An ore contains \(2.296 \%\) of the mineral argentite, \(\mathrm{Ag}_{2} \mathrm{~S}\), by mass. How many grams of this ore would have to be

The given percentage of argentite in the ore is \(2.296\%\). This means that there are \(2.296\) grams of Argentite (Ag₂S) per \(100\) grams of the ore.

In order to convert the given percentage to a fraction, we need to divide the percentage by 100. Therefore, we have: \(\frac{2.296}{100} = 0.02296\) So, \(2.296\%\) equals to \(0.02296\) in fraction.

Let's assume that we require \(x\) grams of argentite (Ag₂S). Let \(y\) be the mass of the ore required to obtain this amount of argentite. We can now apply the fraction we obtained in step 2 to find the relationship between \(x\) and \(y\) as follows: \(\frac{x \,\text{grams of Ag}_2\text{S}}{y\,\text{grams of ore}} = 0.02296\)

To find the mass of the ore (\(y\)) required to obtain \(x\) grams of argentite (Ag₂S), we can rewrite the equation above as: \(y\,\text{grams of ore} = \frac{x\,\text{grams of Ag}_2\text{S}}{0.02296}\) Now, we can plug in the desired mass of argentite to find the mass of the ore needed. For example, if we need \(1\) gram of argentite, we can substitute \(x=1\) and solve for \(y\): \(y\,\text{grams of ore} = \frac{1\,\text{gram of Ag}_2\text{S}}{0.02296} \approx 43.515 \,\text{grams of ore}\) So for this example, approximately \(43.515\) grams of the ore would have to be mined to obtain \(1\) gram of argentite (Ag₂S). This solution can be easily adapted to find the required mass of the ore for any desired mass of argentite by substituting the appropriate value for \(x\).