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3 years ago

Define and explain each trend.

1 answer:

3 0

Atomic radius- a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding cloud of electrons.
Electron Affinity-the electron affinityof an atom or molecule is defined as the amount of energy released or spent when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.
Electronegativity-a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is assigned a value of 4.0, and values range down to caesium and francium which are the least electronegative at 0.7.
Lonization energy- qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation.

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Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance

ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0

3 years ago

2.20 moles Sn to grams

mihalych1998 [28]

261.162 grams. Use the equation n=M÷Mr.

3 0

3 years ago

2NH.

liq [111]

Answer:

N=2

H=6

Explanation:

1.Balance a chemical equation in terms of moles.

2.Use the balanced equation to construct conversion factors in terms of moles.

3.Calculate moles of one substance from moles of another substance using a balanced chemical equation.

The law of conservation of matter says that matter cannot be created or destroyed. In chemical equations, the number of atoms of each element in the reactants must be the same as the number of atoms of each element in the products.

(P.s it could also be where you have to solve it in which you have to simplify it first then solve it.) like adding them all up.

Hope this is the answer. :)

6 0

3 years ago

What part of an Atom undergoes change during radioactive decay

Aloiza [94]

It is clear that the core is subject to change

7 0

3 years ago

What is the answer of these compound

olganol [36]

BaO, Barium Oxide.

Na2SO4, Sodium Sulfate.

CuO, Copper (II) Oxide.

P2O5, Diphosphorus Pentoxide.

HNO3, Nitric Acid.

CO32-, Molecular Formula.

Hope this helps. :)

8 0

3 years ago