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m_a_m_a [10]
3 years ago
11

Which requires more energy to move an electron?

Chemistry
2 answers:
Mandarinka [93]3 years ago
4 0

Answer:

From n=1 to n=2

Explanation:

Electrons in n=1 are strongly attracted to the nucleus and therefore will require great force to overcome the electrostatic force of attraction to displace them from the energy level to another.

The electrostatic force reduces as you progress to the outer energy levels.

MrRa [10]3 years ago
4 0

<u>Answer:</u>

<em>C. from n = 1 to n = 2</em>

<u>Explanation:</u>

A.

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2 )}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(4^2 )}-\frac {1}{(3^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(16 )}-\frac {1}{(9)}]KJ mol^{-1}

\Delta E=-1312[0.0625-0.111]KJ mol^{-1}

\Delta E=64 KJ mol^{-1}

B.

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2 )}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(3^2 )}-\frac {1}{(2^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(9 )}-\frac {1}{(4)}]KJ mol^{-1}

\Delta E=-1312[0.111-0.25]KJ mol^{-1}

\Delta E=182 KJ mol^{-1}

C.

\Delta E=E_{final}-E_{initial}

\Delta E=-1312[\frac{1}{(n_f^2 )}-\frac {1}{(n_i^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(2^2 )}-\frac {1}{(1^2 )}]KJ mol^{-1}

\Delta E=-1312[\frac{1}{(4)}-\frac {1}{(1)}]KJ mol^{-1}

\Delta E=-1312[0.25-1]KJ mol^{-1}

\Delta E=984 KJ mol^{-1}

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How can the speed of solvent molecules be slowed down?
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Explanation:

The speed of molecules increases when temperature is increased as it will result in more number of collisions between the molecules. Thus, there will be  increase in kinetic energy of molecules and increase in the speed of solvent molecules.

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Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib
KengaRu [80]

Answer:

<u>For methanol:</u> Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

<u>For ethanol: </u>Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

<u>For propanol: </u>Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

<u>For methanol:</u>

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g  (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

<u>Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)</u>

<u></u>

<u>For ethanol:</u>

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g  (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

<u>Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)</u>

<u></u>

<u>For propanol:</u>

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

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<u>Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)</u>

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