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Anna11 [10]
3 years ago
9

What is the lowest numbered energy level where a d sublevel is found?

Chemistry
2 answers:
SpyIntel [72]3 years ago
7 0
<span>The answer would be 3.</span>
UkoKoshka [18]3 years ago
4 0

Answer : The lowest numbered energy level where a 'd' sublevel is found is 3.

Explanation : According to the theory of order of electron filling;

The 4s orbitals is observed to have a lower energy than the 3d, and so the 4s orbitals are filled first.

It is known that 4s electrons are lost first during ionisation activity. Therefore, the electrons lost first will come from the highest energy level, furthest from the influence of the nucleus.

So, the lowest energy level where 'd' sub shell will be found is 3d.

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Which of the following chemical equations is correctly balanced?
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According to the collision theory, what is the best explanation for why a higher temperature makes a reaction go faster?
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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
What is the molarity of SO4^2- in a solution prepared by mixing 2.17 g of alum, KAl(SO4)2•12H2O, with 175 mL of water? The molec
vodka [1.7K]

<u>Answer:</u> The concentration of sulfate ions in the solution is 0.0522 M

<u>Explanation:</u>

To calculate the the molarity of solution:, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of alum = 2.17 g

Molar mass of alum = 478.39 g/mol

Volume of solution = 175 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{2.17\times 1000}{474.39\times 175}\\\\\text{Molarity of solution}=0.0261M

The chemical equation for the ionization of alum follows:

KAl(SO_4)_2.12H_2O\rightarrow K^++Al^{3+}+2SO_4^{2-}+12H_2O

1 mole of alum produces 1 mole of potassium ions, 1 mole of aluminium ions, 2 moles of sulfate ions and 12 moles of water

So, concentration of sulfate ions = (2\times 0.0261)=0.0522M

Hence, the concentration of sulfate ions in the solution is 0.0522 M

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3 years ago
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