The relative magnitudes of any atom's first and second ionization energies, the correct answer is A) The second ionization energy is always greater than the first ionization energy.
The amount of energy needed to ionise an atom or ion in its gaseous form is equivalent to removing an electron from it. The initial ionisation energy (IE1) is the amount of energy needed to ionise an atom, whereas the second ionisation energy (IE2) is the amount of energy needed to ionise an atom after the first electron has been ionised.
Because the positively charged nucleus holds the remaining electrons more securely after the first electron is removed, the second ionisation energy is always higher than the initial ionisation energy. Therefore, removing the second electron from the positively charged ion requires more energy than removing the first electron from the neutral atom.
As you move from the second to the third ionization energy and so on, this pattern persists. As the positive charge on the ion increases and the electrons are confined to the nucleus more tightly, the energy needed to remove each additional electron rises.
In conclusion, the rising attraction between the positively charged ion and its remaining electrons causes the second ionization energy to always be higher than the first ionization energy.
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Air bags in automobiles inflate when sodium azide, NaN3, rapidly decomposes to its component elements:2NaN3(s) ----> 2Na(s) +3N2(g).A) How many moles of N2 are formed by the decomposition of 1.60 mol of NaN3?B) How many grams of NaN3 are required to form 7.00 g of N2 gas?
A) To determine the number of moles of N2 formed by the decomposition of 1.60 mol of NaN3, we need to use the stoichiometric coefficients in the balanced chemical equation: 2NaN3(s) ----> 2Na(s) + 3N2(g)
From the equation, we can see that 2 moles of NaN3 produce 3 moles of N2. Therefore, we can set up the following proportion to solve for the number of moles of N2 produced: (3 mol N2 / 2 mol NaN3) x (1.60 mol NaN3) = 2.40 mol N2.Therefore, 1.60 mol of NaN3 will produce 2.40 mol of N2.
B) To determine the amount of NaN3 required to produce 7.00 g of N2 gas, we need to use the molar mass of NaN3 and the stoichiometric coefficients in the balanced chemical equation: 2NaN3(s) ----> 2Na(s) + 3N2(g)
The molar mass of NaN3 is calculated as:
Na: 1 x 22.99 g/mol = 22.99 g/mol
N: 3 x 14.01 g/mol = 42.03 g/mol
Molar mass of NaN3 = 22.99 g/mol + 42.03 g/mol = 65.02 g/mol
From the equation, we can see that 2 moles of NaN3 produce 3 moles of N2. Therefore, we can set up the following proportion to solve for the number of moles of NaN3 required:
2 mol NaN3 / 3 mol N2) x (7.00 g N2 / 28.02 g/mol) = 1.38 mol NaN3 .Therefore, 1.38 mol of NaN3 (or 89.53 g) is required to produce 7.00 g of N2 gas.
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21. Which of these amino acids can be directly converted into a citric acid cycle intermediate by transamination?
A) Glutamic acid
B) Serine
C) Threonine
D) Tyrosine
E) Proline
Glutamic acid is the amino acids which can be directly converted into a citric acid cycle intermediate by transamination. The correct answer is: A)
Amino acids can be converted into citric acid cycle intermediates through transamination, which involves the transfer of an amino group from an amino acid to a keto acid. The resulting products are an amino acid with a keto acid side chain and a new keto acid that can enter the citric acid cycle.
Of the amino acids listed, glutamic acid can be directly converted into a citric acid cycle intermediate by transamination. Specifically, glutamic acid can be transaminated to form alpha-ketoglutarate, which is an intermediate in the citric acid cycle.
Therefore, the correct answer is: A) Glutamic acid.
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You are given 10% hydrochloric acid, 10% sodium bicarbonate, and/or 10% sodium hydroxide solutions to separate a mixture of the following two components. Both substances are soluble in ether. -What is the solvent in 10% NaOH? In 10% NaHCO3?
The solvent in both 10% sodium hydroxide (NaOH) and 10% sodium bicarbonate (NaHCO₃) solutions is water. These percentages indicate that 10% of the solution's weight is the solute (NaOH or NaHCO₃) while the remaining 90% is water. Both substances, NaOH and NaHCO₃, are soluble in the ether as mentioned in your question, but the primary solvent for these 10% solutions is water.
The solvent in both 10% NaOH and 10% NaHCO₃ solutions is water. These solutions are prepared by dissolving the respective chemicals in the water. The solutes (NaOH and NaHCO₃) dissolve in water to form a homogeneous solution. These solutions can be used for separating a mixture of two components that are soluble in ether, as the ether layer can be separated from the aqueous layer containing the dissolved solutes. The choice of the specific solution to use for the separation would depend on the specific properties of the components in the mixture and their solubilities in the different solutions.
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83) How many moles of NF3 contain 2.55 × 10^24 fluorine atoms?A) 1.41 moles NF3B) 4.23 moles NF3C) 12.7 moles NF3D) 7.87 moles NF3E) 2.82 moles NF3
The answer is A) 1.41 moles [tex]NF_{3}[/tex].
The molar mass of [tex]NF_{3}[/tex] can be calculated as follows:
N = 1 x 14.01 g/mol = 14.01 g/mol
F = 3 x 18.99 g/mol = 56.97 g/mol
Molar mass of NF3 = 14.01 g/mol + 56.97 g/mol = 71.98 g/mol
To find the number of moles of [tex]NF_{3}[/tex] , we need to divide the given number of fluorine atoms by the number of fluorine atoms in one mole of NF3:
2.55 × [tex]10^{24}[/tex] F atoms / 3 F atoms per [tex]NF_{3}[/tex] molecule / 6.022 x 10^23 molecules per mole
This simplifies to:
2.55 × [tex]10^{24}[/tex] F atoms / 18.1326 x [tex]10^{24}[/tex] molecules per mole
= 1.407 moles [tex]NF_{3}[/tex] (rounded to three significant figures)
Therefore, the answer is A) 1.41 moles [tex]NF_{3}[/tex]
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