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alexdok [17]
3 years ago
11

Which statement best describes how dynamic equilibrium relates to a saturated solution of sodium chloride? The concentration of

the dissolved particles remains constant. No more solid dissolves in a saturated solution. Each second, the amount of solid sodium chloride that dissolves equals the amount of solid sodium chloride that recrystallizes. Each second, solid sodium chloride dissolves and solid sodium chloride recrystallizes.
Chemistry
2 answers:
Nookie1986 [14]3 years ago
8 0

Answer:

C) Each second, the amount of solid sodium chloride that dissolves equals the amount of solid sodium chloride that recrystallizes.

Explanation:

kati45 [8]3 years ago
4 0
<span>Each second, solid sodium chloride dissolves and solid sodium chloride recrystallizes. </span>
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The answer is B plastic bags made from petroleum.
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A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
spin [16.1K]

Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

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4 years ago
A sample of Neon is in a sealed container held under isothermic conditions. The initial pressure and volume are 2.7 atm and 4.5
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Answer:

The final volume in mL is 7.14 mL or 7.1 mL.

Explanation:

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<em />

<em>2. Plug in values. </em>V_{2} =\frac{(2.7 atm)(4.5 mL)}{(1.7 atm)}  = 7.14 mL

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<span>Photosynthesis Is a Reaction To Make Food</span>
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The pka of hf is 3.2 determine the pkb of hf?
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Well, first we must remember that

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This is because

K_{a}*K_{b}=10^{-14}

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So then

pK_{b}=14-pK_{a}=14-3.2=1.8

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4 years ago
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