=0.284\ lb/in^{3}$. First convert density to $kg/L$.

• $1\ lb = 0.453592\ kg$ and $1\ in^{3}=16.3871\ cm^{3}=16.3871\times10^{- 3}\ L$.
• $

ho=\frac{0.284\ lb}{1\ in^{3}}\times\frac{0.453592\ kg}{1\ lb}\times\frac{1\ in^{3}}{16.3871\times10^{-3}\ L}\approx7.9\ kg/L$.

• Then from $

ho=\frac{m}{V}$, we have $V=\frac{m}{
ho}=\frac{228\ kg}{7.9\ kg/L}\approx28.99\ L\approx29.0\ L$.

An atom is the smallest unit of an element. A solution is a homogeneous mixture. A molecule is a molecular compound made up of two or more atoms held together by chemical bonds.

Answer:

None of the provided options (4.50 L and 0.0116 L) are correct. The correct volume is approximately 29.0 L.

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