Question

1. convert 0.6590 kg/l into g/el using dimensional analysis. show complete work with units and answer with the correct number of significant figures.

Explanation:

Step1: Recall unit conversions

We know that \(1\space \text{Kg} = 1000\space \text{g}\) and \(1\space \text{EL}=10^{18}\space \text{L}\) (assuming EL here is exaliter, a common metric prefix where \(1\space \text{Exa}=10^{18}\)). So we need to convert \(\text{Kg}\) to \(\text{g}\) and \(\text{L}\) to \(\text{EL}\).
The given value is \(0.6590\space \text{Kg/L}\). First, convert \(\text{Kg}\) to \(\text{g}\): multiply by \(1000\space \text{g/Kg}\), and convert \(\text{L}\) to \(\text{EL}\): divide by \(10^{18}\space \text{L/EL}\) (since \(\text{EL}\) is larger than \(\text{L}\), to get per \(\text{EL}\), we divide by the number of \(\text{L}\) in an \(\text{EL}\)).
So the conversion factor for \(\text{Kg}\) to \(\text{g}\) is \(1000\space \text{g/Kg}\) and for \(\text{L}\) to \(\text{EL}\) is \(\frac{1}{10^{18}\space \text{L/EL}}\) or \(10^{- 18}\space \text{EL/L}\).
The formula for conversion will be:
\(0.6590\space \frac{\text{Kg}}{\text{L}}\times\frac{1000\space \text{g}}{1\space \text{Kg}}\times\frac{1\space \text{EL}}{10^{18}\space \text{L}}\)

Step2: Perform the calculations

First, multiply the numerical values and the units separately.
For the numerical part: \(0.6590\times1000\times10^{- 18}\)
\(0.6590\times1000 = 659.0\)
Then \(659.0\times10^{-18}=6.590\times 10^{-16}\) (using scientific notation, and keeping the correct number of significant figures. The original number \(0.6590\) has 4 significant figures, and the conversion factors are exact (1000 and \(10^{18}\) are exact definitions), so the result should have 4 significant figures)
For the units: \(\frac{\text{Kg}}{\text{L}}\times\frac{\text{g}}{\text{Kg}}\times\frac{\text{EL}}{\text{L}}=\frac{\text{g}\cdot\text{EL}}{\text{L}^2}\)? Wait, no, wait. Wait, the original unit is \(\text{Kg/L}\), we are converting to \(\text{g/EL}\). Let's re - express the unit conversion:
We want to go from \(\text{Kg/L}\) to \(\text{g/EL}\). So:
\(\frac{\text{Kg}}{\text{L}}=\frac{\text{Kg}}{\text{L}}\times\frac{1000\space \text{g}}{1\space \text{Kg}}\times\frac{1\space \text{EL}}{10^{18}\space \text{L}}\)
The \(\text{Kg}\) cancels, and \(\text{L}\) in the denominator of the first term and \(\text{L}\) in the denominator of the third term? Wait, no, the third term is \(\frac{1\space \text{EL}}{10^{18}\space \text{L}}\), so when we multiply \(\frac{\text{Kg}}{\text{L}}\times\frac{1000\space \text{g}}{\text{Kg}}\times\frac{\text{EL}}{10^{18}\space \text{L}}\), the \(\text{Kg}\) cancels, and we have \(\frac{1000\space \text{g}\times\text{EL}}{\text{L}\times10^{18}\space \text{L}}=\frac{1000\space \text{g}\cdot\text{EL}}{10^{18}\space \text{L}^2}\)? That can't be right. Wait, maybe there is a typo in the problem, maybe it's \(\text{g/EL}\) where we want to convert \(\text{Kg/L}\) to \(\text{g/EL}\), so actually, the correct way is:
We know that \(1\space \text{EL} = 10^{18}\space \text{L}\), so \(\frac{1}{\text{EL}}=\frac{1}{10^{18}\space \text{L}}\)
And \(1\space \text{Kg}=1000\space \text{g}\)
So \(\frac{\text{Kg}}{\text{L}}=\frac{1000\space \text{g}}{10^{18}\space \text{L}}\times\frac{1}{\text{EL}}\)? No, let's start over.
Let \(x = 0.6590\space \text{Kg/L}\)
We want to find \(x\) in \(\text{g/EL}\)
First, convert \(\text{Kg}\) to \(\text{g}\): \(0.6590\space \text{Kg/L}=0.6590\times1000\space \text{g/L}=659.0\space \text{g/L}\)
Now convert \(\text{L}\) to \(\text{EL}\): since \(1\space \text{EL} = 10^{18}\space \text{L}\), then \(1\space \text{L}=\frac{1}{10^{18}}\space \text{EL}\)
So \(\frac{\text{g}}{\text{L}}=\frac{\text{g}}{\frac{1}{10^{18}}\space \text{EL}}=10^{18}\space \text{g/EL}\)
Therefore, \(659.0\space \text{g/L}=659.0\times10^{18}\space \text{g/EL}\)? Wait, that's the opposite. Wait, no. If \(1\space \text{EL}=10^{18}\space \text{L}\), then to convert from \(\text{L}\) to \(\text{EL}\), we divide by \(10^{18}\). So if we have a quantity per \(\text{L}\), to get per \(\text{EL}\), we need to divide by \(10^{18}\) because there are \(10^{18}\) liters in an exaliter.
Wait, let's take a simple example: if we have \(1\space \text{g/L}\), how many \(\text{g/EL}\) is that? Since \(1\space \text{EL}=10^{18}\space \text{L}\), then in \(1\space \text{EL}\) of volume, the mass would be \(1\space \text{g/L}\times10^{18}\space \text{L}=10^{18}\space \text{g}\) per \(\text{EL}\). Oh! I made a mistake earlier. So the correct conversion is:
If \(y\space \text{g/L}\), then in \(\text{g/EL}\), it's \(y\times10^{18}\space \text{g/EL}\) because \(1\space \text{EL}\) has \(10^{18}\space \text{L}\), so the mass in \(1\space \text{EL}\) is \(y\space \text{g/L}\times10^{18}\space \text{L}=y\times10^{18}\space \text{g}\), so per \(\text{EL}\), it's \(y\times10^{18}\space \text{g/EL}\)
And we had \(y = 659.0\space \text{g/L}\) (from converting \(0.6590\space \text{Kg/L}\) to \(\text{g/L}\) by multiplying by \(1000\))
So now, \(659.0\space \text{g/L}=659.0\times10^{18}\space \text{g/EL}\)
But let's do it with dimensional analysis properly:
\(0.6590\space \frac{\text{Kg}}{\text{L}}\times\frac{1000\space \text{g}}{1\space \text{Kg}}\times\frac{10^{18}\space \text{L}}{1\space \text{EL}}\)
Now, the \(\text{Kg}\) cancels, \(\text{L}\) cancels (the \(\text{L}\) in the denominator of the first term and the \(\text{L}\) in the numerator of the third term), and we are left with \(\frac{\text{g}\times10^{18}}{\text{EL}}\)
Calculating the numerical part: \(0.6590\times1000\times10^{18}=0.6590\times10^{21}=6.590\times 10^{20}\) (using scientific notation, and keeping 4 significant figures as the original number \(0.6590\) has 4 significant figures)

Wait, now I see my earlier mistake. The key is that when converting from per liter to per exaliter, since an exaliter is a larger unit of volume, the number of grams per exaliter will be larger than per liter. So the correct conversion factor for \(\text{L}\) to \(\text{EL}\) is \(\frac{10^{18}\space \text{L}}{1\space \text{EL}}\) (because \(1\space \text{EL}=10^{18}\space \text{L}\)), so when we multiply \(\frac{\text{Kg}}{\text{L}}\) by \(\frac{\text{g}}{\text{Kg}}\) and \(\frac{\text{L}}{\text{EL}}\), the units work out:
\(\frac{\text{Kg}}{\text{L}}\times\frac{\text{g}}{\text{Kg}}\times\frac{\text{L}}{\text{EL}}=\frac{\text{g}}{\text{EL}}\)

So let's recast the steps:

Step1: Convert Kg to g

We know that \(1\space \text{Kg} = 1000\space \text{g}\), so multiply the given value by \(\frac{1000\space \text{g}}{1\space \text{Kg}}\)
\(0.6590\space \frac{\text{Kg}}{\text{L}}\times\frac{1000\space \text{g}}{1\space \text{Kg}} = 659.0\space \frac{\text{g}}{\text{L}}\)

Step2: Convert L to EL

We know that \(1\space \text{EL}=10^{18}\space \text{L}\), so multiply by \(\frac{10^{18}\space \text{L}}{1\space \text{EL}}\)
\(659.0\space \frac{\text{g}}{\text{L}}\times\frac{10^{18}\space \text{L}}{1\space \text{EL}}=659.0\times 10^{18}\space \frac{\text{g}}{\text{EL}}\)

Step3: Express in scientific notation

\(659.0\times 10^{18}=6.590\times 10^{20}\) (since \(659.0 = 6.590\times10^{2}\), so \(6.590\times10^{2}\times10^{18}=6.590\times10^{20}\))

Answer:

\(6.590\times 10^{20}\space \text{g/EL}\)