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stiv31 [10]
3 years ago
15

Each row in the periodic table corresponds to one energy level in an atom. True Or False ?

Chemistry
2 answers:
Nitella [24]3 years ago
8 0
For the answer to the question above asking, e<span>ach row in the periodic table corresponds to one energy level in an atom.

The answer is true.

I hope my answer helped you. Feel free to ask more questions. Have a  nice day!</span>
777dan777 [17]3 years ago
7 0

Answer:

True

Explanation:

Let's take examples. Hydrogen and helium are in the first row because they have a total of one energy levels each due to the configuration of H: 1 and He: 2.

Sodium is in period three because that it has three energy levels carrying atoms. Look at the following configurations

Na:2, 8, 1.  S: 2,8,6 . Both elements are in period/row three and they both have three energy levels

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The mass of C02 contains same number of molecules as there are number of atoms in 2.3 g of sodium
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1- mole of any Substance will have the same number of Molecules.

Molecular weight of CO2= 44

Molecular weight of H20= 18

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8 0
3 years ago
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.

5 0
4 years ago
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