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larisa [96]
3 years ago
10

Consider four elements from Group 7A: fluorine in the second period, chlorine in the third period, bromine in the fourth period,

and iodine in the fifth period. Which element has the largest first ionization energy?
Chemistry
2 answers:
Lyrx [107]3 years ago
7 0
Ionization energy is the energy required to lose an electron and form an ion. The stronger is the attraction of the atom and the electron the higher the ionization energy, and the weaker is the attraction of the atom and the electron the higher the ionization energy. This leads to a clear trend in the periodic table. Given that the larger the atom the weaker the attraction of the atom to the valence electrons, the easier they will be released, and the lower the ionization energy. This is, as you go downward in a group, the ionization energy decreases. So, the element at the top of the group will exhibit the largest ionization energy. <span>Therefore, the answer is that of the four elements of group 7A, fluorine will have the largest first ionization energy.</span>
polet [3.4K]3 years ago
4 0
Plato Users, A) Fluorine
#PlatoFam
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Consider the following reaction at equilibrium. What will happen if SO2SO2 is added to the reaction? 4FeS2FeS2(s) + 11O2O2(g) ⇌⇌
kupik [55]

Answer:

The equilibrium will change in the direction of the reactants

Explanation:

7 0
3 years ago
What is the work done when a force of 5 N is applied to a ball and it moves 8000 cm?
kolezko [41]

Answer:

Explanation:Given;

applied force, F = 8000 N

time of force application, t = 15 s

Work done is given as the product of force and displacement. Since the car is unable to move, then the displacement is zero and the work done is zero.

Work done = Force x displacement

Work done = 8000 N x 0

Work done = 0

Therefore, the work done is zero.

3 0
3 years ago
Describe the results of Ernest Rutherford's gold-foil experiment and explain how his results changed ideas about the distributio
sashaice [31]
Physicist Ernest Rutherford<span> established the nuclear theory of the atom with his </span>gold-foil experiment<span>. When he shot a beam of alpha particles at a sheet of </span>gold foil<span>, a few of the particles were deflected. He concluded that a tiny, dense nucleus was causing the deflections.</span>
4 0
4 years ago
Read 2 more answers
The normal boiling point of ethanol (C2H5OH) is 78.3 oC and its molar enthalpy of vaporization is 38.56 kJ/mol. What is the chan
Sloan [31]

Answer:

The  change in entropy in the system is -231.5 J/K

Explanation:

Step 1: Data given

The normal boiling point of ethanol (C2H5OH) is 78.3 °C = 351.45 K

The molar enthalpy of vaporization of ethanol is 38.56 kJ/mol = 38560 J/mol

Mass of ethanol = 97.2 grams

Pressure = 1 atm

Step 2: Calculate the entropy change of vaporization

The entropy change of vaporization = molar enthalpy of vaporization of ethanol / temperature

The entropy change of vaporization = 38560 J/mol / 351.45 K

The entropy change of vaporization = -109.72 J/k*mol

Step 3: Calculate moles of ethanol

Moles ethanol = mass / molar mass ethanol

Moles ethanol = 97.2 grams / 46.07 g/mol

Moles ethanol = 2.11 moles

Step 4: Calculate  change in entropy

For 1 mol = -109.72 J/K*mol

For 2.11 moles = -231.5 J/K

The  change in entropy in the system is -231.5 J/K

7 0
3 years ago
A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
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