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3 years ago

Which atom’s ionization energy is greater than that of phosphorus (P)? A. Ba B. K C. As D. Cl

Chemistry

1 answer:

WINSTONCH [101]3 years ago

5 0

Answer is: D. Cl (chlorine).

The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).

Barium, potassium and arsenic are metals (easily lost valence electrons), chlorine is nonmetal (easily gain electrons).

Alkaline metals (in this example, potassium) have lowest ionizations energy and easy remove valence electrons (one electron), earth alkaline metals (in this example, barium) have higher ionization energy than alkaline metals, because they have two valence electrons.

Nonmetals (in this example chlorine) are far right in the main group and they have highest ionization energy, because they have many valence electrons.

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3 years ago

Read 2 more answers

A 22.0 mLmL sample of a 1.16 MM potassium sulfate solution is mixed with 14.8 mLmL of a 0.860 MM barium nitrate solution and thi

dezoksy [38]

Answer:

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

Explanation:

Step 1: Data given

Volume of a 1.16 M potassium sulfate solution (K2SO4) = 22.0 mL = 0.022 L

Volume of a 0.860 M barium nitrate solution (Ba(NO3)2 = 14.8 mL = 0.0148 L

The solid BaSO4 is collected, dried, and found to have a mass of 2.57 grams

Step 2: The balanced equation

K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2KNO3(aq)

Step 3: Calculate moles

Moles = volume * molarity

Moles K2SO4 = 0.022 L * 1.16 M

Moles K2SO4 = 0.02552 moles

Moles Ba(NO3)2 = 0.0148 L * 0.860 M

Moles Ba(NO3)2 = 0.012728 moles

Step 4: Calculate the limiting reactant

For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

Ba(NO3)2 is the limiting reactant. It will completely be consumed. (0.012728 moles) . K2SO4 is in excess. There will remain 0.02552 - 0.012728 = 0.012792 moles

Step 5: Calculate moles BaSO4

‬For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

For 0.012728 moles Ba(NO3)2 we'll have 0.012728 moles BaSO4

Step 6: Calculate mass BaSO4

Mass BasO4 = moles BaSO4 * molar mass BaSO4

Mass BaSO4 = 0.012728 moles * 233.38 g/mol

Mass BaSO4 = 2.97 grams

Step 7: Calculate the percent yield

% yield = (actual yield / theoretical yield ) * 100 %

% yield = ( 2.57 grams / 2.97 grams ) * 100 %

% yield = 86.5 %

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

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3 years ago

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suter [353]

Answer:

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