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Allushta [10]
2 years ago
6

For the following pair, indicate which element has the lower first ionization energy: Match the words in the left column to the

appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
1. Given the elements Cl and Ge ,_________ has the smaller first ionization energy.
2. Given the elements Te and Se , __________ has the smaller first ionization energy.
3. Given the elements Ba and Ti , __________ has the smaller first ionization energy.
4. Given the elements Cu and Ag ,___________ has the smaller first ionization energy.

to fill in the blank:

(A) Ag
(B) Ge
(C) Cl
(D) Ti
(E) Cu
(F) Ba
(G) Se
(H) Te
Chemistry
1 answer:
emmainna [20.7K]2 years ago
5 0

<u>Answer:</u>

<u>For 1:</u> The correct answer is Ge.

<u>For 2:</u> The correct answer is Te.

<u>For 3:</u> The correct answer is Ba.

<u>For 4:</u> The correct answer is Ag.

<u>Explanation:</u>

Ionization energy is defined as the energy required to remove an electron from the outermost shell of an isolated gaseous atom. It is represented as E_i

X(g)\rightarrow X^+(g)+1e^-;E_i

Ionization energy increases as we move from left to right in a period. This happens because the atomic radius of an element decreases moving across a period, which increases the effective attraction between the negatively charged electrons and positively-charged nucleus. Hence, the removal of electron from the outermost shell becomes difficult and requires more energy.

Ionization energy decreases on moving from top to bottom in a group. This happens because the number of shells increases as we move down the group. The electrons get added in the new shell. So, the shielding of outermost electrons from the inner ones is more which decreases the attraction between the electrons and the nucleus. Hence, the removal of electron from the outermost shell becomes easy and requires less energy.

For the given options:

  • <u>Option 1:</u>

Chlorine is the 17th element of the periodic table belonging to Period 3 and Group 17.

Germanium is the 32nd element of the periodic table belonging to Period 4 and Group 14.

Hence, germanium will have smaller first ionization energy.

  • <u>Option 2:</u>

Tellurium is the 52nd element of the periodic table belonging to Period 5 and Group 16.

Selenium is the 34th element of the periodic table belonging to Period 4 and Group 16.

Hence, tellurium will have smaller first ionization energy.

  • <u>Option 3:</u>

Barium is the 56th element of the periodic table belonging to Period 6 and Group 2.

Titanium is the 22nd element of the periodic table belonging to Period 4 and Group 4.

Hence, barium will have smaller first ionization energy.

  • <u>Option 4:</u>

Copper is the 29th element of the periodic table belonging to Period 4 and Group 11.

Silver is the 47th element of the periodic table belonging to Period 5 and Group 11.

Hence, silver will have smaller first ionization energy.

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<span>ΔU = -653J + 386J </span>
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<span>ΔU = 322J + 0 </span>
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<span>Thus internal energy increase by 322J</span>
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= 2K^+(aq) + SO4^2-(aq) + Ca^2+(aq) + 2i^-(aq) --->CaSO4(s) + 2K^+(aq) +2 i^-(aq)

cancel the spectator ions that is 2k^+ and 2i^-

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How many grams of Cr can be produced by the reaction of 44.1 g of Cr2O3 with 35.0 g of Al according to the following chemical eq
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Answer:

30.17 grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al.

19.33 grams of the excess react will remain once the reaction goes to completion.

Explanation:

You know:

2 Al + Cr₂O₃ → Al₂O₃ + 2 Cr

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of the compounds participating in the reaction react and are produced:

  • Al: 2 moles
  • Cr₂O₃: 1 mole
  • Al₂O₃: 1 mole
  • Cr: 2 moles

Being:

  • Al: 27 g/mole
  • Cr: 52 g/mole
  • O: 16 g/mole

the molar mass of the compounds participating in the reaction is:

  • Al: 27 g/mole
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  • Al₂O₃: 2*27 g/mole + 3 *16 g/mole= 102 g/mole
  • Cr: 52 g/mole

Then, by stoichiometry of the reaction, the amounts of reagents and products that participate in the reaction are:

  • Al: 2 moles* 27 g/mole= 54 g
  • Cr₂O₃: 1 mole* 152 g/mole= 152 g
  • Al₂O₃: 1 mole* 102 g/mole= 102 g
  • Cr: 2moles*52 g/mole= 104 g

First, you must determine the limiting reagent. The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

Then, for this you must apply the following rule of three: if 54 grams of Al react with 152 grams of Cr₂O₃ by stoichiometry of the reaction, 35 grams of Al with how much mass of Cr₂O₃ will react?

mass of Cr_{2} O_{3} =\frac{35 grams of Al*152 gramsof Cr_{2} O_{3}}{54 grams of Al}

mass of Cr₂O₃= 98.52 grams

But 98.52 grams of Cr₂O₃ are not available, 44.1 grams are available. Since you have less moles than you need to react with 35 grams of Al, the compound Cr₂O₃ will be the limiting reagent.

To determine the amount of Cr that can be produced, you use the amount of limiting reagent available and apply the following rule of three: if by stoichiometry 152 grams of Cr₂O₃ produce 104 grams of Cr, 44.1 grams of Cr₂O₃ how much mass of Cr does it produce?

mass of Cr =\frac{104 grams of Cr*44.1 grams of Cr_{2} O_{3}}{152 grams of Cr_{2} O_{3}}

mass of Cr= 30.17 grams

<u><em>30.17 grams of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al.</em></u>

As Al is the excess reagent, you must first calculate the amount of mass that reacts with 44.1 grams of Cr₂O₃ using the following rule of three: if 152 grams of Cr₂O₃ react with 54 grams of Al, 44.1 grams of Cr₂O₃ with how much mass of Al reaction to?

mass of Al =\frac{54 grams of Al*44.1 gramsnof Cr_{2} O_{3}}{152 gramsnof Cr_{2} O_{3}l}

mass of Al=15.67 grams

With 35 grams being the amount of Al available, the amount of Al that will remain in the reaction after all the limiting reagent reacts and the reaction is complete is calculated by:

mass of excess= 35 grams - 15.67 grams= 19.33 grams

<em><u>19.33 grams of the excess react will remain once the reaction goes to completion.</u></em>

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2 years ago
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