A government laboratory wants to determine whether water in a certain city has any traces of fluoride and whether the concentration exceeds the recommended 4.00 ppm. A 5.00-g sample of this water is found to have 0.152 mg of fluoride. Should the water be declared safe for drinking?

The given sample of water is not safe for drinking.

Explanation:

We are given:

Concentration of fluorine in water recommended = 4.00 ppm

ppm is the amount of solute (in milligrams) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of fluorine in water, we use the equation:

Both the masses are in grams.

We are given:

Mass of fluorine =    (Conversion factor:  1 g = 1000 mg)

Mass of water = 5.00 g

Putting values in above equation, we get:

As, the calculated concentration is greater than the recommended concentration. So, the given sample of water is not safe for drinking.

Hence, the given sample of water is not safe for drinking.

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol