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Alecsey [184]
3 years ago
14

The nth shell holds a minimum number of 2n2 electrons True False Other:

Chemistry
1 answer:
BARSIC [14]3 years ago
8 0

Answer: The statement the  n^{th} shell holds a minimum number of 2n^{2} electrons is true.

Explanation:

The alphabet 'n' denotes the principal quantum number whose value can be equal to 1, 2, 3, and so on but can never be zero.

Basically, 'n' is the number of shell of an atom and total number of orbitals present in a shell is n^{2}.

Each orbital can hold up to a maximum of wo electrons. Hence, the n^{th} shell can hold a minimum number of 2n^{2} electrons.

Thus, we ca conclude that the statement the  n^{th} shell holds a minimum number of 2n^{2} electrons is true.

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Explanation:

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{ \boxed{ \bf{vapour \: density = 2 \times molecular \: mass}}} \\{ \tt{ PV= (\frac{m}{ m_{r}}) RT}} \\ { \tt{3 \times 5.6 =  \frac{11}{m _{r}}  \times 0.0831 \times 273}} \\ { \tt{m _{r} = 14.85 \: g}} \\  \\ { \bf{vapour \: density = 2 \times m _{r}}} \\  = 2 \times 14.85 \\  = 29.7 \: { \tt{g {dm}^{ - 3} }}

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We calculate the amount of carbon fixed as follows.

          455 \times 10^{3} kg/km^{2} \times 4.46 \times 10^{7} km^{2}

            = 2029.3 \times 10^{10} kg

\Delta H_{f}_{C_{6}H_{12}O_{6}} = -1273.1 kJ/mol

\Delta H_{f}_{CO_{2}} = -393.5 kJ/mol

\Delta H_{f}_{H_{2}O} = -285.8 kJ/mol

Hence, total \Delta H reaction will be as follows.

             \Delta H_{rxn} = (\Delta H_{f}_{C_{6}H_{12}O_{6}}) - 6(\Delta H_{f}_{CO_{2}}) - 6(\Delta H_{f}_{H_{2}O})

                           = -1273.1 kJ/mol - (6 \times 393.5 kJ/mol) - (6 \times 285.8 kJ/mol)

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Therefore, calculate the number of moles of fixed carbon as follows.

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                   = 3.786 \times 10^{7} mol km^{-2} year^{-1}

Thus, we can conclude that the annual enthalpy change resulting from the photosynthetic carbon fixation over the land surface is 3.786 \times 10^{7} mol km^{-2} year^{-1}.

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