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Natali [406]
3 years ago
14

What is the number of electrons shared between the atoms in I2 molecule (That is an i not a 1)

Chemistry
1 answer:
labwork [276]3 years ago
4 0
Iodine is one of seven elements that usually form as diatomic molecules. These other "diatomics" are O_2, H_2, F_2, Br_2, N_2, and Cl_2. They typically bond to themselves.

Iodine has seven valence electrons. Using valence shell electron pair repulsion (VSEPR) theory, we can predict the Lewis dot structure for I_2. We see there are fourteen electrons, and we can make a covalent bond, leaving three lone pairs on each atom. The bond replaces two electrons, so the number of electrons shared is two.

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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.

5 0
4 years ago
What are the answers to these?
antiseptic1488 [7]

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Explanation:

Balance the equation

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Always leave hydrogen and oxygen for last.

If you have more than one element left to balance:

Add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation.

Balance the hydrogen atoms next.

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D. The particles are tightly packed together
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Some of the reactants or the products are in the gaseous phase.
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Answer:

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Explanation:

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