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SCORPION-xisa [38]
4 years ago
7

In which reaction is the first ionization energy greatest?1.na + energy → na+ + e−2.k + energy → k+ + e−3.mg + energy → mg+ + e−

4.al + energy → al+ + e−?
Chemistry
2 answers:
SpyIntel [72]4 years ago
5 0
The answer would be the 3. Mg + energy--> mg+ + e- 
Ionization Energy is the amount of energy vital to eliminate the most loosely bond electron from an atom in the gas stage. Since Magnesium has a higher first energy which is 738 kj/mol, is the energy needed to remove the outermost, or highest energy.
natta225 [31]4 years ago
3 0

Answer: Option (d) is the correct answer.

Explanation:

The energy required to remove the most loosely bound electron from a neutral gaseous atom is known as ionization energy.

As the atomic size of an atom increases then there will be less force of attraction between the nucleus and valence electrons of the atom. Hence, with lesser amount of energy the valence electrons can be removed.

As a result, ionization energy will decrease.

Now, potassium is larger in size than sodium. Hence, ionization energy of potassium is less than sodium.

And, Na, Mg and Al are all period 3 elements. It is known that across a period there is an increase in ionization energy of the elements due to decrease in their atomic size.

Hence, aluminium being the smallest will have the greatest ionization energy.

Thus, we can conclude that the reaction Al + energy \rightarrow Al^{+} + e^{-} will have the greatest first ionization energy.

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I hope it helps

Explanation:

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4 years ago
A sample of an unknown biochemical compound is found to have a percent composition of 45.46 percent carbon, 7.63 percent hydroge
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Answer:

Formular = C₅H₁₁NO₃

Explanation:

The empirical formular is the simplest formular of a compound can have.

We use the steps below to obtain the empirical formular;

Step 1: Obtain the mass of each element present in grams. Element % = mass in g = m.

Carbon = 45.46% = 45.46g

Hydrogen = 7.63% = 7.63g

Nitrogen = 10% = 10g

Oxygen = 100% - (45.46% + 7.63% + 10%) = 36.31% = 36.31g

Step 2: Determine the number of moles of each type of atom present.

Molar amount (M) = m/atomic mass

Carbon = 45.46 / 12 = 3.7883

Hydrogen = 7.63 / 1 = 7.63

Nitrogen = 10 / 14 = 0.7143

Oxygen = 36.91 / 16 = 2.3069

Step 3: Divide the number of moles of each element by the smallest number of moles. Smallest = 0.7143

Carbon = 3.7883 / 0.7143 = 5.3035

Hydrogen = 7.63 / 0.7143 = 10.67

Nitrogen =  0.7143 / 0.7143 = 1

Oxygen = 2.2693 / 0.7143 = 3.1770

Step 4: Convert numbers to whole numbers

Carbon = 5

Hydrogen = 11

Nitrogen = 1

Oxygen = 3

Formular = C₅H₁₁NO₃

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