Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Let's start with calculating the concentration of the fluoride ion with the provided data:
[tex]\%m/v\text{ F=}\frac{20.8\text{ g}}{1.37*10^4L}=1.5182\text{ *10}^{-3}\text{ g/L}[/tex]To convert to ppm, we have to remember the formula:
[tex]ppm=\frac{mass\text{ of substrate \lparen mg\rparen}}{mass\text{ of the solution \lparen kg\rparen}}[/tex]But, as the dissolvent is water and we can assume that the density is 1 g/ml, the same as 1 kg/L, we can calculate the ppm by dividing the mg of F- into the provided volume of water:
[tex]ppm=\frac{(20.8*1000)\text{ mg F}}{1.37*10^4\text{ L sln}}=1.5182\text{ ppm F}^-\text{ }[/tex]So, the answer is 1.5182 ppm F-.
For the second part, we only have to notice that ppm is the same (in this case) that mg/L. So, it means that the level is acceptable as it is 1.5182 mg/L, which is lower than 2 mg/L.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
