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Iteru [2.4K]
3 years ago
8

Where are the most active nonmetals located on the periodic table

Chemistry
2 answers:
natulia [17]3 years ago
5 0

Answer:

Group 7

Explanation:

Elements in the group 7 of the periodic table are known as the halogens. They are the most active/reactive nonmetals because they have just one electron to receive in order to achieve there octet configuration. They react easily with elements in group 1 (in the same proportion) since group 1 elements have just one electron in there outermost shell. Halogens easily form acids with hydrogen by sharing the only electron hydrogen has with its own valence electron via covalent bonding. They however undergo ionic bonding with the remaining members of group 1. They also react with metals in group 2, 3 and 4 in different proportion.

AURORKA [14]3 years ago
4 0

Answer:

The most active nonmetals belong to the halogen family, which sits to the left of the noble gases on the right side of the periodic table. The halogens are so reactive that they are never found in nature by themselves. The elements fluorine, chlorine, bromine, iodine and astatine make up the halogen family.

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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
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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

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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}

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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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