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Sergeeva-Olga [200]
3 years ago
5

A government laboratory wants to determine whether water in a certain city has any traces of fluoride and whether the concentrat

ion 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?
3.04 ppm < 4.00 ppm, safe to drink

7.60 ppm > 4 ppm, safe to drink

30.4 ppm > 4 ppm, unsafe to drink

30,400 ppm > 4 ppm, unsafe to drink
Chemistry
2 answers:
vladimir2022 [97]3 years ago
6 0
"<span>30.4 ppm > 4 ppm, unsafe to drink" is the one among the following choices given in the question that shows that the water should be declared unsafe for drinking. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that the answer has helped you.</span>
forsale [732]3 years ago
6 0

Answer: The water is unsafe to drink because 30.4 ppm > 4 ppm.

Explanation: The recommended concentration of fluoride in the water is 4.00 ppm. If the concentration of fluoride increases the recommended limit, the water is unsafe to drink and vice-versa.

Parts Per Million (ppm) is the measurement of the concentration of the solution. It is expressed in mg/kg, which means:

ppm=\frac{\text{Weight of solute (in mg)}}{\text{Weight of the sample ( in kg)}}

We are given a sample of 5.00 grams and the solute is fluoride which has a weight of 0.152 mg.

Weight of the sample = 5.00 grams = 0.005 kg     (Conversion factor: 1 kg = 1000g)

Putting values in ppm equation, we get:

ppm=\frac{0.152mg}{0.005kg}=30.4mg/kg=30.4ppm

As the ppm value is more than the recommended value. Hence, the water is unsafe to drink.

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Explanation:

It is given that vapor pressure of liquid iodomethane is 40.0 mm Hg. So, if we calculate the vapor pressure according to the given values and if its value will be greater than the the given vapor pressure of iodomethane then it means that some of the vapors has converted into liquid state.

As the given values are as follows.

      P_{1} = 72.0 mm Hg,       T_{1} = 404 K

      P_{2} = ? ,               T_{2} = 249 K

As volume is constant so, according to Gay-Lussac's law pressure is directly proportional to temperature.

                      P \propto T         (at constant volume)

or,                    \frac{P}{T} = k

Therefore, the formula to calculate the value of P_{2} is as follows.

              \frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}

            \frac{72.0 mm Hg}{404 K} = \frac{P_{2}}{249 K}

                 P_{2} = 44.37 mm Hg

As calculated vapor pressure is more than the given vapor pressure. Hence, the liquid will convert into gas.

As a result, no condensation will occur and only vapors of iodomethane will be present.

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A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t
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Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x  (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is  initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for  weak acids for the equilibria:

protonated lidocaine + H₂O   ⇆  lidocaine + H₃O⁺

                      protonated lidocaine          lidocaine        H₃O⁺

Initial(M)               0.22                                       0                  0

Change                   -x                                      +x                 +x

Equilibrium          0.22 - x                                  x                    x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] =  x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸  = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22  - x  ≈ 0.22

and solve for x. The pH  will then  be the negative log of this value.

8.71 x 10⁻⁷  = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

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3 years ago
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