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Yanka [14]
3 years ago
5

Elements are arranged in groups by similar atomic structure on the periodic table. This allows for an element's properties to be

predicted based on general periodic trends. One of these trends, ionization energy, increases up a group and to the right along a period and can be defined as the _______. A) energy required to add an electron B) energy required to excite an electron C) energy required to remove a core electron D) energy required to remove a valence electron
Chemistry
2 answers:
REY [17]3 years ago
5 0

Answer:

D

Explanation:

svlad2 [7]3 years ago
4 0

D) energy required to remove a valence electron

Explanation:

The ionization energy is the energy required to remove a valence electron from an element.

Different kinds of atoms bind their valence electrons with different amount of energy.

  • To remove the electrons, energy must be supplied to the atom.
  • The amount of energy required to remove the an electron in the valence shell is the ionization energy or ionization potential.
  • The first ionization energy is the energy needed to remove the most loosely bound electron in an atom in the ground state.
  • The ionization energy measures the readiness of an atom to loose electrons.

Learn more:

Ionization energy brainly.com/question/5880605

#learnwithBrainly

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

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Answer:

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Explanation:

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