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harina [27]
2 years ago
8

The sum of IE₁ through IE₄ for Group 4A(14) elements shows a decrease from C to Si, a slight increase from Si to Ge, a decrease

from Ge to Sn, and an increase from Sn to Pb.(a) What is the expected trend for IEs down a group?
Chemistry
1 answer:
Anna007 [38]2 years ago
4 0

The correct answer is IE decreases down the group.

On moving down the group, the size of the element increases due to the increase in the number of shells. The element with smallest size has less the attraction of the nucleus on the valance electron. It needs more energy to remove an electron from its valance shell. Hence, the IE decreases down the group.

Why does ionization energy increase down the group but decreases going across a period?

Because there are more protons with time, the ionization energy rises. As a result, there will be more attraction because the nuclear charge has increased.

Even if there is stronger attraction, one should be aware that the shielding effect and distance from the nucleus remain largely constant. The same primary quantum shell contains all of the valence electrons, which explains this.

Therefore, while distance from the nucleus and the shielding effect stay fairly constant, an increase in nuclear charge causes an increase in attraction and increases the energy required to remove an electron.

The ionization energy drops with each group. This is because the outside electrons acting as a shield or screen for the nucleus make the attraction between them weaker and make it easier for them to be withdrawn.

To learn more about ionization energy refer the link:

brainly.com/question/28305735

#SPJ4

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NH4NO3 + Na3PO4 → (NH4)3PO4 + NaNO3
Papessa [141]

Based on the equation of the reaction and the data provided,

  • NH4NO3 is the limiting reactant
  • mass of Na3PO4 left is 29.375 g
  • 18.75 g of (NH4)3PO4 is produced
  • 31.875 g of NaNO3 is produced

<h3>What are limiting reactants?</h3>

A limiting reactant is a reactant which is used up in a reaction after which the reaction stops.

In the given reaction:

3 NH4NO3 + Na3PO4 → (NH4)3PO4 + 3 NaNO3

The limiting reactant is determined from the stoichiometry of the eqaution.

Moles of reactant = mass/molar mass

Molar mass of NH4NO3 = 80 g/mol

Molar mass of Na3PO4 = 165 g/mol

Molar mass of (NH4)3PO4 = 150 g/mol

Molar mass of NaNO3 = 85 g/mol

From the equation of the reaction, 240 g (3 × 80) of NH4NO3 is required to react with 165 g of Na3PO4

There are only 30.0 g of NH4NO3 reacting with 50.0 g of Na3PO4

30 g of Na3PO4 will react with 30 × 165/240 = 20.625 g of Na3PO4

Therefore, NH4NO3 is the limiting reactant

Na3PO4 is the excess reactant

mass of Na3PO4 left = 50 - 20.625

mass of excess reactant left = 29.375 g

moles of NH4NO3 in 30 g = 30/80 = 0.375 moles

3 moles of NH4NO3 produces 1 mole of (NH4)3PO4

0.375 moles of NH4NO3 will produce 0.375 × 1/3 = 0.125 moles of (NH4)3PO4

mass of 0.125 moles of (NH4)3PO4 = 0.125 × 150

mass of (NH4)3PO4 produced = 18.75 g of (NH4)3PO4

3 moles of NH4NO3 produces 3 moles of NaNO3

0.375 moles of NH4NO3 will produce 0.375 moles of NaNO3

mass of 0.375 moles of NaNO3 = 0.375 × 85

mass of NaNO3 produced = 31.875 g of NaNO3

Learn more about limiting reactant at: brainly.com/question/24945784

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2 years ago
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