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Sergeeva-Olga [200]
2 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]2 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]2 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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Rank the boiling points of the following molecules from highest to lowest. butanone diethyl ether butane and butanol.
Gekata [30.6K]

Answer:

From highest to lowest:

butanol: 117.7 degree Celsius

butanone: 79.64 degree Celsius

diethyl ether: 34.6 degree Celsius

n-butane: -0.4 degree Celsius

7 0
2 years ago
A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net
Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0
2 years ago
Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces
Shalnov [3]

Answer:  The concentration of excess [OH^-] in solution is 0.017 M.

Explanation:

1. Molarity=\frac{moles}{\text {Volume in L}}

moles of HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles

1 mole of HCl give = 1 mole of H^+

Thus 0.019 moles of HCl give = 0.019 mole of H^+

2. moles of Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles

According to stoichiometry:

1 mole of Ba(OH)_2 gives = 2 moles of OH^-

Thus 0.012 moles of Ba(OH)_2 give = 2 \times 0.012=0.024 moles of OH^-

H^++OH^-\rightarrow H_2O

As 1 mole of H^+ neutralize 1 mole of OH^-

0.019 mole of H^+ will neutralize 0.019 mole of OH^-

Thus (0.024-0.019)= 0.005 moles of OH^- will be left.

[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M

Thus molarity of [OH^-] in solution is 0.017 M.

4 0
2 years ago
Read 2 more answers
How does the arrangement of atoms in a mineral relate to its properties?
MakcuM [25]
The arrangement of atoms in a mineral can change its physical and chemical properties. 
Diamonds and coal are both made of carbon, however, their chemical and physical properties are very different. 

6 0
2 years ago
Read 2 more answers
Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4
Ghella [55]

Answer:

The K sp Value is  K_{sp}=7.40

Explanation:

From the question we are told that

   The of KClO_3 is = 122.5 g/ mol

    The mass of KClO_3 dissolved is m_s = 4.0g

    The volume of solution is  V_s = 12mL = 12*10^{-3}L

The number of moles of KClO_3 is mathematically evaluated as

           No \ of  \ moles  \ = \frac{mass }{Molar \ mass}

Substituting values

                                  = \frac{4}{122.5}

                                  =0.0327\ moles

Generally concentration is mathematically represented as

         concentration = \frac{No \ of \ moles}{volume }

For KClO_3        

               Z= \frac{0.0327}{12*10^{-3}}

                              =2.72 \ mol/L

The dissociation reaction of KClO_3  is

         KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}

The solubility product constant is mathematically represented as

                   K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }

Since there is no ionic reactant we have

                  K_{sp} = [k^+] [ClO_3^-]

                          = Z^2

                          = 2.72^2

                          K_{sp}=7.40

                         

5 0
2 years ago
Read 2 more answers
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